# I Polynomial ideals

#### Wledig

This time my struggle is with ring ideals. Book still won't provide examples, so I'm again trying to come up with some of my own. I figured {0,2} might fit the definition as an ideal of $\mathbb{Z/4Z}$ since it is an additive subgroup and $\forall x \in I, \forall r \in R: x\cdot r, r\cdot x \in I$. But I'm not sure if I'm missing something here or, if I'm correct, how to generalize this result. Biggest problem though has been coming up with polynomial ideals, if someone could provide an explicit example I'd be really grateful.

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#### fresh_42

Mentor
2018 Award
This time my struggle is with ring ideals. Book still won't provide examples, so I'm again trying to come up with some of my own.
There should be plenty found on the internet: matrix rings, polynomial rings, number rings or any algebra, which are rings, too.
I figured {0,2} might fit the definition as an ideal of $\mathbb{Z/4Z}$ since it is an additive subgroup and $\forall x \in I, \forall r \in R: x\cdot r, r\cdot x \in I$. But I'm not sure if I'm missing something here or, if I'm correct,...
Yes, $2\mathbb{Z}/4\mathbb{Z}$ is an ideal in $\mathbb{Z}/4\mathbb{Z}$.
... how to generalize this result.
One way is the above: $I,J \trianglelefteq R$ and $J\subseteq I$ then $I/J \trianglelefteq R/J$ or more generally by the use of the isomorphism theorems.
However, I would start without factor, resp. quotient rings. Just two sources found on Google by "examples of ideals rings":
http://www.maths.usyd.edu.au/u/de/AGR/CommutativeAlgebra/pp22-51.pdf
http://www-history.mcs.st-and.ac.uk/~john/MT4517/Lectures/L5.html
You get probably better results, if you demand the ring to be PID or Euclidean, with zero divisors, matrix ring or similar, means you probably don't know many other distinctions yet.
Biggest problem though has been coming up with polynomial ideals, if someone could provide an explicit example I'd be really grateful.
To begin with, you could take an arbitrary polynomial $f(x) \in \mathbb{R}[x]$ and consider the ideal generated by $f$, which means all polynomials, which are divisible by $f$. E.g. for $f(x)=x^2$ you get $I=\mathbb{R}x^2 + \mathbb{R}x^3 +\ldots$ and $\mathbb{R}[x]/I \cong \mathbb{R} + \mathbb{R}x$. Now you can start to consider
• more complex $f$
• other fields as $\mathbb{R}$
• ideals generated by more than one polynomial, i.e. $I=f(x)\cdot \mathbb{R}[x] + g(x)\cdot \mathbb{R}[x]$
In any case you should repeat long division of polynomials.

#### Stephen Tashi

Biggest problem though has been coming up with polynomial ideals.
Consider the set of all polynomials that vanish at x = 3. The sum of two such polynomials is a polynomial that vanishes at x = 3. The product of such a polynomial with another arbitrary polynomial is a polynomial that vanishes at x = 3.

Intuitively, there is a similarity between that situation and the situation with the integers where the property of "is divisible by 3" can be used to form ideals.

It would be interesting to see examples of ideals where the defining property of the ideal didn't resemble the concept of "is divisible by". If $I$ is a (right) ideal and $P(x)$ is the property $x \in I$ then
$P(a)$ and $P(b)$ implies $P(a+b)$
$P(a)$ implies $P(ax)$.
What's an example of a property $P$ that isn't analogous to "is divisible by" ?

#### Wledig

To begin with, you could take an arbitrary polynomial $f(x) \in \mathbb{R}[x]$ and consider the ideal generated by $f$, which means all polynomials, which are divisible by $f$. E.g. for $f(x)=x^2$ you get $I=\mathbb{R}x^2 + \mathbb{R}x^3 +\ldots$ and $\mathbb{R}[x]/I \cong \mathbb{R} + \mathbb{R}x$. Now you can start to consider
• more complex $f$
• other fields as $\mathbb{R}$
• ideals generated by more than one polynomial, i.e. $I=f(x)\cdot \mathbb{R}[x] + g(x)\cdot \mathbb{R}[x]$
In any case you should repeat long division of polynomials.
That clarified things quite a bit, but I get the feeling that my problem with polynomial ideals is that they all seem infinite. Is that true? Looking for a more explicit example, I somehow assumed that the polynomial ideal had to be finite. Dumb mistake, since there's nothing in the ideal definition that prohibits it from being infinite.

