Other Should I Become a Mathematician?

[George Jones]

one of my most brilliant friends, Fabrizio Catanese, is chaired professor at Bayreuth, and like many Italian mathematicians of his generation, note he does not even have a phd at all.

he also formerly held the Gauss chair at Gottingen.

Physicist/mathematician/author Freeman Dyson never bothered to get a phd.

 

[mathwonk]

Barry Mazur supposedly does not even have a BA. [The story goes that at the time MIT had an ROTC requirement which he declined to complete.]

 

[interested_learner]

There is a Nobel Prize winner in Physics (Kirby) who only has a master's degree and you are right Dyson has only a BA. But these guys are exceptional. Mazur has a Phd.

 

[mathwonk]

here another exceptional one - Andrew Gleason, who I believe contributed to the solution of hilberts 5th problem, and is a professor emeritus at harvard, reportedly has no phd.

 

[mathwonk]

here are my day one algebra notes, what think you?

8000 fall 2006 day one.

Introduction:
We will begin with the study of commutative groups, i.e. modules over the integers Z. We will prove that all fin gen abelian groups are products of cyclic groups. In particular we will classify all finite abelian groups. Then we will observe that the same proof works for modules with an action by any Euclidean or principal ideal domain, and generalize these results to classify f.g. modules over such rings, especially over k[X] where k isa field. This will allow us to deduce the usual classification theorems for linear operators on a finite dimensional vector space, since a pair (V,T) where T is a linear operator on the k space V, is merely a k[X] module structure on V.

For completeness sake we recall some familiar definitions.

A group is a set G with a bjnary operation GxG--->G which satisfies:
(i) associativity, a(bc) = (ab)c, for all a,b,c, in G;
(ii) existence of identity: there is an element e: ea = ae = a for all a in G.
(iii) existence of inverses: for every a in G, there is a b : ab = ba = e.

A subgroup of G is a subset H ⊂ G which is also a group with the same operation. H has the same identity as G, and the inverse for any element in H is its inverse in G.

A group G is commutative, or abelian, if also
(iv) ab = ba for every a,b, in G.

Remarks:
We will study mostly commutative groups in the first part of the course, and we will usually write them additively instead of multiplicatively, thus we write the identity as 0 . Two advantages of commutative groups are the following, which will make more sense shortly: if G (commutative) has elements a,b, such that na = 0 = mb, where n,m are positive integers, and if p = lcm(n,m), then G has an element c such that pc = 0. Also the subset of elements a of G such that na = 0 for some integer n>0, the set of elments of "finite order", is a subgroup of G. Thus it is easier to understand the "orders" of the elements of a commutative group. Also it is easier to construct the "coproduct", sometimes called the "direct sum", of a family of commutative groups.

Blanket assumption, all groups are assumed commutative until we say otherwise.

Important Examples: i) the set Z of integers is a group for addition; ii) if n is an integer, the multiples of n, form a subgroup nZ ⊂ Z; iii) the rationals form a subgroup of the reals for addition Q ⊂ R; the positive rationals form a multiplicative subgroup of the positive reals Q+ ⊂ R+; S1 = the multiplicative group of complex numbers of length one, is also called the circle group.

It is efficient to use only a few elements of a group to represent all others, and the number of elements so needed also helps measure the size of the group.
A subset S ⊂G generates G if there is no subgroup containing S except G, equivalently if every non zero element of F can be written as a finite linear combination n1a1 + ...+nkak, where all ai are in S and the ni are integers. If G is written multiplicatively it means all elements except e can be written as a finite product ∏ aini, where where all ai are in S and the ni are integers.

Examples: {1} generates Z, as does {-1}. The empty set generates the trivial group {0}. The interval (0,d) generates (R,+) if d>0. The set of positive primes generates Q+.

We say G is finitely generated, or fin gen, or f.g., if there is a finite set of generators for G.
We proceed now to the classification of all fin gen abelian groups. The relevant concepts we will use are products, quotients, isomorphisms, and other linear maps.

Fundamental constructions (on abelian groups):
I) Products: Given an indexed family of (always abelian) groups {Gi}I, form the cartesian product set ∏IGi of all functions from the index set I to the union of the groups Gi and where the vaue of a function at i lies in Gi. We define the operation pointwise on functions, i,.e. we multiply or add the values of the functions. If the set I is finite of cardinality n, this means the elements are ordered n tuples if elements, one from each Gi, and we add them componentwise, like vectors. The identity is the function whose value at every i is the identity of Gi.

II) Coproducts: This is the same construction as above, except that the function must have the value 0 except possibly at a finite number of indices. Hence it is exactly the same in case the index set I is finite. It is denoted by an upside down product or by a summation sign, ∑ Gi. Obviously the coproduct of a family of groups is a subgroup of their product.

If all the groups Gi are equal to the integers Z, we call their coproduct a "free abelian group" on the set I, i.e. a group of form ∑I Zi.
We also write Zn for the product (or sum) of n copies of Z. The standard basis vectors
{ei = (0,...,0,1,0,...0) where the 1 is in the ith place, for i =1,...,n}, generate Zn.
Other commonly encountered products groups are Rn, S1xR, and S1x S1= the torus group.

III) Quotients: If H ⊂ G is a subgroup, define the quotient group G/H as the set of equivalence classes of elements of G for the relation x ≡ y iff x-y belongs to H. Write [x] for the equivalence s of x and add by setting [x]+[y] =[x+y], after checking this is independent of choice of representative elements of the classes.

The basic quotient group is Z/nZ, the additive group of integers "mod n".

When we define isomorphism, we will see that the circle group is isomorphic to a quotient group S1 ≅ R/Z. and also S1xS1 ≅ (R/Z)x(R/Z) ≅ (RxR)/(ZxZ). The interchange of quotients and products is more subtle than it may appear here, and will play a crucial role in the proof of the fundamental theorem we are seeking. The fact that renders the interchange easy here is that each factor in the denominator is a subgroup of a factor group in the numerator. When this is not the case the problem is more difficult.

One way to construct a finite abelian group is to form a product (Z/n1)x(Z/n2)x...x(Z/nk). Our goal is to show that these examples give essentially all finite abelian groups. To make the phrase "essentially all" precise, we must define how we will compare two groups, and when we will say two groups are essentially the same.

A map of groups f:G-->H (abelian or not) is a homomorphism, sometimes called simply a map, if for all a,b, in G, we have f(ab) = f(a)f(b), or additively, if f(a+b) = f(a) + f(b).
It follows that f(0) = 0 , and f(-x) = -f(x).

The set of homomorphisms from G to H is denoted Hom(G,H). When G,H are abelian it is also an abelian group under pointwise addition, [but it is not even a group if H is not abelian].

Examples of homomorphisms: The inclusion of a subgroup H ⊂ G is a homomorphism; The map
G--->G/H taking an element x to the class [x] is a homomorphism; The ith projection ∏IGi--->Gi taking a function to its value at i, that is taking a vector to its ith component, is a homomorphism. The injection Gi--->∑I Zi taking an element x of Gi to the function having value x at i and value 0 elsewhere, is a homomorphism. [This puts x in the ith component of a vector and 0's elsewhere.] The map R--->S1 taking t to e^(2πit) is a homomorphism.

Important invariants of a homomorphism:
To understand homomorphisms we focus on what goes to 0, and what things get "hit" by it.
If f:G-->H is a homomorphism of groups (abelian or not), the subset kerf = {x in G : f(x) = 0} is called the kernel of f; it is a subgroup of G. The subset Im(f) = {y in H: y = f(x) for some x in G} = the image of f, is a subgroup of H.

The quotient H/Im(f), defined for abelian groups only, is the cokernel of f.

An isomorphism is a homomorphism with an inverse homomorphism. I.e. a homomorphism f:G-->H , is an isomorphism if there is a homomorphism g--->G such that fog = id(H), and gof=id(G).

How to recognize an isomorphism:
A homomorphism f:G-->H is an isomorphism if and only if it is bijective,
if and only if kerf = {0} and Im(f)= H.

How to define homomorphisms:
1) To define a homomorphism into a product G--->∏IGi is equivalent to defining one homomorphism G--->Gi into each Gi. I.e. Hom(G, ∏IGi) ≅ ∏IHom(G,Gi), via the map taking the homomorphism
f:G---> ∏IGi to the family of compositions πiof where πi is the projection ∏IGi --->Gi .

2) To define a homomorphism out of a coproduct ∑ Gi--->H, is equivalent to defining one map out of each summand Gi--->H, i.e. Hom(∑Gi,H) ≅ ∏IHom(Gi,H) via the map taking f:∑ Gi--->H to the family of compositions foßi where ßi is the injection Gi--->∑ Gi.
3) To define a map from Zn--->Zm by 1) and 2), it suffices to define mn maps Z--->Z, i.e,. to give mn integers, in the form of an mxn matrix, where the ith column is the image under the map of the ith standard basis vector ei = (0,...,0,1,0,...0).

4) To define a map out of a quotient G/H--->K, is equivalent to defining a homomorphism f:G---K such that f(H) = {0}, i.e. Hom(G/H,K) ≅ Hom((G,H), (K,{0})) (maps of pairs), via the map taking
f:G/H--->K, to the composition foπ:G--->K, where π:G--->G/H is the projection.

Examples: The map R--->S1 sending t to e^(2πit) induces an isomorphism R/Z--->S1, via the correspondence in 3) above.
The maps Z--->(Z/riZ), induce a map Zn--->(Z/r1Z)x...x(Z/rnZ) which induces an isomorphism
Zn/[(r1Z)x...x(rnZ)]--->(Z/r1Z)x...x(Z/rnZ).
If f:G--->H is a surjective homomorphism, it induces an isomorphism G/kerf--->H.

Our first main theorem is the following.
Theorem: If G is any finitely generated abelian group, then there exist integers n,m≥ 0, and a sequence of integers r1,...,rm ≥ 2, such that G ≅ Zn x (Z/r1Z)x...x(Z/rmZ). Moreover these r's can always be chosen so that r1| r2|...,|rm-1|rm, i.e. each one divides the next one, and if this is done, all the integers are uniquely determined by the isomorphism class of )G. We call n the rank of G and the integers r1,...,rm the torsion coefficients, or invariant factors. Thus G is completely determined by the sequence (n, r1,...,rm). If n=m=0, G ={0}.

Exercises: if Tor(G) = { x in G such that there is an integer n>0 with nx = 0} then Tor(G) is a subgroup of G, called the torsiion subgroup. [This is false if G is not abelian.]

Cor: If G ≅ Zn x (Z/r1Z)x...x(Z/rmZ), then Tor(G) ≅ (Z/r1Z)x...x(Z/rmZ), and G/Tor(G) ≅ Zn . Thus the torsion part of G is a uniquely defined subgroup of G, and the free part of G is a uniquely defined quotient group.

Proposition: If G is a fin gen abelian group, there is a homomorphism f:Zn--->Zm , such that G ≅ coker(f) = Zm/f( Zn).

