## AGCA: HBT Proof and a sweet Bijection.

### August 27, 2010

It’s becoming a bit more difficult to write these, with grading and homework taking up much of my time, and I no longer am officially "taking" AGCA for credit.  Nonetheless, it’s an extremely interesting topic, and I’m going to go along with it for as long as I can.  I should have noted in the previous post: a ring $R$ will always mean "commutative, with unit."

Last time, we stated the Hilbert Basis Theorem.  Let’s state this again.

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

Proof. This is pretty straightforward, but there’s a really clever trick.  It’s one of those "pick a minimal element, show it’s not minimal" deals.

Alright, so, we want to prove $R[x]$ is noeth given that $R$ is noeth.  So, let’s use that finitely generated ideal definition.  We want to show that some arbitrary ideal $I\subseteq R[x]$ is finitely generated.  Then let’s choose $f_{1}\in I$ to be a polynomial of least degree.  There is clearly one — if there is a constant, then $I$ is the entire ring, if not, then we have some linear, quadratic, or some kind of polynomial.  There are going to be many polynomials of least degree, but just pick one.

Now, let’s let $f_{2} \in I - (f_{1})$ be of least degree; that is, we want to pick $f_{2}$ such that it is in the ideal $I$ but not the ideal generated by $f_{1}$.  This should look familiar: it’s how we chose new basis elements when making a basis for a vector space.

Continue this process.  We then have a sequence of functions

$f_{1}, f_{2}, f_{3}, \dots$

Now, here’s a clever part.  If we have picked all possible functions after a finite number of picks (that is, if $I - (f_{1}, f_{2}, \dots, f_{n}) = \{0\}$), then this ideal is finitely generated by those functions.  This is nice, ’cause it proves what we want.  But let’s suppose that we have an infinite number of functions.  We’re gonna show that we get a contradiction.

For each function, let’s write them like this:

$f_{j}(x) = a_{j,n}x^{n} + a_{j,n-1}x^{n-1} + \cdots + a_{j,0}$

where $j$ is just our index of what function we’re at.  Okay, so now, note that $a_{j,n}$ is the coefficient of the highest term.  Note also that these coefficients are in $R$.  So, what do you think we’re gonna do?  $R$ is noeth, so let’s make a sequence of initial (highest power) coefficients.  Also, since I don’t care about the particular power that they’re at, let’s just call them $b_{j} = a_{j,n}$ for the $j$-th function’s leading coefficient.  Okay?  Okay.  Just to get rid of the $n$ power part.  Now we have

$b_{1}, b_{2}, \dots$

is a sequence.  Now, consider the ideal $B$ generated by all these things.

$B = (b_{1}, b_{2}, \dots )$

BUT WAIT, this ideal is in $R$ and so it is finitely generated since $R$ is noeth!  Yessss.  Without loss of generality, let’s say that it is generated by

$B = (b_{1}, b_{2}, \dots, b_{n})$

by potentially reordering these elements, and picking the elements of the least ordered polynomials that generate this ideal.  Let’s let $f_{i}$ correspond to $b_{i}$.  Then we claim that

$I = (f_{1}, \dots, f_{n})$

Why?  Let’s consider $f_{n+1}$.  We have, by this ideal $B$, that the initial coefficient $b_{n+1} = \sum_{i=1}^{n} c_{i}b_{i}$ for some coefficients $c_{i}\in R$.

If $f_{m+1}\notin I$, when we have that its degree must be greater than every other $f_{i}$ that generates the ideal.  Now we define

$\displaystyle g(x) = \sum_{j=1}^{n}b_{j}x^{deg(f_{n+1}) - deg(f_{j})}f_{j}$.

Notice that $f_{j}$ at the end there.  Now, what is this?  Well, it doesn’t really matter, but just notice that the leading coefficient of $g$ is the same as the leading coefficient of $f_{n+1}$ (why?  check this.  really.  it took me a little while to figure out).

Note, then, that since the leading coefficients are the same, $(g-f_{n+1})(x)$ is a polynomial of degree less than $f_{n+1}$ and since $g\in (f_{1}, \dots, f_{n}) = I$, we have that $(g - f_{n+1})\in I$ implying $f_{n+1}$ in $I$.  This shows that, for all other functions $f_{k}$ for $k > n$, we have that they are generated by the first $n$ functions.