What's an example of a property $P$ that isn't analogous to "is divisible by" ?
Hm, I think you got me with this one. Looking up matrix rings, I could generalize my result with $2\mathbb{Z}/4\mathbb{Z}$, by just taking the matrix with entries from it which is clearly an ideal of the ring of matrices with entries from $\mathbb{Z}/4\mathbb{Z}$. Still, I'm guessing the notion of divisibility is still implicit here?

#### fresh_42

Mentor
2018 Award
That clarified things quite a bit, but I get the feeling that my problem with polynomial ideals is that they all seem infinite. Is that true? Looking for a more explicit example, I somehow assumed that the polynomial ideal had to be finite. Dumb mistake, since there's nothing in the ideal definition that prohibits it from being infinite.

Hm, I think you got me with this one. Looking up matrix rings, I could generalize my result with $2\mathbb{Z}/4\mathbb{Z}$, by just taking the matrix with entries from it which is clearly an ideal of the ring of matrices with entries from $\mathbb{Z}/4\mathbb{Z}$. Still, I'm guessing the notion of divisibility is still implicit here?
No. For one thing, there are finite fields $\mathbb{Z}_p = \mathbb{Z}/p\mathbb{Z}$ and we can build the ring $\mathbb{Z}_p[x]/x^n\mathbb{Z}_p[x]$ in which case we would have polynomials of highest degree $n-1$ with coefficients in a finite field, so everything here is finite. But the other thing is, that finiteness in this context is rather irrelevant, as long as you do not use it for coding theory or other areas in which finite fields play a role. If we allow e.g. rational coefficients, then of course we have infinitely many, the constant polynomials alone are already infinite.

Mathematically interesting are finitely generated ideals, i.e. $I=a_1R +\ldots +a_nR$ or the question whether a somehow differently defined ideal is finitely generated or not. In principal ideal domains, ideals are generated by only one element, that is $I=f(x)\mathbb{Q}[x]+g(x)\mathbb{Q}[x] \trianglelefteq \mathbb{Q}[x]$ can always be written as $I=h(x)\mathbb{Q}[x]$.

Matrix rings offer many examples, however usually not commutative. So all starts with the requirements of your ring. E.g. consider
$$I= \left\{ \,\begin{bmatrix} a&b \\ 0&c \end{bmatrix}\, \right\} \trianglelefteq \mathbf{M}_{2,2}(\mathbb{Q})$$
Is $I$ an ideal and does it be connected to divisibility? As long as we stay close to numbers, and polynomials are still close for several reasons, divisibility is all around - in the end because ideals and rings were defined for the reason to investigate numbers.

#### mathwonk

Homework Helper
Polynomials in one variable over a field, like the ring of integers, are as fresh said, what is called a "principal ideal domain", i.e. every ideal in those rings is generated by one element. So in those rings all ideals are of the form "all elements divisible by some one given generator". If you jazz up to polynomials in 2 variables, suddenly you have ideals that have more than one generator. Generators of ideals are somewhat analogous to generating sets of vector subspaces, i.e. given elements u1,....,un of a ring R, the ideal they generate consists of all linear combinations r1.u1+...+rn.un for all choices r1,...,rn of elements of R.

E.g. the ideal in k[x,y] generated by x and y, consists of all polynomials in the two variables x,y such that every term is divisible by either x or y. So in a sense the concept of divisibility is central to the idea of an ideal. Every ideal has a set, possibly infinite, of generators and the ideal consists of all finite sums of elements A1,...,An such that each Aj is divisible by one of the generators. In particular all polynomial ideals except (0) are infinite since if f is one ≠ 0 element, so is x^n.f for all n. In fact the polynomial ring has the property that every ideal can be generated by a finite set of generators. Such rings are called "noetherian" after emmy noether, the author of the famous symmetry theorems in mathematical physics, although the fact that polynomial rings are noetherian is due to hilbert.

It is interesting that many ideals, but by far not all, can be described in terms of "vanishing at" a given set. E.g. the ideal generated by x and y in k[x,y] is exactly the ideal of all polynomials vanishing at the origin (0,0). However the ideal generated by x^2, xy, y^2, i.e. the "square" of that ideal, does not have such a description, although one can try to describe it as polynomials vanishing at (0,0) together with certain of their derivatives, i.e. polynomials vanishing "twice" at (0,0). There are infinitely many different ideals, generated by x^n, x^n-1.y,....,x.y^n-1, y^n, that also vanish precisely at (0,0), but with increasing "order of vanishing".

for a general theory of this topic, consult any modern book on algebraic geometry, such as Mumford's "redbook". There is a one one correspondence between all ideals of k[x,y] and all "closed affine subschemes" of the affine plane k^2. A more elementary book is Algebraic Curves by William Fulton.

