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 V(J) denotes the set of points in k^n where all elements of J vanish.

(Nullstellensatz) If k = \bar k and J \subset A = k[x_1,..,x_n] is a proper ideal then

a) V(J) \ne \emptyset and

b) I(V(J)) = \sqrt J.

The key to proving this result is

(Noether Normalization) Let B be a finitely generated k algebra.  Then there are algebraically independent elements y_1, .., y_m \in A\ m\ge 0 such that B if a finitely generated module over k[y_1,.., y_m].

The proof given here rests on the following

Claim: If f \in J \subset A is nonzero then there are z_1,..., z_{n-1} \in A such that x_n \in A/J is integral over k[z_1,...z_{n-1}]/J. [1]

PROOF: let \alpha_i be unspecified (for the moment) constant and set z_i = x_i - \alpha_i x_n.  Then

f(z_1 + \alpha_1 x_n, ..., x_n) \in J (1)

note a term x_n is plugged into every variable so if f has degree d then there is necessarily a term x_n^d.  Write f = \sum_{i=0}^d f_i w/ f_i homogeneous, then (1) becomes

f_d(\alpha_1,..., \alpha_n,1)x_n^d + lower order terms in $latex x_n$. (2)

Assuming k is infinite [2], which is fine for me because algebraically closed fields are always infinite, the constants can be chosen so f_d(\alpha_1,..., \alpha_n,1) \ne 0

e.g. use induction on the number of variables, then writing f_d = \sum_i b_i x_1^i by induction hypothesis there are \alpha_2, .., \alpha_{n-1} such that some b_i(\alpha_2, ..., \alpha_{n-1}) \ne 0, 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 n where B = k[x_1,..., x_n]/I for some ideal; the case n = 0 being clear.  If I = 0 there is nothing to prove.  Otherwise there is a f \in I nonzero and the claim applies.  So x_n is integral over B' = k[z_1,..., z_{n-1}]/I and B = B'[x_n].

Applying the inductive hypothesis B' if finite over k[y_1,..., y_m] and B is finite over B' hence also over k[y_1,.., y_m] QED.

Now back to the Nullstellensatz

PROOF: a) follows easily from 

Prop. The maximal ideals of A are all of the form (x_1 - a_1, ..., x_n - a_n).

PROOF:Let m be a max ideal. Noether Normalization \Rightarrow A/m is finite over k[y_1, .., y_m] but its also a field so m = 0.  So A/m is finite over k, in particular algebraic so A/m \cong k.  Let a_i be the image of x_i in A/m; then x_i - a_i \in m QED. 

From the prop J is contained in a maximal ideal (x_1 - a_1, ..., x_n - a_n) and then (a_1, ..., a_n) \in V(J).

b)  Suppose f \in A is such that f(P) = 0 \forall P \in V(J). Consider J' = (J, fy - 1) \subset A[y].  V(J') = \emptyset \Rightarrow 1 \in J'. so 

1 = \sum_i c_i g_i + c_o (fy-1) 

with c_i \in A[y], g_i \in J.  If m is the highest power of y appearing in the c_i then I can write

f^m = \sum_i c'_i(x_i,fy)g_i + c'_o(fy-1)

plugging in fy = 1 shows I(V(J)) \subset \sqrt J, the other inclusion is obvious QED.

 

 

 

 [1]Alternatively, if I let y_i represent the image of x_i in A/J then A/J = k[y_1,...,y_n] and I can restate the claim: if f \in A is nonzero and f(y_1,...,y_n) = 0 then there are z_i \in A/J such that y_n is integral over k[z_1,..., z_{n-1}.
 [2] This claim holds for arbitrary fields but the proof is more difficult; see Reid’s book.

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