Thus, $I$ is finitely generated.  Thus $R[x]$ is noetherian.  Whew.  $\Box$.

Now that we have this, let’s talk about some ways to talk about AAS.  Let’s define two things before we go on, and these are two extremely important things in commutative algebra.  I kind of poo-poo’d them the first time I took it, and I was screwed for the rest of the course.  They’re arguably as important as the noetherian definition.  You remember what an algebraic set is, right?  For the next section, we’re going to make all of our standard ideals called $J$ instead of $I$ because of this kind of crappy notation.

Definition: Let $I(V)$ be the ideal of all functions vanishing on $V$.  In other words, $I(V) = \{f\in k[x_{1}, \dots, x_{n}] | f(x)=0 \mbox{ for all } x\in V\}$.

Definition: Let $V(J)$ be the zero-locus in ${\mathbb A}^{n}$ on which the functions in $J$ vanish.  In other words,  $V(J) = \{x\in {\mathbb A}^{n}\, |\, f(x) = 0 \mbox{ for all } f\in S\}$.

So, remember, we have $V(J)$ is an ideal of POINTS and $I(V)$ is an ideal of functions.  This is something I never remember, so I made up the acronym "IFVP = I Function Very Poorly" to remember "I is for Functions", "V is for Points."

We’re gonna talk a bit more about what these things have to do with anything in the next post, where we post an extremely important due to Hilbert called the "Nullstellinsatz" or "Zero-Point Theorem."  But let’s just talk about some things dealing with the things we have.

## Bijections and Stuff.

Given an affine algebraic set in ${\mathbb A}^{n}$, we have that this corresponds to some ideal in $k[x_{1}, \dots, x_{n}]$ in a pretty cool way: namely, we have that for any AAS (let’s call it $X$), we have that we can find $I(X)$ (with the notation above) and this is a way to associate $X$ with an ideal in $k[x_{1}, \dots, x_{n}]$.  Also remember what $I(X)$ gives us: "I function very poorly", so $I(X)$ is for functions, verifying its inclusion in the polynomial ring.

In the same way, given some ideal $J\in k[x_{1}, \dots, x_{n}]$, we associate an AAS in ${\mathbb A}^{n}$ in the following way: we simply take $V(J)$ and associate them this way.  Recall that "I function very poorly", so $V(J)$ corresponds to a Point in affine space.

In fewer words, there are mappings

$X \rightarrow I(X)$

$V(J) \leftarrow J$

but these aren’t particularly nice mappings.  We have no idea if they’re one to one, onto, or what.  And, in fact, many times they are not.

It would be really nice to be able to have some kind of bijection between Affine Algebraic Sets which are points, and Ideals in the ring of polynomial functions.

And this is the motto for algebraic geometry, essentially.  We really want such a mapping, so that we can go back and forth.  We know a lot about points, and we could spread some of that to ideals, and vice versa.

Okay, let’s end with some basic properties of $V$ and $I$.

## Basic Properties

Properties: If $X\subseteq Y$, then $I(X) \subseteq I(Y)$.

Properties: If $J\subseteq K$, then $V(K) \subseteq V(J)$.  Notice the "switch" when we apply $V$ to these ideals.

These are easy enough to prove by yourself, so I leave them to you.  One that I’ll be proving next time is:

Properties: $V(I(X)) = X$.

Which is pretty nice.  Now, the main point of the Hilbert Null theorem is essentially to trick people.  Well, maybe not, but it’s still kind of fun.  So, you ask, "Well, if $V(I(X)) = X$, then what is $I(V(J))$?"  If you answered $J$, then you’d be WRONG.  This is the main point of the theorem.  It turns out that it’s a set larger than $J$, but still relatively easily describable: it’s called the "radical of $J$" and it’s simply every element in the ring such that some finite power of that element is in $J$.  Yeah, we’ll see why that works next time.

And why do we care about this?  Because once we have this, there is a bijection between these "radical" ideals and Affine Algebraic Sets in ${\mathbb A}^{n}$!  Nice.