Another good undergraduate book is Ideals, Varieties and Algorithms, by Cox, Little and O'Shea. In a principal ideal domain where each ideal consists of all elements divisible by some one element, one checks whether and element belongs to that ideal by dividing it by that element and seeing whether there is no remainder. In a polynomial ring of one variable this is achieved by the division algorithm. In a polynomial ring of more variables, belonging to an ideal generated by say f,g,h means deciding if an element is a sum of terms each of which is divisible by one of f,g, or h. So one needs a generalized division algorithm suited to division by more than one divisor. In modern times algorithms to decide this have been developed, based on the theory of "grobner bases" and this is discussed in the book of Cox, Little and O'Shea. this allows one to use computers to decide whether a given polynomial does or does not belong to the ideal generated by a given finite set of polynomial generators.

The geometric point of view already gives a lot of insight, since every algebraic plane curve defines an ideal of polynomials vanishing identically on that curve. But as suggested above, there are infinitely different ideals all having the same vanishing locus. If we restrict to "radical" ideals, those such that whenever a power of a polynomial belongs to the ideal then the polynomial itself already does so, there is a one one correspondence between algebraic plane curves and radical ideals of k[x,y], at least when k is an algebraically closed field. More generally, if we define the radical of an ideal as the ideal of all elements such that some power of the element belongs to the original ideal, then for k algebraically closed, two ideals I, J in k[x,y] have the same vanishing locus in k^2, if and only if they have the same radical.

So in a very real sense, ideals of k[x,y] are a generalization of the set of all plane curves. I.e. the first invariant of an ideal is what is its zero locus? that does not determine the ideal unless the ideal is radical (and k is alg closed) but it is a start.

This is a fascinating and highly complex area of study.

By the way, for rings such as Z and Z/n, there are no ideals other than subgroups, so the concept is not so new there, but considering subgroups as ideals can still give rise to new insights.

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#### Wledig

That was a beautiful rundown mathwonk, I really appreciate your post. I think Mumford's book is still beyond my level, since I was reading the preface and came up with the following:
"The hope was to make the basic objects of algebraic geometry as familiar to the reader as the basic objects of differential geometry and topology: to make a variety as familiar as a manifold or a simplicial complex."
I really better get familiar with those first! I'm still not that sharp with differential geometry and topology. Still, I think I'll benefit a lot from Fulton's book, thanks for the suggestion.

#### mathwonk

Homework Helper
just as integers are studied by means of their prime factorizatiomns, so also are ideals studied by means of "prime ideals". If we restrict to radical ideals, then a radical ideal in k[x,y], (k algebraically closed), is prime if and only if it is not the intersection of two other ideals. By definition an ideal I is prime if and only if whenever ab belongs to I then eother a or b must belong to I. In k[x,y], every radical ideal is an irredundant intersection of a unique finite collection of prime ideals.

An ideal in k[x,y] is prime if and only if its zero locus consists of only one "piece", i.e. cannot be written as the union of two other zero loci of ideals. such loci arte called irreducible. so prime ideals in k[x,y] (k always algebraically closed), correspond to irreducible algebraic subsets of the plane.

when we look at the zero locus of an ideal, intersecting ideals corresponds to taking the union of their zero loci. So writing a radical ideal as an intersection of prime ideals is equivalent to decomposing its zero locus as a union of irreducible algebraic sets. Moreover larger ideals vanish simultaneously on smaller sets, so in fact maximal ideals (always prime) correspond to points of the plane, while non zero and non maximal prime ideals, i.e. prime ideals of "height one" correspond to irreducible curves. The prime ideal (0) in k[x,y] corresponds to the whole plane.

the radical ideal (xy) in k[x,y] is the intersection of the two prime ideals (x) and (y). correspondingly, the zero locus of (xy), is the union of the zero loci of (x) and of (y), i.e. the union of the y axis and the x axis, both irreducible.