Proposition: If A is the matrix of f:Zn--->Zm , and if B is a matrix obtained by elementary row and column operations from A, then the cokernels Zn/A(Zm) ≅ Zn/B(Zm), are isomorophic.

Proposition: Every matrix A of integers can be reduced by elementary row and column operations, to a diagonal matrix B.

Corollary: The theorem is true.

 

[mathwonk]

do you think i am going too fast, too slow? [for some reason, all isomoprhism and equivalence relation symbols came out as subsets. when i saw the first one, which really was a subset i was encouraged briefly.]

 

[mathwonk]

this is for a first year grad course by the way, but many schools teach this in undergrad even sophomore year at places like Brandeis in the 1960's.

 

[mathwonk]

apparently i was going way too fast. i need to do the case of products of 2 groups before doing the case of arbitrary indexed families. and do some detailed examples. ill do that friday.

heres some homework problems, show that defining a homomorphism from AxB to G (abelian groups) is the same as defining one A-->G and another oen B--->G. then show that defining a homomorphism G--->AxB is the same as defining one G---<>A and another one G--->B.

then show that the multiplicative group C* of non zero compelx numbers is isomorphic to the product group RxS^1, where R is the aditive group of reals and S^1 is the multiplicative circle group of complex numbers of length one.

then show that C* is aLSO ISOMORPHIC TO THE PRODUCT R*xS^1 where R* is the multiplicative group of positive real numbers.

then find a homomorphism from RxR --->C* that has kernel equal to {0}xZ. [is that right?]

 

[mathwonk]

hey guys and gals, i want to brag on and advertise my colleague here at uga, Valery Alexeev, who will speak next week at the ICM, in madrid, international congress of matehmaticians. here is the schedule of talks in algebraic geometry for the first day of sectional talks.


Jaroslaw Wlodarczyk, Purdue University, West Lafayette, USA
Algebraic Morse theory and factorization of birational maps Abstract
Chair: Phillip Griffiths

Lawrence Ein, University of Illinois at Chicago, Chicago, USA and University of California at Irvine, Irvine, USA
Invariants of singularities of pairs Abstract
Chair: Phillip Griffiths

Valery Alexeev, University of Georgia, Athens, USA
Higher-dimensional analogues of stable curves Abstract
Chair: Phillip Griffiths

Tom Graber, University of California, Berkeley, USA
Rational curves and rational points Abstract
Chair: Óscar García-Prada

Tomohide Terasoma, University of Tokyo, Tokyo, Japan
Geometry of multiple zeta values Abstract
Chair: Óscar García-Prada

Jun-Muk Hwang, Korea Institute for Advanced Study, Seoul, Korea
Geometric structures arising from varieties of minimal rational tangents Abstract
Chair: Luc Illusie

Tom Bridgeland, University of Sheffield, Sheffield, UK
Derived categories of coherent sheaves Abstract
Chair: Luc Illusie

Yuri Tschinkel, Georg-August Universität Göttingen, Göttingen, Germany and Courant Institute, New York University, USA.
Geometry over nonclosed fields Abstract
Chair: Ignacio Sols

Jean-Benoît Bost, Université Paris-Sud, Orsay, France
Evaluation maps, slopes, and algebraization Abstract
Chair: Ignacio Sols

 

[ircdan]

Hey mathwonk,

I just read all the way up to the part where it says "Fundamental constructions (on abelian groups):". Anyways my point is, I have not formally studied algebra at school(I start a course in it in a few days), and I was able to understand it, so I think that's a good indication maybe that up to that point the notes are very good. I take it the prerequisites for this course are for students who have had 1-2 semesters of undergraduate algebra.

 

[mathwonk]

Thank you very much ircdan. The next part, the fundamental constructions, may be over done by using arbitrary index sets. just starting with a product of 2 groups and a coproduct of two (abelian) groups.

see if you can work out the special case of the definitions just for two factors, from what is given there.

the main goal of the notes is to prove the 4 statements about how to define homomorphisms, and then start using them.

let me know what you think when you get there. yes the prereqs are what you surmised.

 

[mathwonk]

those day one notes by the way were about right for the first two weeks of the course.

here is some advice on how and where to study for the PhD prelim exsms in algebra at UGA:

Algebra PhD prelim remarks for students.

It is difficult to give a single reference where one can find proofs, examples, applications, and intuitive explanation of all topics. This makes it incumbent on the student to seek this multi level understanding by combining several sources. In general Dummit and Foote excels in illustrative examples, but the proofs there may sometimes not be complete or may not be the simplest. Of course this is somewhat a matter of taste. Smith’s notes (mathwonk) focus more on proofs and have fewer examples and exercises. DF spends roughly 600 pages on the graduate prelim material while Smith covers it in about 350 pages. Hungerford is another good reference, as well as Artin, Lang, and Van der Waerden. Smith’s webnotes on linear algebra are extremely brief (15 pages, including the real spectral theorem and canonical forms), but may be useful as a summary review for someone who has already studied the material.

The citations below are mainly to Dummit - Foote and Smith for “graduate” material, and Artin for “undergraduate” material. Artin is also an excellent source for much of the graduate material, especially groups, free groups, canonical forms of matrices, fields and Galois theory.

Graduate material:
Group theory:
For group theory, the first 6 chapters (220 pages) of DF are excellent, with the following comments. They define isomorphisms as bijective homomorphisms, which to modern minded people is a theorem, an isomorphism always being technically a homomorphism with a homomorphism inverse. The proofs there of Sylow theorems and simplicity of A(n) for n > 4 look good. The proof of the most fundamental isomorphism theorem, and the proof of the Jordan Holder theorem, one easy, one hard, are both left to the exercises. The proof JH is seldom tested on the prelim, but proofs are always of interest. Of course doing proofs as exercises is beneficial if one can do them.

The discussion in DF of the meaning and applications of the fundamental theorem of finite abelian groups, and of cyclic groups, looks excellent, but the proofs of the theorem given in chapters 6 and 12 are not constructive. The constructive proof, needed to find actual decompositions, is left as exercises #16-#19 in chapter (12.1). Note that in applying the theorem to normal forms of matrices, DF appeal to the constructive version in their exercises.

All the definitions and proofs of these theorems on groups are given in Smith’s web based lecture notes for math 843 part 1. There is more than one proof of Sylow’s theorems, a complete proof the only simple groups of order between 60 and 168 have prime order, plus an introduction to the categorical and functorial point of view, in only 80 pages, but far fewer examples and exercises than in the first 220 pages of DF.
The best source for free groups may be Artin, the last 3 sections of chapter 6. In particular he gives there an algorithm due to Todd and Coxeter, which let's one calculate some properties of a finite group presented as a quotient of a free group.

Rings:
All the necessary material on rings is covered in DF, chapters 7,8,9 (roughly 115 pages).
Smith’s notes treat rings in a somewhat scattered way, in sections 13, 14, of 843 -2 , and the first 5 sections of 844-1, (43 total pages), mainly as needed to treat Galois theory. The main results are proved, but the most elementary facts are assumed as similar to those for groups. Gauss’s important theorem on unique factorization of polynomials is proved twice however in Smith for clarity, once for Z[X], and again for R[X], R any ufd.

Comments:

1. DF again define a ring isomorphism as a bijective homomorphism, rather than giving the categorical definition as a homomorphism with an inverse homomorphism. (Of course this matters less in algebra than in topology, algebraic geometry, and analysis, where it actually gives the wrong notion, but it seems a bad habit to acquire.)

2. The fundamental isomorphism theorems for rings are not fully proved in DF, but are easy and useful exercises. They are not even stated in Smith’s notes, who takes the attitude they should be obvious by this point. DF is more systematic in the first few sections of chapter 7, to state again most of the elementary facts about homomorphisms, ideals and quotient rings, analogous to those for groups. Smith has little discussion of basic ring properties. One crucial condition DF inexplicably do not even state however (but Smith does), is when a ring map f, out of a ring R, induces a ring map out of a quotient R/I, namely whenever f(I) = {0}.

3. Zorn’s lemma is clearly used in DF to find maximal ideals (p.254), and clearly stated in the appendix. The proof that Zorn follows from Ax. Ch. is not given, but this is never tested on prelims, and seldom needed by non logicians. It can be found in Lang if desired. Smith also discusses extensively in 844 - 1, how to use Zorn to find maximal ideals, to construct algebraic field closures, and why Zorn is unnecessary in countable or noetherian rings.

4. The definition of Euclidean domain is not universally agreed on. The one in Smith’s notes (p.27, section 5, 844-1), includes the condition that the norm of a product N(ab) is at least as large as that of a factor N(a), which DF do not assume. DF can still deduce that a Euclidean domain is a p.i.d. and hence a u.f.d., but the argument for the existence of an irreducible factorization is harder, making the proof that Z and k[X] are ufd’s (p.289, p.300 DF) unnecessarily difficult. This can make those cases longer to prove on a prelim exam.

An interesting topic covered in DF p.277, but not in Smith is a necessary criterion for a ring to be a Euclidean domain, allowing DF to give a simple example of a non Euclidean p.i.d.

5. The section 8.2 on p.i.d.’s in DF looks good but the history may be faulty. They give a trivial result about the equivalence of pid’s and rings with “Dedekind - Hasse” norms, attributing one direction (p.281) to Greene in 1997. This result appears already in Zariski - Samuel vol.1, p.244, in 1958, and is so easy it was surely known to Dedekind.

6. The nice remark in DF, p.305, that a polynomial ring over a ufd, in infinitely many variables is still a ufd, is easy, but not always noted (cf. exercise in Smith 844-1, section 5).

7. The statement of Eisenstein’s criterion in DF p.309, is unnecessarily weak, since it is stated only for monic polynomials, although the proof of the non monic case is the same. Smith’s notes and most other books have the full version. There is also a much stronger version not often seen, using a Newton polygon argument on the graph of the points, one for each term of the polynomial, whose (s,t) coordinates are the exponents of X and of some prime factor of the coefficients, due to G. Dumas, in Van der Waerden, vol. 1, 2nd edition, p.76.

8. The example of the cyclic product structure of (Z/nZ)* is an exercise in DF, and is explained in more detail in Smith (844 -2, section 18).

9. DF includes a nice introduction to Grobner bases and division algorithms for polynomials in several variables, 9.6.

Modules over pid’s and Canonical forms of matrices.
The treatment in DF in sections 10.1, 10.2, 10.3, and 12.1, 12.2, 12.3, appears to be excellent (77 pages) but the constructive proof of the decomposition is given only as exercises 16-19, in 12.1 of DF. The treatment in Smith 845-1 and 845-2, (also 77 pages), includes a detailed discussion of the constructive proof.

Field theory/ Galois theory.
DF in chapters 13,14, seem to give an excellent treatment of field and Galois theory, (about 145 pages). Smith treats it in 843-2, sections 11,12, and 16-21 (39 pages), 844-1, sections 7-9 (20 pages), 844-2, sections 10-16, (37 pages).Undergraduate linear algebra

Determinants: M. Artin, Algebra, chapter 1.3, 1.4, 1.5; DF chapter 11.4.

