I procrastinated posting this It's probably because the ending is so trippy.

First, what is Hilbert's Nullstellensatz?

Let be an algebraically closed field (a thing where you can add, subtract, multiply, divide by everything except zero, and find roots of all nonconstant polynomials), let be the polynomial ring you get from adding indeterminates to , and let be an ideal of this polynomial ring. We dumped some of the terminology in the last post already so I won't repeat that.

Then , where:

1. means the set of zeroes of , i.e. the points of that give 0 when you plug them into any element of
2. means the ideal of polynomials that are zero on all the points of the
3. is the radical of , which means all elements that are some th root of an element of .

It is "easy" to see that : If , we want to prove is zero wherever all of is zero. But some is in , and wherever is zero, is also zero, since fields are nice enough that you can't multiply nonzero things to get zero.

The other direction is more annoying. But having proven Zariski's lemma, we're already more than half done. Our last intermediate step is Weak Nullstellensatz: If is an algebraically closed field and is a proper ideal in the polynomial ring , then , the set of points at which all elements of are zero, is not empty.

Proof: Since is a field, it's Noetherian, so is Noetherian, so we can pick a maximal ideal that includes .

Then (via ideal preliminary) is a field. Furthermore we can view as a subfield of this. Now we apply Zariski's lemma to see that is module-finite over .

In fact, we claim that this means is isomorphic to . This part relies on the algebraic closedness of .

Suppose that we have some (treated as a field extension of ); we want to prove it is actually in . Then the module-finiteness also implies is integral over (by Lemma 2 of the Zariski post darn I'm forgetful, although in my defense, that lemma was optional for my cobbled-together proof). But if is the root of a polynomial with coefficients in , then we can factor into linear factors merely in because is algebraically closed. So we have is zero and is a field, which means one of those linear factors is zero and , as desired.

Okay. So , so in the field , for each there's a residue that it's in the same class as. So .

But the ideal is maximal, because by "plugging in" in any element of you can reduce it to something in , which differs from it by an element of . Therefore, adding any other element to would reduce it to cover the entirety of . So is maximal, yet and is proper, so .

This means that and we are done.

— end proof —

Now we can tackle the hard direction of Hilbert's nullstellensatz. Quick recap, it states that if be an algebraically closed field and is an ideal of the polynomial ring , then . We now want to prove that , or that if then .

To do this, we add another indeterminate to the polynomial ring to get . And we consider the ideal generated by . Yes, this is trippy. The star doesn't mean anything, I just haven't found an excuse to mean a star superscript in a while.

But this is useful because at any point where are all zero, that means is also zero which means is , so never completely evaluates to zero at any point. Therefore is zero and by the contrapositive of the weak nullstellensatz, . Therefore contains 1.

In other words we have an equation where each is an element of .

Divide everything by a high power of until there's no more of it in the numerator so you get something like:

where each is an element of , where is a magical indeterminate.

Then set to get the result:

And magically, we're done!

Okay, I don't know if you feel like these plugging in variables is fishy the same way I do, but we can do it unfancily like this:

• treat as a subring of
• take the homomorphism of that to where or , and supposing
• clearing denominators in the equation to make it involve and feature on the right side
• saying "And magically, we're done!"

Really.

I should be several chapters ahead of where I am right now.