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