Eigenvalues, eigenvectors, Artin, chapter 4.

Cayley Hamilton: DF, chapter 12.2, Artin, ex.15, p.153-4, Smith 845-2, section 10.
Remark: It is hard to find the most elementary proof of Cayley Hamilton in modern books. It follows immediately from Cramer’s rule, Artin, page 29. I.e. by Cramer we have for a matrix A, with characteristic matrix (XI-A), that adj(XI-A).(XI-A) = f(X)I, where f(X) is the characteristic polynomial of A. Thus f(X)I viewed as a polynomial with matrix coefficients, is divisible from the right by the linear polynomial (XI-A). It follows immediately from the (trivial) non commutative root - factor theorem, that X=A is a “right root” of the polynomial f(X)I, in particular that f(A) = 0.

Canonical forms for matrices, Artin, chapter 12, plus references above.

Linear groups, Artin, chapter 8.1; this whole chapter is excellent, and one of the few elementary introductions available, but apparently the definitions of these groups is about all that is required for prelims.

Dual spaces, DF, chapter 11.3.A much more sophisticated discussion is in Smith 845-3, section 13.

Finite dimensional spectral theorem, Artin: chapter 7, Smith 845-2: section 11.

References: Dummit - Foote, Smith, Artin.
Supplementary references: Hungerford, Van der Waerden, Milne.

 

[JasonRox]

The best training is to read the greatest mathematicians you can read. Gauss is not hard to read, so far as I have gotten, and Euclid too is enlightening. Serre is very clear, Milnor too, and Bott is enjoyable. learn to struggle along in French and German, maybe Russian, if those are foreign to you, as not all papers are translated, but if English is your language you are lucky since many things are in English (Gauss), but oddly not Galois and only recently Riemann.

So, are you saying pick up their books?

Like, what Gauss wrote in paper?

 

[mathwonk]

gauss has a nice book called disquisitiones arithmeticae it costs about 80 bucks unfortunately, but is in libraries. the good side is that anything you read in gauss will be abiout 100 times as valuable as what you read in my notes or some other book by most other schnooks.

if you are a wine drinker compare to the difference between a glass of good wine and a glass of bali hai or ripple.

 

[JasonRox]

gauss has a nice book called disquisitiones arithmeticae it costs about 80 bucks unfortunately, but is in libraries. the good side is that anything you read in gauss will be abiout 100 times as valuable as what you read in my notes or some other book by most other schnooks.

if you are a wine drinker compare to the difference betwen a glass of good wine and a glass of bali hai or ripple.

Interesting.

I always though they would be unreadable for some reason.

 

[J77]

I came across this link http://wuphys.wustl.edu/~katz/scientist.html and many others from PhDs.org. What are your opinions about his experiences/advice? Also, is it very competitive to become a professor? Are there major differences in becoming a professor at an undergraduate only institution versus becoming a professor at both a undergraduate and graduate institution? Which is more competitive?

Thanks

I like that article - thought it was a bit tongue-in-cheek at the start, but then went a bit mental towards the end

Just to add to mathwonk's reply about professorships...

I don't know about the system in the US too much - I presume a full professor is like getting a chair in the UK. However, I've come across some confusion where everyone that teaches in the US calls themselves a professor, and even friends who have done postdoc get to call themselves Assistant profs.

Anyway, in the UK, if you're good enough - ie. written a lot of top quality papers, got in the grants, you can make professor. This generally comes about by the university having a chair available (sometimes associted with eg. head of department), or you applying for a prmotion and them recognising your ability. (You can also apply externally for a chair.)

It's a bit harder in other places I've been - Germany, Belgium, Netehrlands - there, there is generally one professor per group, ie. at the top of the group. You can become a very senior scientist but never reach that top level - until the big guy retires. This causes a lot of good academics to leave their established uni and move elsewhere - incl. US and UK - looking for that higher title.

Tat's my take.

 

[J77]

Actually, reading some of Katz's other rants - I don't like this guy too much: "I am a homophobe, and proud."

Also, misinformed: "The first campaign in the war against terrorism is nearly over [2002], and has been a great success."

Back to the maths...

 

[JasonRox]

If being a professor takes you away from doing any maths, then why become a professor to do math? That makes no sense.

Am I better off becoming an Accountant/Lawyer and then just doing mathematics on my free time or something? I'll make better money too. I can work by a university and keep up with conferences.

Note: Becoming a lawyer would take an extra three years of school and becoming a certified Accountant would take maybe an extra one or two. I've done 2 years of accounting already before pursuing mathematics.

 

[CPL.Luke]

keep in mind that becoming a lawyer or accountant takes up a lot of time, you can't just go into one of those fields and get easy money. You also spend 2-3 years of schooling with very few classes in higher mathmatics, and so will be distanced from the subject.

Whereas if you become a professor you've probably had 2-3 extra years of math training when you would have been otherwise trained to work as an accountant. As a professor you would also be working on higher mathematics with your students and in all likelihood auditing other mathematics classes in order to further your own work.

to sum it all up if you move in a direction other than mathematics the odds of you being able to work on higher mathematics goes down alot.

 

[JasonRox]

keep in mind that becoming a lawyer or accountant takes up a lot of time, you can't just go into one of those fields and get easy money. You also spend 2-3 years of schooling with very few classes in higher mathmatics, and so will be distanced from the subject.

Whereas if you become a professor you've probably had 2-3 extra years of math training when you would have been otherwise trained to work as an accountant. As a professor you would also be working on higher mathematics with your students and in all likelihood auditing other mathematics classes in order to further your own work.

to sum it all up if you move in a direction other than mathematics the odds of you being able to work on higher mathematics goes down alot.

I don't think becoming an Accountant will take much of my time though. It's just around the corner for me now.

I understand your point though, but I'm not worried about being behind 1-2 years in the study of mathematics considering you will be working for 30-50 years. 2 years is barely noticeable if at all.

 

[pivoxa15]

Interesting.

I always though they would be unreadable for some reason.

I have always thought the same and a big part of that reason is their different use of notations and convention of writing maths. This can put people off very easily. But offcourse another more fundalmental reason is that if you are an undergraduate than you might not have enough mathematical maturity and/or knowledge to read it even if their notation was modern.

 

[CPL.Luke]

Jason remember that working in mathematics is not the same as learning mathmatics, if your at the level where you could do real work in mathematics now than by all means you would be fine in your education. However if you aren't at that level yet than those extra years could be critical.

 

[mathwonk]

In the US getting hired in the first place is a bit of a crap shoot, but after that works out, getting promoted is a matter mostly of hanging in there and doing your job for 5-10 years, after which promotion to full professor is usual.

not everywhere of course: at harvard they wait for you to get a fields medal somewhere else then hire you away to harvard.fermat of course set a good example of a jurist by trade who did outstanding math in spare time, but fermat could be hard to emulate.

 

[JasonRox]

In the US getting hired in the first place is a bit of a crap shoot, but after that works out, getting promoted is a matter mostly of hanging in there and doing your job for 5-10 years, after which promotion to full professor is usual.

not everywhere of course: at harvard they wait for you to get a fields medal somewhere else then hire you away to harvard.


fermat of courtse set a good example of a jurist by trade who did outstanding math in spare time, but fermat could be hard to emulate.


if you want to be a professor you need to have a high tolerance, preferably enjoyment, for teaching students. to emulaten gauss, the ticket these days would probably be to kiss up to someone rich like ross perot and try to get a life long stipend to do research.

but i doubt ross perot respects math that much. where in the US of A would one find a rich person intelligent enough to value pure research? apparently harvard found one, landon clay, and leeched him to the bone to get funding for the clay institute.

sorry for the cranky rude tone, it is friday night and i am kicking back.

Cranky? Not at all. You made very good points regardless of the tone.

I plan on doing Graduate School in Mathematics before considering anything because you never know what can happen. Maybe I'll land a job. Maybe I'll land a great article. Maybe I'll land great connections with big projects ahead, which is also important although it probably wouldn't pay me anything.

I'll have to wait and see I guess.

 

[mathwonk]

basic graduate algebra (in mid level US school)

here is the detailed optomisticm syllabus for my one semester rgad slgebrfa course, including brief explanation, logical connections, and motivation for all topics.

Goals of 8000 course: To prepare students to use the basic tools of commutative and non commutative algebra, and pass the PhD alg. prelim.

Proposed content of course:
We will treat commutative topics first, generalizing vector spaces. The fundamental concept is “linear combinations”.

I. Linear and commutative algebra
i) Abelian groups
First we treat abelian groups, representing them as cokernels of maps between free abelian groups, especially finitely generated ones using matrices. Modifying familiar techniques for matrices, elementary row and column operations, let us diagonalize integer matrices and prove the fundamental structure theorem, giving explicit models of all finitely generated abelian groups as products of cyclic groups.
Key properties peculiar to abelian groups include the ability to form quotients by modding out any subgroup, the existence of an element whose “order” is the l.c.m. of the orders of any two given elements, and especially the ability to define maps out of free abelian groups by defining them arbitrarily on the basis elements. In particular a finite abelian group has elements of every order dividing the annihilator of the group, and subgroups of every order dividing the order of the group.

ii) commutative rings R
Then we discuss rings more general than the integers, making each important theorem about integers into a definition.
We call a ring “entire” or “integral domain” or “domain”, if there are no zero divisors, a Euclidean domain if it admits a Euclidean algorithm (e.g. polynomials in one variable over a field, Gaussian integers), a “p.i.d.” if also every “ideal” (subgroup closed under multiplication by R) has a single generator (e.g. local rings of smooth curves), and a u.f.d. if it also admits unique factorization into irreducibles (e.g. polynomials in any number of variables over a field or over any u.f.d.).
R is “noetherian” if every ideal has a finite set of generators (e.g. any quotient ring of any polynomial ring over a field or any noetherian ring). In a noetherian domain factorization into irreducibles is always possible, but maybe not uniquely - uniqueness is equivalent to the existence of gcd’s for any pair of elements.
The rational root theorem leads to calling a domain R “integrally closed” or “normal” if every root of a monic polynomial which lies in its fraction field, already lies in R (e.g. the coordinate ring of an affine hypersurface which is smooth in codimension one).
A “dedekind domain” is a normal domain where (as in Z) every proper prime ideal is maximal, (e.g. the affine ring of a smooth curve, or the ring of integers in a number field; algebraic number theory is more difficult apparently because these rings do not always have the stronger properties listed previously).

iii) R modules
Since our study of abelian groups used crucially the multiplication of group elements by integers, analogous to multiplying vectors by scalars, we define “R modules” as abelian groups which allow multiplication by elements of a ring R. For rings R which share those properties of Z used in the proof of the fundamental structure theorem, we obtain analogous theorems for R modules.
Every finitely generated module over a noetherian ring is the cokernel of a matrix. Every matrix over a Euclidean domain can be diagonalized by elementary row and column operations, and invertible secondary operations suffice to diagonalize matrices over a p.i.d. Hence we get structure theorems for finitely generated modules over all p.i.d.’s, and an algorithm for computing the decomposition over a Euclidean domain.
Presumably there are some theorems for modules over normal and dedekind domains, but I do not know what they are.

iv) Canonical forms of linear operators
Applications of these theorems include the important case of finitely generated “torsion” modules (analogous to finite groups) over k[X] where k is a field, since these are equivalent to pairs (V,T) where V is a finite dimensional k - vector space, and T is a k linear transformation.
The structure theorem above gives as a result, the rational and Jordan canonical forms for T, (with certain hypotheses), as well as diagonalization criteria for the matrix of T.
We point out analogies between the order of a finite group and the characteristic polynomial of T, and between the annihilator of a finite group and the minimal polynomial of T. The Cayley Hamilton theorem falls out too.
This completes the first half of the course.

