# The Nullstellensatz

A friend recently asked me to think about the nullstellensatz. I didn’t remember the proof so I looked it up and it turns out its not so hard to prove. This follows Miles Reid book: Undergraduate Commutative Algebra. Note here denotes the set of points in where all elements of vanish.

(Nullstellensatz) If and is a proper ideal then

a) and

b) .

The key to proving this result is

(Noether Normalization) Let be a finitely generated algebra. Then there are algebraically independent elements such that if a finitely generated module over .

The proof given here rests on the following

Claim: If is nonzero then there are such that is integral over . [1]

PROOF: let be unspecified (for the moment) constant and set . Then

(1)

note a term is plugged into every variable so if has degree then there is necessarily a term . Write w/ homogeneous, then (1) becomes

lower order terms in $latex x_n$. (2)

Assuming is infinite [2], which is fine for me because algebraically closed fields are always infinite, the constants can be chosen so

e.g. use induction on the number of variables, then writing by induction hypothesis there are such that some , so plugging in I’m left with a nonzero polynomial in one variable etc.

Note (2) gives a relation of integral dependence QED.

PROOF: use induction on where for some ideal; the case being clear. If there is nothing to prove. Otherwise there is a nonzero and the claim applies. So is integral over and .

Applying the inductive hypothesis if finite over and is finite over hence also over QED.

Now back to the Nullstellensatz

PROOF: a) follows easily from

Prop. The maximal ideals of are all of the form .

PROOF:Let be a max ideal. Noether Normalization is finite over but its also a field so . So is finite over , in particular algebraic so . Let be the image of in ; then QED.

From the prop is contained in a maximal ideal and then .

b) Suppose is such that . Consider . . so

with . If is the highest power of appearing in the then I can write

plugging in shows , the other inclusion is obvious QED.

[1]Alternatively, if I let represent the image of in then and I can restate the claim: if is nonzero and then there are such that is integral over .

[2] This claim holds for arbitrary fields but the proof is more difficult; see Reid’s book.

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- Published:
- June 22, 2010 / 1:18 am

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- alg. geo.

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