# 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.