II. Non commutative algebra: groups and field extensions
The basic concept in non abelian group theory is “conjugation”, studying the extent to which the action sending y to a^-1 y a is non trivial.

i) Groups, Existence of subgroups, Subnormal towers
The first goal is to understand something about the elements and subgroups of a given group, just from knowing its order.
We begin the study of possibly non abelian groups, motivated by a desire to understand not just individual matrices, but groups of matrices, as well as symmetries of both geometric and algebraic objects, e.g. field extensions. Fresh difficulties here include the fact that not all subgroups can be modded out to form quotient groups, i.e. not all subgroups are “normal”, and the related problem that not all subgroups can be kernels of homomorphisms, so it is harder to find non trivial maps between groups, hence harder to compare groups.
Even non normal subgroups are harder to find, as it is no longer true that there are subgroups of all orders dividing the order of a finite G, nor elements whose order equals the lcm of the orders of two given ones. Hence an initial problem is finding the orders of elements in a given G and the orders of subgroups. We recover some general results by restricting to prime power divisors. I.e. Sylow: for every prime power p^r dividing #G, there exist subgroups of G of order p^r and elements of order p.
As a substitute for the product decomposition of a finite abelian group into cyclic groups, we have the concept of a “simple” group (only trivial normal subgroups), and of a “subnormal” tower for G in terms of simple constituents which are uniquely determined by G.

ii) Free groups, Group actions
After getting some handle on the elements and subgroups of a given group, we seek to construct homomorphisms of groups.
Defining a homomorphism out of G is most naturally accomplished by finding an “action” of G on a set S, which yields a homomorphism of G into Sym(S) the group of bijections S--->S or “symmetries” of S.
A non abelian group is characterized by the fact that some conjugation actions are non trivial. Letting G act on its Sylow subgroups by conjugation or translation, can provide non trivial homomorphisms and non trivial normal subgroups. These actions are also used in the proof of the Sylow theorems.
From this perspective, the structure theorems for linear transformations give unique standard representatives useful for computing the orbits of conjugation actions on GLn(k). This let's us understand GL3(Z/2) = collineations of the 7 point plane, the next interesting simple group after the icosahedral rotation group A(5).
Free abelian groups, from which homomorphisms to abelian groups are easy to define, must be replaced by “free [non abelian] groups”, which allow easy homomorphisms to all groups, but whose structure is much harder to understand. Thus although every finite group G is the quotient of a free group by a (free) subgroup, this information is harder to use as the presentation of G by a map between free groups is harder to simplify, since matrices are inapplicable. M. Artin’s discussion of the Todd - Coxeter algorithm is presumably relevant here.

iii) Semi direct products
To classify even small non abelian groups, we need some standard examples and some standard constructions. Basic examples include symmetric groups and dihedral groups. Most non abelian groups do not decompose as direct products, but we learn to recognize those which do. We introduce a more general “semi direct” product, and show that many small groups do decompose as “semi direct” product of still smaller groups, and learn to recognize when a group has such a decomposition. Semi direct products of abelian groups can be non abelian. Dihedral groups D(2n) are semi direct products of two cyclic groups, Z/2 and Z/n.

iv) Application to Galois groups of field extensions
Galois honed the tool of group theory to analyze the structure of finite field extensions, to decide when a polynomial with roots in an extension of k, could be expressed in terms of various “nth roots” and field operations, i.e. whether the a sequence of linear operations and forming powers, can be inverted by a sequence of rational operations and taking roots.
The answer is given by analyzing a subnormal tower of a subgroup G of all field automorphisms of the full root field, namely those which leave the coefficient field point wise fixed. Solvability of the polynomial is equivalent (over Q say) to a subnormal tower for G having only abelian simple constituents.
As a technically convenient device, we will construct a universal algebraic extension of a given base field, its “algebraic closure”, using Zorn’s lemma. Then we can always work within this one field and with quotients of its Galois group.

v) Examples, computations of Galois groups
We practice computing a few small Galois groups, including some of form D(2n), S(n), and we recall the structure of a subnormal tower for these groups, relating it to solvability of polynomials. In particular we deduce Abel’s theorem that a general, polynomial of degree 5 or more is not solvable by radicals. Then we discuss some fields associated to the cyclotomic polynomials X^n - 1, whose Galois groups are abelian.

Remarks:
Although Galois groups are certain finite matrix groups, most books seem to ignore that, although presumably our knowledge of matrices could help study them sometimes.

This marks the end of the prelim syllabus, and probably the end of the course. It seems ambitious to expect to reach here, but if we get here early, I would enjoy treating the topic of specialization of Galois groups, and the allied topic of monodromy groups from geometry, e.g. Galois groups of finite extensions of functions fields of one or more variables.

 

[mathwonk]

is this useful?

 

[mathwonk]

update on class lectures

the notes for "day one" have occupied 8 lectures so far and I plan to spend the 9th lecture winding up the uniqueness part of the theorem, which i omitted in day one.

it goes roughly as follows:if p divides n then (Z/nZ)/p(Z/nZ) is iksomorphic to Z/pZ, but if p does not divide n, that quotient is isomorphic to {0}.

It follows that if G = Z/n1 x Z/n2 x ...xZ/nk, where each ni divides n(i+1), then p^(r-1)G/p^rG is isomorphic to (Z/p)^s where s is the number of factors Z/ni sucvh that p^r divides ni.

this proves uniqueness, although if you are like me you will need to spend a few pages proving this in detail. (I spent about 8).

 

[mathwonk]

perhaps it is time to write up day two?

 

[mathwonk]

ok here is the homework after i finished the details of day one:

8000 HW 4. due friday sept. 15, 2006.

I want you to carry out the proof of the decomposition theorem for finitely
generated modules over a Euclidean domain, imitating the proof for abelian
groups, as follows:

All rings are assumed commutative and to have identity. fix a ring R.
Recall basic examples of R modules are:

i) ideals I in R. An ideal I is generated by elements x1,...,xm of I, iff
every element of I is a linear combination a1x1+...+amxm where the ai are
in R. Conversely given any elements of R, the set of all such linear
combinations does give an ideal. An ideal is "proper", i.e. not equal to
R, iff it does not contain a unit, iff it does not contain 1.

ii) quotients of R: Given a ideal I in R, the quotient group R/I is also
an R module. (If I is proper then R/I is also a ring.) If I contains a
non zero element y, then R/I is a torsion module. I.e. any element [x] of
this module is annihilated by multiplication by y.

iii) Direct sums and products. If {Mi} is any collection of R modules,
their direct product is the R module ½ Mi consisting of all indexed
families {xi} of elements where for each i, xi belongs to Mi. We add and
multiply componentwise, like vectors with the entries xi. Their direct sum
is the submodule … Mi where in each family {xi} only a finite number of
entries xi are non zero. Note, if I is a finite index set, then the sum
and product construction give the same module.

for example, R^m = R x...xR, m times, is an R module, called a free R
module. (It has a well defined rank too, i.e. m, but we do not know this
yet.)

iv) A module M is called cyclic if it has one generator, i.e. for some x in
M, we have M = Rx. for example R is cyclic with generator 1, and every
quotient R/I is cyclic with generator [1].Goal: We want know for which rings R we can still prove that every finitely
generated R module M is isomorphic to a finite product (i.e. sum), of
cyclic quotient modules. I.e. when is every fin.gen. M isomorphic to some
(R/ I1 x...x R/ Im), as in the case of R = Z ?

Approach: All we have to do is perform proof analysis on our previous
proof and see what properties of Z were actually used. The key property
was using the Euclidean algorithm, i.e. the division algorithm, to
diagonalize a matrix, plus the fact that every submodule of R^m was finitely
generated, to get the matrix in the first place. So we make the euclidean
property of Z a definition. Oh also we used that Z has extremely few zero
divisors.Definition: A ring R is a "domain" if the only zero divisor is 0, we will
also say R has "no" zero divisors.

Definition: A domain R is called "Euclidean" if there is a notion of
"size", i.e. a function | |:R - {0}--->Z, whose values are bounded below,
say by 0, such that after division, the remainder has smaller size. I.e.
given a,b, in R with b not zero, there exist q,r, in R such that a = bq +
r, and either r=0, or |r| < |b|.Definition: A map of R modules f:M--->N is a homomorphism, or simply R
mopdule map, iff it is a group map and preserves multiplication by scalars,
i.e. f(ax) = af(x) for all a in R all x in M.

Assume R is a Euclidean domain.
Problem 1: Prove every ideal I in R is "principal", i.e. is a cyclic module.
(hint: If I contains non zero elements, choose one x of smallest size,
and prove x divides all the other elements of I.) conclude that either I =
{0}, or I is isomorphic to R.

problem 2: Prove every submodule of R^m is finitely generated, and in fact
isomorphic to R^n where n is at most equal to m. hint: read the proof for
Z.

problem 3: If M is any R module, and x1,...,xm are any elements of M, prove
there is a unique R module map f:R^m--->M taking ei to xi , where e1 =
(1,0,...,0), e2 = (0,1,0,...,), etc...
Show moreover f is surjective if and only if the {xi} generate M and
injective iff the {xi} are linearly independent over R.

problem 4. If M is a finitely generated R module, say with m generators,
prove M is isomorphic to the cokernel of an R module map f:R^n--->R^m, where
f can be given by an m by n matrix of elements of R.

problem 5. Prove every m by n matrix over R can be diagonalized by
invertible row and column operations. hint: use induction on the "size" of
the upper left entry of the matrix instead of on the number of prime
factors.

problem 6: prove that if f:M--->N is any R module map, and g:M--->M and
h:N--->N are isomorphisms, then hfg:M--->N has kernel and cokernel
isomorphic to those of f.

problem 7: Prove: R^m/(Ra1e1x...xRamem), where a1,...,am are elements of R,
is isomorphic to R/Ra1 x...x R/Ram. problem 8: Conclude that if M is a finitely generated R module, with
a set of generators x1,...,xm, of minimal cardinality, then M is isomorphic
to some product
of form R/Ra1 x...x R/Ram, where some ai may be zero, but none are units

is this feasible for anyone?