Since a nested sequence of irreducible algebraic sets in the plane can have at most three elements, consisting of a point and a curve through that point and the whole plane, a sequence of nested prime ideals in k[x,y] has length at most 3, such as (0) contained in (f) where f is an irreducible polynomial of degree ≥ 1, contained in (x-a,y-b), where (a,b) is some point on the curve f=0.

an arbitrary proper algebraic subset of the plane is a finite union of say n points and m irreducible curves. The ideal of all polynomials vanishing simultaneously on thjis set is the intertsection of the corresponding n maximal ideals defining the points and the m principal prime ideals defining the curves.

it goes on and on, this game of using geometry to visualize the algebra of ideals, and using algebra to make precise calculations in geometry.

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Homework Helper
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#### mathwonk

Homework Helper
Here is another nice simple example that still reveals how much more complex general ideals are than just radical ideals. The simplest rings are the so called principal ideal domains, where every ideal consists of all multiples of some one given generator. Of course multiplying a generator by a "unit" (i.e. invertible element) gives another generator, since units divide everything. So in a domain, every non zero principal ideal has as many generators as there are units in the ring. So in a sense, the more units there are, the fewer ideals there are, since in the presence of a lot of units, each principal ideal has many generators.

The most familiar principal ideal domains are the integers and the polynomial ring k[X] in one variable over any field k. (In k[x], the ideal generated by f and g is also generated by their gcd.) The ideals in the integers correspond one to one with the non negative integers, since the only units are 1 and -1, so every ideal has exactly one non negative generator. In the polynomial ring k[X] the non zero ideals correspond one to one with the monic polynomials, since the units are exactly the non zero constants, so every every non zero polynomial generates the same ideal as some unique monic polynomial.

But there is another nice even simpler principal ideal domain, namely k[[X]], the ring of formal power series over a field k. In this ring, every power series with non zero constant term is a unit, which is a nice exercise, whose coefficients are solvable for by recursion. It follows that every non zero power series looks like some power of X multiplied by a unit. In particular every ideal is principal, and genertated by some non negative power of X. Thus the ideals are in one to one correspondence with the non negative integers, namely the power of X occurring the generator. I.e. every ideal has form (X^n) for some n ≥ 0.

Moreover there is exactly one prime power series, up to units, namely X itself. It follows there is only one prime ideal, (X), the multiples of X. This is a maximal ideal and since every maximal ideal is prime this is also the only maximal ideal. A ring with exactly one maximal ideal is called a "local" ring. So the large number of units in this ring is the reason there are so few ideals, but there are still a lot more ideals than there are "points" since the only point observable here is the origin, corresponding to the unique maximal ideal (X). (X) is also the only radical ideal here.

exercise: check that f = X + X^2 + X^3 +..... also generates the maximal ideal (X), (as does any power series whose lower order non zero term is linear in X). In particular this f is also a prime element, associate to X.
That's all for now.

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#### Wledig

That might be a bit too much to grasp at once, but hopefully I'll get to it one day. Thanks again for the insight and the suggestions, mathwonk. I'll make sure to look into Reid's book.

#### mathwonk

Homework Helper
reid also has an undergraduate intro to commutative algebra that might be even more apt. best wishes. hard to hold back on this topic after studying it for over 50 years. apologies for overdoing.

#### mathwonk

Homework Helper
by the way, inverting power series involves a very important and elementary example, the so called "geometric series". I.e. the geometric series allows one to write the fraction 1/(1-x) as a power series, and in fact using this one can invert any power series with non zero constant term. But it is more elementary for the general case, say in case the constant is 1 (if not just divide through by the non zero constant term), to write (1+ ax + bx^2+....)(1+ux + vx^2 +...) = 1, and multiply out, and solve for the u,v,... coefficients in terms of the a,b,...

E.g. one gets 1 + ax + ux + vx^2 + aux^2 + bx^2 +..... = 1, so a+u = 0, so u = -a. then one gets v+ au + b = 0, and we know u = -a, so v -a^2 + b = 0, so we can solve for v = a^2 -b. just keep going,.... notice that each time the one new coefficient we want to solve for occurs with coefficient 1, so it is easy to solve for. In fact this seems to show that even for power series over a ring, the power series is invertible if and only if the constant term is a unit in the coefficient ring. E.g. the series 1-x is invertible over the integers, as we know from the geometric series.

I first saw the geometric series in 8th grade, so it is not too early or advanced for you now to learn it. (We had a visiting lecture that day from a college student who was learning about teaching middle school, and that was almost the only day all year I learned anything new.)

"Polynomial ideals"

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