## AGCA: Introduction to Affine Algebraic Sets, Ideals, and Noetherian Rings.

### August 24, 2010

Okay, first, I’m going to abbreviate Algebraic Geometry + Commutative Algebra as AGCA, since, you know, it’s a lot to say.  Second, these posts (with the title AGCA) may require a bit more mathematical maturity and abstract algebra than the others.

This is following the lectures of Professor Vitter, and, unfortunately, is only going to be a shallow copy with, most likely, more typos.  My intention is to learn the material by teaching, and to clear up and expand on the proofs given.  I do not pretend to be anywhere near as good a lecturer or mathematician, and this post is essentially “for entertainment purposes only.”  Reader beware.

First and foremost, for this post you must know what a field is.  If you don’t know and you don’t want to take fifteen seconds to look on wikipedia, then just think about ${\mathbb R}$.  Good?  Good.

Definition: If $k$ is a field, then $k[x_{1}, x_{2}, \dots, x_{n}]$ is called the ring of polynomials in n-variables with coefficients in $k$.

This is exactly what it sounds like.  Each element is a polynomial in $n$ variables; in other words, each term contains a coefficient from $k$ followed by varying powers of $x_{1}, x_{2}, \dots, x_{n}$.  So, for example, an element of $k[x,y]$ is $2xy^{2} + 3x^{3}y + 5x + 1$.

The one that we’re going to most commonly use is $k[x]$ which is simply all polynomials in $x$.  So, for example, $x^{2} + 2x + 1$, or $x^{500} + 22$.  Any polynomial in $x$ will work here.  Note that these cannot be power series, and must have a finite number of terms.

Definition: Given a field $k$, we may form the affine n-space denoted ${\mathbb A}^{n}_{k} = k^{n}$, or, in other words, affine n-space is simply $n$ direct products of $k$.

For example, ${\mathbb R}^{n}$ is real affine n-space.

Now, an extremely important idea in commutative algebra is the notion of an ideal.  I don’t want to spend a whole ton of time explaining what this is, but, generally, it’s something very similar to the notion of “multiples” in the integers.  Specifically, we have

Definition: An ideal of a ring $(R, +, \cdot)$ is a subgroup of $R$ under $+$, and for each $x\in I$ and for all $r\in R$, we have that $x\cdot r \in I$.

Ideals are generally written $(x)$ where $x$ is the generator of this ideal, or $(a,b,c, \dots, d)$ if there are many generators, or simply $I$.

The prototypical example of ideals come from the integers.  For example, if we consider the multiples of 3 in the integers, we note that these form a subgroup under addition of the integers (which are a ring with multiplication and addition) and for every number $n$ in the integers, we have that $3n$ is a multiple of 3, and is therefore in our subgroup.  This last property is essentially the one that distinguishes this set as an ideal as opposed to a subgroup.  There are numerous other examples, though.  For any general field, we have that the only proper ideal is $\{0\}$ (why?).

Okay, now we’re getting to the meat and potatoes of this post.  It’s extremely important in algebraic geometry to consider the set of zeros of certain polynomials.  I mean, we do that in algebra and calculus all the time, so why not in algebraic geometry?  So, we define this crazy looking thing.

Definition: $X\subseteq {\mathbb A}^{n}$ is called an Affine Algebraic Set (AAS) if there exists some functions $f_{\alpha} \in k[x]$ such that $X = \{p\in {\mathbb A}^{n}\, |\, f_{\alpha}(p) = 0, \forall \alpha \in \Lambda\}$.

Where $\Lambda$ is just some larger indexing set.  What is this saying?  We have that $X$ is an AAS if we have some crazy functions $f_{\alpha}$ such that every point in $X$ is a zero of every single one of those functions.  So, for example, if we have the functions $f_{1}(x) = x^{2} - 1$ and $f_{2} = x - 1$ in ${\mathbb R}[x]$ then the associated affine algebraic set for these would just contain the element $\{1\}$, since this is the only zero common to both.  Notice that our element $\{1\}$ is actually sitting in the real affine 1-space ${\mathbb A}^{1}_{\mathbb R} = {\mathbb R}$.