 

[mathwonk]

remark on jobs: maybe useless: if you stick to your love, and go with what you want, there WILL be a job for you.
try not to be foolish, but also to push for your dreams. there is a god, and she/he loves you, and will tilt in your favor if you are pure of heart and pursue your love.

 

[JasonRox]

there is a god, and she/he loves you, and will tilt in your favor if you are pure of heart and pursue your love.

Prove it.

Anyways, regardless of the existence of God, the wise choice is to pursue your dreams and happiness shall come simply by the sake that you are chasing your dreams.

 

[mathwonk]

touche. i agree. i have also received unexplainably wonderful assistance at times in my quest, which may be simply because there is a unity in the world that responds when we accommodate to its rhythm. this universal harmony may be my god. but i am drifting from the topic.

 

[J77]

remark on jobs: maybe useless: if you stick to your love, and go with what you want, there WILL be a job for you.

Yep - I've always just gone with the flow.

''Career Plan'' are not words in my vocabulary

 

[mathwonk]

Day two of algebra: elementary ring theory

Rings and ideals
1) An ideal I of a ring R is a subgroup closed under multiplication by R.
An ideal is proper if and only if it does not contain a unit, iff it does not contain1. If I is a proper ideal of R, the quotient group R/I has a natural ring structure, such that R--->R/I is a ring map.
Two elements a,b of a ring are associates if a = ub, where u is a unit. An element u is associate to 1 if and only if u is a unit. If a divides b, every associate of a divides b. An associate of a zero divisor is a zero divisor.
An ideal I is generated by a subset {xi} of I, if each element of I is an R linear combination of a finite subset of the {xi}. If the same finite subset of elements {xi} can be used for every element of I, I is finitely generated. An ideal is principal if it has 1 generator.
A ring is noetherian, if every ideal is finitely generated, and
principal if every ideal is principal.
(Hilbert) If R is noetherian, so is R[X].

2) A ring R is a domain, if it has no zero divisors except 0; i.e. if xy = 0 implies at least one of x or y = 0.
An ideal I is prime if ab in I implies at least one of a or b is in I, iff R/I is a domain. An element x is prime iff the principal ideal Rx = (x), is prime, i.e. if x divides ab only when x divides at least one of a or b..
In a domain, two elements are associates iff they divide each other, iff they generate the same ideal.
A principal domain is a pid (principal ideal domain).

3) R is a field if each non zero element is a unit. A unit is not a zero divisor so a field is a domain.
A proper ideal I is maximal iff it is not contained in another proper ideal. A ring is a field iff the only proper ideal is {0}. An ideal I is maximal iff R/I is a field, so every maximal ideal is prime.
Every ideal I of any ring is contained in some maximal ideal.

3) A domain R is Euclidean, if there is a function | |:R-{0}---> {non neg integers}, s.t. for b ≠ 0, and any a, there exist q,r such that a = qb + r, where r = 0, or |r| < |b|. I.e. if b does not divide a, the remainder r is smaller than b. Euclidean domains are pid's.
A Euclidean domain R is strongly Euclidean, if for all a,b, |a| ≤ |ab|, and equality holds if and only if b is a unit.

4) An element x of a domain is irreducible if x is not zero, not a unit, and when x = bc, either b or c is a unit. A prime element of a domain is irreducible, but not vice versa.

5) A greatest common divisor, or gcd, of two elements x,y, in a domain, is an element z such that z divides both x and y, and if any element w divides x and y, then w divides z. An associate of a gcd of x,y, is a gcd of x,y. In a pid, any generator of the ideal (x,y) generated by x and y, is a gcd of x,y.

6) A domain is factorial, a unique factorization domain, or u.f.d., if each non zero, non unit, is a product of irreducible elements, and whenever x = ∏bi, = ∏cj, with all bi, cj irreducible, there is the same number of b's and c's, and after renumbering, each bi is associate to the corresponding ci.
In a ufd, x is prime if and only if x is irreducible.
Two elements of a ufd, have a gcd given by the product of the greatest common prime power factors in their prime factorizations.
Two elements of a ufd are called relatively prime if 1 is a gcd.
All pid's are ufd's. In a pid, x,y are rel. prime iff the ideal (x,y) = (1) = R.
(Gauss) If R is factorial, so is R[X].

7) If R is a domain, its field of fractions ff(R) = {a/b: a,b, are in R, b ≠ 0: a/b = c/d if ad=bc}. This field contains R, and is contained in every field containing R. Hence a ring is a domain if and only if it is contained in a field.

8) If S is a multiplicatively closed subset of R, with S containing 1 but no zero divisors, we form the fraction ring R(S) by allowing as denominators only elements of S, with the usual equivalence.
In the ring R(S), elements of S become units, and a ring map R--->T such that the image of every element of S is a unit in T, extends uniquely to a ring map R(S)--->T.
If S is the set of all non zero-divisors, R(S) is the total ring of fractions of R = ff(R), if R is a domain.
An important example is S = R-P where P is a prime ideal. Here we write R(P) for R(R-P).

9) If a ring F contains a domain R, x in F is integral over R, if x satisfies a monic polynomial over R. A domain R is integrally closed, or normal, if the only elements x in ff(R) integral over R are elements of R, (rational root theorem).
Ufd's are normal.

10) The Krull dimension of R is the maximal length -1 of a strict chain of proper prime ideals. A field has Krull dimension zero, Z has dimension one. A domain which is not a field has Krull dimension one if every non trivial prime ideal is maximal, e.g. a pid.

13) A domain is a Dedekind domain if it is noetherian, normal, and Krull dimension one. P.i.d.'s and rings of integers in numbers fields are Dedekind.

Relations among these properties:
Strongly Euclidean implies Euclidean implies principal implies u.f.d. implies normal. No implications can be reversed, but a one dimensional noetherian ufd is a pid, and if R is Dedekind R(P) is a pid for every prime ideal P. I.e. a Dedekind domain R is locally principal.

Modules: generalizing ideals, vector spaces, abelian groups
If R is a ring, and M an abelian group, a (left) R module structure on M is a ring map s:R--->End(M) = {group homomorphisms M---M}, where End(M) is a (usually non commutative) ring with composition as multiplication. Using this we can multiply elements of M by elements of r, by definition rx = (s(r))(x). Then (ab)x = a(bx), (a+b)x = ax +bx, 1x = x, and a(x+y) = ax+ay.

We mostly assume R is commutative. EndR(M) is the subring of End(M) of those group homomorphisms which commute with multiplication by elements of R. EndZ(M) = End(M), but EndR(M) is usually smaller than End(M). The homomorphisms in EndR(M) are called R module maps, similarly for HomR(N,M) in Hom(N,M).

If R--->End(M) is an R module structure, the kernel of
R--->End(M) = ann(M) = {elements r in R: rx = 0 for all x in M}, an ideal of R. An R module structure is faithful if ann(M) = {0}. For any R module structure, M has a natural induced faithful R/ann(M) module structure. M has a faithful R module structure if and only if R is isomorphic to a subring of End(M), and M has an R module structure iff some quotient R/I is isomorphic to a subring of End(M).

Ex. A finitely generated abelian group has no Q module structure.
An ideal of R is just a submodule of R.

14) A submodule is generated by elements {xi} of M, if each element of M is an R linear combination of a finite subset of the {xi}. If the same finite subset of elements {xi} can be used for every element of M, M is finitely generated. M is noetherian if every submodule is finitely generated. R is noetherian if it is a noetherian module. M is cyclic if it has one generator. Thus an ideal is cyclic iff it is principal. A cyclic module is isomorphic to R/I for some ideal I. In a p.i.d. a non zero ideal I is isomorphic to R as modules.

Exericses
15) If 0--->A--->B--->C--->0 is an exact sequence of R module maps, B is noetherian iff A and C are noetherian. (Hint: A,C fin. gen. implies B is also.)

16) If R is noetherian, R^n is a noetherian module.

17) If R is noetherian and M fin gen R module, M is noetherian.

18) R/P is a domain iff P is a prime ideal, and a field iff P is maximal.

19) A prime element of a domain is irreducible.

20) An element x of a domain R, is irreducible iff (x) is maximal among all principal ideals of R.

21) A Euclidean domain is p.i.d.

22) In a strongly Euclidean domain, a non zero non unit is a product of one or more irreducibles.

23) If k is a field, k[X] is strongly Euclidean where |f| = deg(f).

24) In a pid, a gcd of x,y is a generator of the ideal (x,y), hence any gcd is a linear combination of x,y.

25) R is a domain iff R[X] is a domain.

26) If {Ij} is a linearly ordered set of proper ideals in R, i.e. if for any two ideals Ij and Ik, one is contained in the other, their union is a proper ideal.

27) In a ufd R, if a,b are rel prime, and a divides xb, then a divides x.

Rings with unique factorization.
Assume all rings are domains.
Existence of factorization is easy by induction in any strongly Euclidean domain, and it also follows from the noetherian condition.

Definition: A partially ordered set has the ascending chain condition, or ACC, if strictly increasing sequences of elements are finite in length.

Lemma: The set of ideals in a ring R satisfies ACC wrt inclusion if and only if each ideal is finitely generated.
proof: If I is not finitely generated, let a1 be any element of I. Then (a1) does not equal I, so there is an element in I - (a1), say a2. Then (a1,a2) is strictly larger than (a1) but not equal to I, so we have a chain of two ideals (a1) contained in (a1,a2). Then we can choose another element a3 of I - (a1,a2) and then we have a strictly increasing chain of three ideals: (a1) in (a1,a2) in (a1,a2,a3). Continuing, we obtain an infinite sequence of strictly increasing ideals, contradicting ACC.
If finite generation holds, we claim no infinite weakly increasing sequence of ideals, I1, I2, I3,..., is strictly increasing. Take the union I of all the ideals, itself an ideal, hence finitely generated, say by x1,...,xn. Then all xi are in some one of the ideals in the chain, say IN. The remainder of the ideals in the sequence contain the generators of the union I, hence all the rest of the ideals are all equal to I and to IN. So the infinite sequence of ideals is not strictly increasing. QED.

Lemma: In a noetherian domain R, e.g. a p.i.d., a non zero non unit can is a finite product of irreducible elements.
proof: We will show if some non zero, non unit x, has no such expression, then there is an infinite strictly ascending chain of ideals in R.
Note: if x is a product of elements each of which is itself a product of irreducibles, then x is also a product of irreducibles.
Now we get a contradiction as follows. Since x is not a product of irreducibles, it is not irreducible, and not a product of elements which are themselves products of irreducibles. Hence x = a1b1, is a product of non units, where at least one factor, which we call a1, is not a product of irreducibles. Since b1 is not a unit, a1 does not belong to the ideal (x), so the ideal (a1) is strictly larger than (x). Then a1 is a product a1 = a2b2 of non units, where at least one, which we call a2, is not a product of irreducibles. Then (a2) is strictly larger than (a1).
Continuing, we get a strictly increasing sequence of ideals (x), (a1), (a2),..., contradicting the ACC. QED.

