The Hilbert Basis theorem is probably one of the easiest-to-state theorems that I know of in commutative algebra.  The last time I posted about it, I really butchered the proof; not that it was long, but it doesn’t really do anything for me.  Reading back on it now, it doesn’t seem at all intuitive to me.  The proof came through a long line of telephoning: the professor was reading from his notes, I was copying from the board, and then I was copying from my notes.  Now that I have a bit more time, I’d like to go through the proof again, but this time I’d like to motivate the theorem and proof.  Not just because the proof is a common proof-type (there are a ton of proofs that go a similar way in the commutative algebra book I’m going through) but because it’s not nearly as difficult as it looks at first glance.

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Hilbert’s Nullstellensatz.

October 15, 2010

This serious sounding title is apt for this post, because the Nullstellensatz is the big time.  This is one of the "big" results in algebraic geometry.  Before we dive into the theorem, though, let’s motivate this a little bit. 

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A small tangent, because I enjoyed this so much in class.

Let’s call a polynomial of n variables a homogeneous polynomial if every term has the same degree.  Recall that we compute the degree of a term by summing the powers of each variable in the term.

For example 3x_{1}^{2}x_{2}^{4}x_{3}^{4} has degree 10.  For another example, 2x_{1}^{3}x_{2}^{3} has degree 6.  So, an example of a homogenous polynomial in three variables is

f(x_{1}, x_{2}, x_{3}) = 2x_{1}^{3}x_{2}^{1} + 5x_{1}^{2}x_{3}^{2} + x_{3}^{4}

and this polynomial is of degree 4.

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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."

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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.

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