Note that it is entirely possible that functions do not share zeros.  Then that kind of stinks, and the associated affine algebraic set for those is the null set.

For AAS, it is equivalent to consider the ideal generated by the set of $f_{\alpha}$‘s.  Why?  Because every element in the ideal will have the same zeros as one or more of these $f_{\alpha}$‘s.  Think about this for a bit, and try to prove it to your self.  Therefore, we have if

$I = (f_{\alpha})_{\alpha\in\Lambda}$

then this is an ideal in $k[x_{1}, \dots, x_{n}]$ depending on what variables are used in the $f_{\alpha}$‘s.

Using this definition of the ideal $I$ we can use the alternate definition:

Alternate Definition: $X$ above is an AAS if we have that $X = \{p\in {\mathbb A}^{n}\, |\, f(p) = 0, \forall f\in I\}$.

Good.  This is saying the exact same thing: $X$ is made up of points that are the zeros of every polynomial in the ideal $I$ above.

Alright, now, before we move onwards, we want to define one last thing.  This is a biggie.  If you only remember one definition from this post, then remember this one because it comes up quite a bit.  SERIOUSLY.  Remember this next thing.

Definition (VERY IMPORTANT): A ring $R$ is noetherian if one of the two following things hold:

1. Every ideal $I$ in $R$ is finitely generated.  That is, for every ideal $I$, there are some elements $a_{1}, \dots, a_{n}\in R$ such that $b\in I$ if and only if $b = \sum_{i=1}^{n} c_{i}a_{i}$ for some $c_{i}\in R$.
2. For every ascending chain of ideals $I_{1} \subseteq I_{2} \subseteq \cdots$ where each of the $I_{i}$ are ideals of $R$, we have that there exists some $n$ such that for all $i > n$ we have $I_{i} = I_{n}$.  That is, the chain essentially “stops” and we keep having the same ideal over and over again.  This is called the “ascending chain condition.”

Note that the term “Noetherian” is generally in uppercase, but I will keep it lowercase.  I feel that this is much more of a compliment to Emmy Noether! Now, it’s not entirely trivial to prove that these two statements are equal, so let’s just do that quickly.

Proof (The equivalence of those two things in the definition above):

$(1 \Rightarrow 2)$.  Let’s assume 1, and let’s consider a chain of ideals $I_{1} \subseteq I_{2} \subseteq \cdots$, and then let’s define $I = \bigcup_{i=1}^{\infty}$.  Prove to yourself that $I$ is in fact an ideal of $R$.  By assumption, it is finitely generated.  Therefore, $I = (a_{1}, \dots, a_{n})$ and so we must have some $m$ such that all of the $a_{i}$‘s are in $I_{m}$, as they are in the union, and these are nested sets.  Therefore, pick the ideal of the least index such that each of those $a_{i}$‘s are in it.  Every ideal after that is equal to $I$.  This proves the first part.

$(2 \Rightarrow 1)$.  Suppose that we have some ideal $I$ and we want to show that this is finitely generated.  Well, let’s consider some $a_{1}\in I$.  If $(a_{1})$ isn’t all of $I$, pick an element that is in $I$ but is not in $(a_{1})$.  Call this element $a_{2}$.  Continue doing this.  We have the chain

$(a_{1})\subseteq (a_{1}, a_{2}) \subseteq \cdots$

and so we must have, by assumption, that this chain eventually stops.  Prove to yourself that it must stop at $I$.  Therefore, because it stops in a finite number of subsets, we have a finite generating set. $\Box$.

So, what?  What kind of things are noetherian?  The next post will focus on this, but I want to mention the main theorem that we’ll be proving next time: the hilbert basis theorem.

Theorem (Hilbert Basis Theorem): If $R$ is noetherian, then so is $R[x]$.

This is a really powerful theorem.  In particular, it also implies that $R[x,y] = (R[x])[y]$ is noetherian, and, therefore, so is $R[x_{1}, \dots, x_{n}]$!  Neat.

Next time, we prove the HBT and give some examples of some noetherian things.