Corollary: In a pid, every non zero, non unit, factors into irreducibles.

So much for existence of irreducible factorizations. Now for uniqueness criteria

Lemma: In a domain R, let x = ∏xi = ∏yj where all xi and yj are prime elements. Then we claim there is the same number of x's as y's, and after renumbering, each xi is associate to the corresponding yi.
proof: Since x1 divides the left side hence also the right, by induction on the definition of prime element, x1 must divide some factor yj on the right. But since all prime elements of a domain are irreducible, then x1 is associate to yj. renumbering the y's we may assume yj is y1 = ux1, where u is a unit. Then after canceling x1 on both sides, and replacing y2 by its associate uy2, we are done by induction on the number of factors occurring on the left. QED.

Lemma: In a pid R, every irreducible element is also prime.
proof: If x is irreducible, then the ideal (x) is maximal among all principal ideals. But since R is a pid, then (x) is maximal, hence also prime, so x is a prime element. QED.

Corollary: Every p.i.d. is a u.f.d.
proof: Since a pid is noetherian, this follows from the two previous lemmas. QED.

More generally any noetherian ring in which all irreducibles are prime is a ufd.

Remark: This result is not reversible, since most ufd's are not principal. The pid's are in fact only the one dimensional (noetherian) ufd's, since we saw in the previous proof that every non zero prime ideal in a pid is maximal. There are noetherian ufd's of arbitrary finite dimension k[X1,...,Xn], and even non noetherian ufd's k[X1,...,Xn,...], of infinite dimension.

We will not take time right now to prove the fact that a noetherian domain is a ufd if and only if any two elements have gcd, but it is not so hard.

Corollary: We can diagonalize a matrix over any p.i.d., by invertible matrix operations, but not elementary row and column operations. I.e. if M is an m by n matrix, with entries in a pid R, there exist invertible matrices A,B over R, such that AMB is diagonal, in the sense that all entries xij, with i≠j, are zero.
proof: Using the same procedure as with integer matrices, it suffices by induction to show that we can arrange for the upper left entry of M to divide all the other entries in the first row and column. We get to use the elementary row and column operations, but we will supplement them by an additional invertible matrix multiplication which does not arise from a product of elementary matrices.
Recall the key step was to show that we can replace a first row of M containing [a b * * * ...], where a does not divide b, by [ d c * * * ...] where d is a proper divisor of a. Since a has only a finite number of proper divisors in a ufd, hence in any pid, this process can only be repeated a finite number of times. Hence eventually, the upper left entry will divide the adjacent entry, and by interchanging columns, also any other entry in the first row. Doing the same for the first column, we eventually get an upper left entry that divides all other entries in both the first row and column, and can be used to replace all other entries by zeroes. Then induction allows the matrix to be reduced to diagonal form.
Now to accomplish this, use the fact R is a pid, to make the key replacement by an invertible matrix operation as follows. If a does not divide b, then gcd(a,b) = d has strictly fewer prime factors than a, and d can be written as a linear combination d = ax+by, where after dividing through by d, we see that the gcd of x,y is 1. Hence we can write 1 = zx + wy. This let's us construct a matrix B with first two rows [ x -w 0 0 0 ... 0] , and [ y u 0 0 0 ...0]. This matrix multiplies our original one from the right to yield upper left entry ax+by = d = gcd(a,b).
Moreover this matrix can be completed to an invertible one B, since the 2 by 2 determinant in the upper left corner is 1. I.e,. we just put zeroes in all the rest of the first two column entries, and put an identity matrix in the bottom right corner.
QED.

Corollary: A finitely generated module N over any p.i.d. R, is isomorphic to a product of cyclic modules R^r x R/(x1) x ...x R/(xs), where no xi is a unit or zero, and each xi divides xi+1. Moreover, the ideals (xi) are uniquely determined by the isomorphism class of N, as well as the integers r and s.
proof sketch: As before, if N has m generators, we map R^m onto N, and since R is noetherian, the kernel of this map is finitely generated, so we can map some R^n onto this kernel, thus realizing N as the cokernel of a map f:R^n --->R^m, hence as the cokernel of a matrix M. Then diagonalizing the matrix by invertible operations as above, does not change the isomorphism class of the kernel and cokernel. But for a diagonal m by n matrix with diagonal entries z1,...,zm, dividing each other, (some of the early ones possibly equal to 1, and some of the last ones possibly equal to 0), the cokernel is easily shown to be isomorphic to the product R/(z1) x ...x R/(zm). Then we delete the factors at the beginning with z = 1, since they are {0}, and put the ones at the end with z = 0 at the beginning, since they are = R. Then letting the x i's be the z's that are different from 1 and 0, we have our decomposition. The uniqueness of the xi is no easier, but no harder, than before. For r, see below. QED.

To prove uniqueness of the rank r, we will use Zorn's lemma to produce some maximal ideals, and then appeal again to the well definedness of dimension of a finite dimensional vector space over a field.

Recall Zorn's lemma:
In any partially ordered set S, a "chain" or "linearly ordered subset" {xi}I, is a subset such that any two elements xi, xj are comparable.

Zorn's lemma says: If each chain in S has an upper bound in S, then S contains some "maximal" elements, i.e. elements which are not less than any other comparable element.

We assume Zorn's lemma, which follows from the axiom of choice. [A proof is in the appendix to Lang, Algebra.]

Corollary: Every ring (with identity) contains maximal ideals.
proof: We have already checked that the union of a linearly ordered collection of proper ideals is a proper ideal, and thus an upper bound for the collection. We are finished by Zorn. QED.

Now assume that N is an R module and I is an ideal of R. Then define IN as the submodule of N generated by all products rx where r is in I and x is in N. This equals all R linear combinations of form ∑ aixi where the ai are in I and the xi are in N.

Exercise: For a product of R modules, N = N1 x ... x Ns, we have IN ≈ IN1 x ... x INs.

Cor: If I is a maximal ideal of R, then R^n/IR^n ≈ (R/I) x ... x (R/I) ≈
K x ... x K, where K is the quotient field R/I.

Cor: If R^n ≈ R^m, for any ring R, then n = m.
proof: If R^n ≈ R^m, then K^n ≈ K^m, and by vector space theory, n = m. QED.

This implies the rank r of a finitely generated module over a pid is well defined.

Lemma: If R is a ufd, and K = ff(R) is its fraction field, we can put a fraction a/b in "lowest form" by canceling common prime factors top and bottom until top and bottom are relatively prime.
proof: Obvious.

Lemma: In a ufd, if a,b are rel. prime, then so are all positive powers a^n, b^m.
proof: Obvious.

The last result mimics the rational root theorem.

Proposition: If R is a ufd, and r/s in ff(R), is a root in lowest terms of a polynomial aoX^n + ...+an = 0, then r divides an and s divides a0.
proof: Substituting in and multiplying out denominators, gives
r^na0 + sr^(n-1)a1 + ...+s^(n-1)ran-1 + s^nan = 0, so s divides r^na0 and r divides s^nan. Since positive powers of s and r are relatively prime, s divides a0 and r divides an, as claimed. QED.

Corollary: Every ufd is integrally closed (in its field of fractions).
proof: If a polynomial mover R is monic, a root r/s from ff(R) in lowest terms, has denominator dividing 1, hence s is a unit of R, and hence the root r/s belongs to R. QED.

 

[mathwonk]

I just learned you can only post 20,000 characters.

 

[mathwonk]

the rest of the story

heres the rest of day two:

Proposition(Hilbert): If R is noetherian, so is R[X].
proof: If I is any ideal of S = R[X] we want to find a finite number of generators for I. Consider the set J of all leading coefficients of elements of I, and check that J is an ideal of R hence finitekly generated say by a1,...,an. Then for i = 1,..,n, choose an element fi of I that has leading coefficient equal to ai. Let r be the maximum of the degrees of the polynomials f1,...fn. If f is any polynomial in I of degree ≥ r, by multiplying the fi by suitable poiwers of x, we obtain polynomials gi of the same degree as f, and whose leading coefficients generate the ideal of all leading coefficients of elements of I. Hence there is an R linear combination of the gi which ahs the same degree and the same leading coefficient as does f. This linear combination is also an R[X] linear combination of the fi. Thus we have for some polynomial coefficients hi, that ∑ hifi - f has lower degree than f. Repeating this we eventually can lower the degree of f until it is less than r. I.e. for some polynomial coefficients ki we get that ∑ ki fi - f belongs to I and also to S(r) = the module of polynomials in R[X] of degree less than r. Since the module S(r) is generated over R by 1,X,...,X^(r-1), it is finitely generated over R, hence noetherian as R module, hence certainly also as R[X] module. Thus I intersect S(r) is a finitely generated R[X] module, so we can choose a finite number of R[X] generators t1,...,tm for it. Then ∑ ki fi - f = ∑ -witi for some polynomials wi, and f = ∑ ki fi + ∑ witi. Hence the finite set {ki, tj} generates I over R[X]. QED.

Proposition(Gauss): if R is a ufd, so is R[X].

Definition: A polynomial in R[X] is called primitive if 1 is a gcd of its coefficients.

Lemma(Gauss): The product of two primitive polynomials is primitive.
proof: f is primitive iff for every prime element p of R. f remains non zero in (R/p)[X]. If f and g are primitive, both remain non zero in (R/P)[X]. Since p is a prime element of R, (R/P) is a domain, and so is (R/p)[X]. Thus fg is also non zero in (R/p)[X] for all primes p of R, so fg is also primitive. QED.

With this lemma, it is a straight forward but slightly tedious exercise to deduce that R[X] is a ufd. from the fact that K[X] is a ufd, where K is the fraction field of R.

Exercise: (i) If f is any polynomial in R[X], there is some d in R such that f/d is a primitive polynomial in R[X]; d = gcd of coefficients of f, is unique up to associates.
(ii) if f is a polynomial in K[X] not in R[X], there is some d in R such that df is a primitive polynomial in R[X], and d = lcm of denominators of coefficients in lowest terms of f, is unique up to associates.
(iii) The units of R[X] are precisely the units of R.

Lemma: If f is primitive in R[X] then f is irreducible in R[X] iff it is so in K[X].
proof: Asume f irred. in K[X], and that f = gh with g,h in R[X]. Then either g or h, say g, is a unit in K[X], hence a non zero element of K. But g belongs to R[X], so g is a non zero element of R. Since f is primitive, and h belongs to R[X], g is a unit in R, and f is irreducible in R[X].
If f is irred. in R[X], assume f = gh, with g,h in K[X], and neither is in R[X], then after multiplying out by the product of the lcm's of the lowest terms denominators of g,h, we have a primitive polynomial on the right but not the left, a contradiction. If both g,h belong to R[X], by hypothesis one is a unit in R[X], hence also a unit in R and K and K[X], and we are done. So we may assume g is in R[X] and h is not. Then for appropriate d,e in R, as in the exercise above, we have f = gh = (d/e)(g/d)(eh), where f, (g/d), and (eh), are primitive in R[X]. Hence d and e are associates, and we have f = (g/e)(eh), where (g/e) and (eh) are both in R[X], hence one is a unit in R[X], i.e. in R. Thus either g or h is a unit in K, hence K[X], and f is also irreducible in K[X]. QED.

Lemma: If R is a ufd, then every primitive non zero non unit f in R[X] factors into irreducible elements of R[X].
proof: By hypothesis f has degree > 0. Since K[X] is a ufd, f = ∏ gi where all gi are irreducible in K[X]. If all gi belong to R[X] then they are primitive hence irreducible by the previous result and we are done. As before, some of them must belong to R[X] or else we find a non unit d of R such that df is a product oif primitive polynomials of R[X], a contradiction. In any event, we again find elements d,e of R such that f = (d/e)∏hi where the hi are primitive elements of R[X] each a non zero R multiple of the corresponding gi , hence each hi still irreducible in K[X]. Then each hi, being primitive, is also irreducible in R[X], and we are done. QED.

Corollary: If R is a ufd, every non zero non unit f in R[X] factors into irreducible elements of R[X].
proof: If f is not primitive, write f = cg where g is primitive in R[X] and c is a non unit in R. Factor f as above into porimitive irreducibles, and factor c into irreducibles in R. These are still irreducible in R[X]. QED.

Lemma: If f is primitive in R[X], g any element of R[X], and if f divides g in K[X], then f already divides g in R[X]. In fact if g = fh, with h in K[X], then h is in R[X].
proof: Assume g = fh, with g in R[X], and h in K[X]. If h is not in R[X], as above there is some c in R, not a unit, with cg = f(ch), where f and ch are primitive in R[X], but cg is not, a contradiction. QED.

Corollary: An irreducible element of R[X] is prime if R is a ufd.
proof: Assume f is irreducible in R[X] and hence primitive, and that f divides gh, with g,h in R[X]. Since K[X] is a ufd, and f is still irreducible in K[X], f is prime in K[X], so f divides either g or h in K[X]. Since g,h are in R[X], the previous result shows that f divides one of them in R[X]. Hence f is prime in R[X]. QED.

Corollary: If R is a ufd, so is R[X].
proof: We have shown R[X] has factorization into irreducibles, and that every irreducible in R[X] is prime. But factorization into primes is always unique. QED.

 

[mathwonk]

basic algebra 1, the best intro to abstract alg

The following review of nathan jacobson's superb book seems to have been deleted from amazon for some reason:

This book has been reviewed by many people from several perspectives. The low reviews seem to come from students who could not follow it. This is not the fault of the author.

This book is by an expert algebraist who has rewritten his earlier introduction to algebra from the experience gained after 20 years as a Yale professor. It contains correct insightful proofs, carefully explained as clearly as possible without compromising their goal of reaching the bottom of each topic.

Other books say that one cannot square the circle with ruler and compass because it would require solving an algebraic equation with rational coefficients whose root is pi, and after all pi is a transcendental number. But Jacobson also proves that pi is a transcendental number, so as not to leave a logical gap. Naturally the burden on the student is somewhat higher than if he is merely told this fact without proof.

It is true that Dummit and Foote have included many more examples, and discussed them at extreme length, producing a book of over 900 pages, whereas Jacobson's book is less than 500 pages, hence cannot include as many words. But Jacobson's words are sometimes far better chosen, as he apparently understands the material at greater depth than those authors.

Nonetheless, Jacobson has made a sincere, and I thought very successful effort, to write his 2 volumes on 2 different levels of sophistication, the first being back - bendingly clear and painstakingly organized as to the true logic of the subject.

As an example of the difference between the books, on page 288 of DF, lines 6-15, they claim to prove existence of irreducible factorization in a pid, by showing no increasing sequence of principal ideas exists, but do not in fact demonstrate that failure of factorization does lead to an increasing sequence of principal ideals. They seem to think saying "and so forth" (Line6) is a proof. They mislead the reader to believe that it is obvious that an infinite tree beginning from one vertex, with either 2 or no edges down from each previous vertex, must contain an infinite path.

This is true but requires proof. Hungerford gives a correct proof on page 138 but one that works only for a pid. Jacobson on the other hand gives essentially the same proof. but more clearly and with more elementary assumptions, and still making it work more generally, in any noetherian ring, coincidentally also on page 138.

After choosing DF for my beginning graduate algebra course, I discovered the superiority ( from my perspective) of Jacobson, and wondered in amazement how such a great work could have been allowed to go out of print. After reading these reviews I understand. The readers who ignorantly criticize the experts have eventually managed to veto the use of their works in classes. This makes the market share fall, and the books cease to exist. We have been obliged recently to remove Jacobson from our list of PhD references, in spite of its excellence, because it is out of print. This is a real disservice to our PhD students seeking to understand the material they will need to use.

To be honest, we should admit that easier books exist, but they are easier at the cost of omitting to get to the bottom of the matter at hand, and at spending hundreds more pages doing so. I find DF rather tedious reading myself, as do some of my beginning students, but I still learn many small things there for which I am grateful.

For instance it is nice to have the little theorem (uncredited there but due perhaps to Kaplansky?) that a domain in which all primes are principal is a pid. (DF, ex. 6 p. 283). It follows that a "one dimensional" ufd, is a pid, and in fact that is a very simple characterization of pids. Thus although it is very nice that DF do give enough information to deduce this simple characterization of a pid, they do not seem to observe it themselves anywhere that i can find. The characterization they do give, that a ufd which is a "Bezout domain" is a pid, seems less natural.

Also their history seems flawed. On page 281 they define a "Dedekind Hasse norm" and credit John Greene writing in the Math Monthly in 1997, with noticing it is fully equivalent to being a pid. To me this is merely silly. The explicit characterization appears essentially the same on page 244 of Zariski Samuel, Cor. 2, in their 1958 book Commutative Algebra, and the direction DF attribute to Greene is so trivial, it was certainly obvious to Dedekind.

I like the clarity of DF and the multitude of examples, but I notice the mistakes, and the lack of global insight into the subject that rolls off Jacobson's fingertips. By all means read DF for help entering the subject, they have done a wonderful service for many beginning students and even old instructors like me. But please be aware that there exist strong students for whom Jacobson's book offers insights not to be found in DF at all. I.e. if you can read Jacobson, you will come away with a better understanding than from other books like DF.

A prime example is in the treatment of finitely generated modules over a pid. This absolutely basic topic is treated rather abstractly in DF, page 460, and concretely only in an exercise, page 470. By contrast, Jacobson gives the concrete treatment in the text, very clearly showing how to diagonalize a matrix, pages 176-186, and immediately using it to apply to canonical forms of matrices, pages 186-202.

DF also use diagonalization of matrices to do their applications to matrices, page 480ff, (indeed there is no other equally practical method), but unlike Jacobson they have not previously explained the solution of that exercise. I find it frustrating when authors choose a useless abstract proof that they themselves find unsuitable for applications later, and then merely cite an exercise for justification of its correctness. A student who reads this book is being given a treatment with logical omissions, to make his road easier but incomplete.

Average students, which is most of us, have the right to learn a subject, but we should not have the right, and we are unwise to try, to vote the best books out of existence simply because we cannot understand them. Please, aspire to understanding the deeper treatment in Jacobsons book. Let's put our copy of Jacobson away and save it, if we cannot yet read it.

To call it a bad book, is simply showing our own limited understanding. Clearly it is not the first book for everyone, but it is still perhaps the best, most masterful treatment of the material in existence to my knowledge at the upper undergraduate - graduate level, for the student who aspires to real mastery and understanding.

I am very unhappy with this sort of situation which prevents me and other seekers after knowledge, even now from finding a copy of Jacobson's volume 2, at an affordable price. They have been driven from the market by people who do not appreciate them for what they are, the best works on the topic, and who are uninterested in being raised to a higher level.

When I was a student I could read Shafarevich but not Hartshorne. Better students than I who read Hartshorne were nonetheless greatly benefited. More power to them. I still have my Hartshorne and hope to master it sometime.

I also hope someday to master Jacobson, whose volume 2 used to look extremely abstract and unapproachable to me. I wish I had bought one back then when it was for sale.

This review is being written today because I came to this site seeking a copy of Jacobson vol 2, and am very frustrated at not finding any affordable ones still out there. Perhaps their owners appreciate them too much to release them.

Best wishes to all.

 

[mathwonk]

more day two, but it probably won't load properly: oh boy a lot of symbols in the proof just disappeared. oh well, you can look in van der waerden for the proof.

Corollary: if k is a field or ufd, then k[X1,...,Xn] is a ufd.
proof: By induction, since k[X1,...,Xn] is isomorphic to k[X1,...,Xn-1][Xn].

Corollary: If k is a field or ufd, then k[X1,...,Xn,...] is a ufd.
proof: A given polynomial involves only a finite number of variables, and cannot factor into a product of factors which involve other variables. Hence a polynomial in k[X1,...,Xn] which is irreducible there is also irreducible in k[X1,...,Xn,...]. This proves existence of irreducible factorizations. But these rings also have the same units, so a polynomila irreducible in k[X1,...,Xn,...] is also irreducible in k[X1,...,Xn]. Thus if f is in k[X1,...,Xn], any two irreducible factorizations of it in k[X1,...,Xn,...] actually belong to k[X1,...,Xn], hence are equivalent there and also in k[X1,...,Xn,...]. QED. Corollary: (Eisenstein): Assume R is a ufd, f in R[X] is a polynomial of positive degree n, and p is a prime element of R that divides every coefficient ai of f with i < n, but not the leading coefficient an, and that p^2 does not divide the constant term a0 of f.
Then f is irreducible in K[X], where K = ff(R).
proof:
If f were reducible over K , the factors would have degree at least one, and we have shown the factors can be chosen in R[X]. (f = c(f).f0, if f = gh, with g,h, in K[X], we have f0 = c(f)-1g.h, and we showed then that f0 = g1h1 where g1 and h1 are the primitive versions of c(f)-1g, and h. Then f = c(f)f0 = c(f)g1h1, is a factorization of f in R[X].)
Thus f = gh, where g,h are in R[X] and not constants.

If we reduce mod p, we get [f] = [g][h], in (R/(p))[X] = [c]X^n, by hypothesis, where [c] is not [0].

Thus both [g] and [h] have non zero leading terms of degree < n. We claim both [g] and [h] have zero constant terms.

If [g] say, has non zero constant term, then multiplying the constant term of [g] by the lowest degree term of [h], gives a term of degree = deg([h]) in [f]. But the lowest degree term of [f] has degree n > deg([h]). This contradiction shows both constant terms of [g] and [h] are zero in R/(p).
That implies p divides the constant terms of both g and h, hence
p^2 divides the constant term of f, in contradiction to the hypothesis.

[Even simpler, although (R/p)[X] is not a ufd, X is a prime element of this domain since the quotient by X is R/p a domain. Factorization by any prime elements is always unique in any domain, so the factors [g] and [h] must be associate to monomials of positive degree in X, hence neither has a constant term.] QED.

Rational root theorem:
If R is a ufd, f = <sum> ciXi ,i = 0,..,n, is a non constant polynomial in R[X], and r/s is a root of f in lowest terms in K, then r divides c0 and s divides cn.
proof: Substitute X = r/s into f(X) = 0 and multiply out the bottoms, getting anrn+ san-1rn-1+...+sn-1a1r + sna0 = 0. Then r divides every term except the last hence also that one. But r and s are relatively prime, so r divides a0. Similarly, s divides an. QED.

Definition: If R is a domain with fraction field K, an element of K is called integral, over R if it is a root of a monic polynomial in R[X].
If R is a domain that contains every element of K which is integral over R, we call R "integrally closed" (in its field of fractions) or "normal".

Corollary: Every ufd is integrally closed, i.e. normal.

The valuation associated to a prime.
Let p be a prime in a ufd R, with fraction field K. define a function
vp:K*--->Z from non zero elements of K to the integers, as follows: if x = a/b with a,b in R, then write a = cpr, and b = dps, where p does not divide either c or d, and define vp(x) = vp(a/b) = vp(a)-vp(b) = r-s. Thus vp(x) is "the number of times p divides x", i.e. the number of times it divides the numerator minus the number of times it divides the denominator. Thus an element x of K is in R iff vp(x) >= 0 for every prime p in R.
In geometry we think of the primes p as points, the elements x as functions, and the functions x with vp(x) < 0 are the ones with poles at p, while those with vp(x) > 0 have zeroes at p. The absolute value of vp gives the order of the zero or pole at p.
Note that the exponent i of Xi is nothing but the valuation vX(Xi) = i, determined by the prime element X of R[X]. So we are in some sense looking at f as a polynomial in the two variables X and p.

(Eisenstein-Dumas): Assume R is a ufd, and f = <sum> aiXi is a polynomial of degree n over R with a0 != 0. Graph the integer lattice points (i,vp(ai)) in the plane ZxZ, and connect up the "first" and "last" points, (0,vp(a0)) and (n,vp(an)), by a line segment L. If the following two conditions hold:
(i) All intermediate lattice points (i,vp(ai)) for 0<i<n, lie on or above L, and
(ii) gcd(n, vp(an)-vp(a0)) = 1,
then f is irreducible over K = ff(R) = fraction field of R.

Corollary: (Eisenstein).
proof: Here we have a line segment L which has height <= 1 everywhere on the interval [0,n], and by hypothesis all intermediate points have height >=1. Moreover, vp(a0) = 1, vp(an) = 0, so gcd(n,-1)=1.

Corollary:(reverse Eisenstein): If p is prime and divides all ai for i >0 but not a0, and p^2 does not divide an, then f is irreducible over K.
proof: Here the line segment L goes from (0,0) to (n,1) instead of from (0,1) to (n,0), hence has the same slope, and all the intermediate lattice points are still above it.

Recall the usual root - factor theorem implies that a polynomial of degree <= 3 with no root over a field, is irreducible. Here is a related result.
Corollary: Let q be a prime integer, and consider f(X) = X^q -c, where c lies in a ufd R. If c has no qth root in R, then f is irreducible over K = ff(R).
proof: Since c has no qth root in R, there is some prime factor p of c such that p does not divide vp(c). Thus gcd(p,vp(a0)) = 1. Since there are only the two extreme lattice points, and vp(aq) = 0, we are done.

Corollary: Irreducibility of polynomials in two variables:
If n, m are relatively prime, then X^n - Y^m is irreducible in k[X,Y], where k is a field. E.g. X^2-Y^3 is irreducible in k[X,Y].
proof: Regard k[X,Y] as the polynomial ring k[Y][X] over the ufd k[Y], where Y is a prime element. Note X^n -Y^m is primitive.

Corollary: If a != 0,1, then Y^2 - X(X-1)(X-a) is irreducible in k[X,Y].
proof: Usual Eisenstein applies to this monic hence primitive polynomial, for the prime element X, in the ring k[X][Y].

proof of Dumas criterion:(Van der Waerden, 2nd ed. vol.1, page 76) For a polynomial f of degree n in R[X], and any prime p in R, graph the set of lattice points defined by the valuation vp. and choose the line L: { }, with , relatively prime integers, and , so that it contains at least 2 lattice points, but none lie beneath it. Then the slope of L = - , is a fraction in lowest terms.
We think of the linear function as assigning a weight to the lattice point coming from the monomial , where p does not divide c. To say that all lattice points lie on or above the line L, means we have chosen as the lowest weight that occurs, and we have chosen our weighting function so that at least two lattice points attain this lowest weight.
Now choose 2 distinct lattice points and on the line L, with as small as possible, and as large as possible, i.e. choose the leftmost and rightmost lattice points on the line L. Then we have = = , so . Since = 1, then divides , and divides . I.e. = m , so = -m , where m = gcd( , ) >= 1.

Lemma: With notation as above, if f = gh, where g,h are polynomials in R[X], then deg(g) = m1 +r1, and deg(h) = m2 +r2, where all of m1,m2,r1,r2 are >= 0, m1+m2 = m, and r1+r2 = n - m .

Remark: This lemma will do the job, since m > 0 implies at least one mi > 0, so at least one factor g or h has degree >= . But under the hypotheses of the theorem, = n, so one of the factors has the same degree as f, and the other must be a constant, hence a unit in K. Thus f would have to be irreducible over K.

proof of lemma:
Now assume f = gh, over R[X]. When we multiply a monomial of g by a monomial of h, the weights add. Then we have to combine terms according to like powers of X, the sum of two monomials with the same degree in X, will have weight equal to that of the monomial of smaller weight. So the weight of the monomial in X of degree d in f, will be the smallest of the weights of all products of monomials of g, h whose degrees add to d.
Now among all monomials of g let the least weight which occurs be , and among all monomials having this least weight, choose one with smallest degree in X, say , and one of highest degree in X, say . Then the same argument showing = m , implies = m1 , for some m1 >= 0. (This time may = 0, so m1 may equal 0.)
Now consider h, let be the least weight of any term of h, and let it occur for the monomial of lowest degree in X, and also for the monomial of highest degree in X. Then again = m2 , for some m2 >= 0.
Then among all products of monomials of g,h, the least weight which occurs is , and the least degree in X of such a monomial product is , i.e. this occurs for the product . When this monomial is combined with other monomials of the same degree in X, the weight of the resulting monomial is as observed above, the smallest weight which occurs, namely .
All other monomials of this least weight must have higher degree in X. Hence this monomial is the one of lowest degree in f, which ahs lowest weight. In particular = , the lowest weight of any monomial in f, and also = = the lowest degree in X, of a monomial of lowest weight in f. Similarly, = .
Hence m = = - ( ) = = m1 +m2 = (m1+m2) , whence m = m1+m2, as claimed. Hence some mi > 0.
Now since g contains the monomial , we know deg(g) >= . Since >= 0, we get deg(g) >= >= = m1 , i.e. deg(g) = m1 +r1, r1 >= 0. Similarly, deg(h) = m2 +r2, for some r2 >= 0. Since n = deg(f) = deg(g)+deg(h) = m1 +r1 + m2 +r2 = m +r1+r2, we have r1+r2 = n-m .
QED lemma, and hence also Dumas criterion.

 

[mathwonk]

having just prepared my lectures on canonical forms of matrices and read both jacobson and sah on the topic, i think i like sah better,

so here is another top recommnendation of an abstract algebra book:

chi han sah, abstract algebra. this is a really fine book. at harvard in about 1963, sah taught the undergraduate algebra course, which apparently passed up the graduate algebra course about thanksgiving and never looked back. this book is the result of that and other courses sah taught in that period. it is superb.

 

[mathwonk]

here are some used copies of sah beginning at $10.

http://www.abebooks.com/servlet/SearchResults?sts=t&an=sah&y=10&tn=abstract+algebra&x=57


i myself own two copies, one for home, one for the office.

 

[courtrigrad]

If one had the option to take either abstract algebra or real analysis, which one should he take? Also, mathwonk what would you recommend a freshman math major do over the summer?thanks

 

[JasonRox]

If one had the option to take either abstract algebra or real analysis, which one should he take? Also, mathwonk what would you recommend a freshman math major do over the summer?


thanks

That's like asking should I take Physics or Psychology.

I would simply go with which ever is of interest to you. For myself, I found both to be interesting. If you really liked Calculus, Analysis would be something to seriously consider.

 

[mathwonk]

as jason implied, both topics are important so you might want to pick based on which has the better teacher, or which you enjoy more.over the summer also you need to do what you want to, but if you want more math, there are vigre, REU, and other summer research experiences. also there is summer school, but you might want to go to the mountains and store up natural energy for the school year.

 

[balletomane]

I've found this thread extremely helpful. Thanks. :)

My question is a little strange: how can I know if I've learned something well enough and how can I improve my retention?

For example, I took calculus in high school and did well. Less than two years later I wouldn't be able to set up a double integral without looking in a book first. How can I keep this from happening, given that I have limited time to review old material.

Also, I am taking abstract linear algebra (we're using Axler) and finding it very difficult (in particular because I haven't had linear algebra before). Is it just me or is the learning curve really steep?

 

[mathwonk]

the learning curve in math is always steep, as the way of thinking and the objects of thought are not found in real life.

forgetting time is very short for this reason, and so even specialists like me forget their own research in 3-6 months.

so always review and think about the topic. try to get beyond reading and on to doing. the best way to review is without the book, trying to reproduce previously learned material. you will find you can reconstruct a good bit of it, the rest you need to review. anything you learn well the first time never needs to be relearned, but this reveals that much material is not well learned the first time.

hang in there, what other fools have done, you can do, as the great sylvanus p. thompson said.

 

[domhal]

Interesting reading. I too had been wandering recently about retention of details from courses past - in particular how it seemed like I had forgotten too much!

As an example I was looking over some analysis recently (I took analysis most recently in the Spring of this year) and had to show that the set of all limit points of E is closed. Feels like it took far too long to do!

Nice suggestion Mathwonk, I'll see how that goes.

 

[balletomane]

Thanks so much, Mathwonk.

 

[mathwonk]

one thing i do myself, is i always try to reproduce material without, looking at any books or notes. that's how i see how much i have understood. this year i am teaching grad algebra and i try every day to write down the lecture without consulting any book or even my own notes from the previous course. i do not always succeed and when i do not, i make a point of noting what it was i overlooked.

 

[courtrigrad]

so basically if one can write down what he remembers perfectly, he is an expert in that area?

 

[CPL.Luke]

^ pretty much, if you are able to reproduce all material on command perfectly without any reference, then you are obviously very knowledgeable on the subject. And if you are very knowledgeable about a particular subject, then doesn't that make you an expert?