Commutative Algebra/Hilbert's Nullstellensatz

Zariski's lemma

Definition 24.1 (Finitely generated algebra):

Let $R$ be a ring. An $R$-algebra $A$ is called finitely generated, iff there are elements $a_{1},\ldots ,a_{n}\in A$ such that $R[a_{1},\ldots ,a_{n}]$ is already all of $A$; that is $A=R[a_{1},\ldots ,a_{n}]$.

$A$ being a finitely generated $R$-algebra thus means that we may write any element of $A$ as a polynomial $p(a_{1},\ldots ,a_{n})$ for a certain $p\in R[x_{1},\ldots ,x_{n}]$ (where polynomials are evaluated as explained in chapter 21).

Lemma 24.2 (Artin–Tate):

Let $R\subseteq S\subseteq T$ be ring extensions such that $R$ is a Noetherian ring, and $T$ is finitely generated as an $S$-module and also finitely generated as an $R$-algebra. Then $S$ is finitely generated as an $R$-algebra.

Proof:

Since $T$ is finitely generated as an $S$-module, there exist $u_{1},\ldots ,u_{n}\in T$ such that $T=\langle u_{1},\ldots ,u_{n}\rangle$ as an $S$-module. Further, since $T$ is finitely generated as $R$-algebra, we find $v_{1},\ldots ,v_{m}\in T$ such that $T$ equals $R[v_{1},\ldots ,v_{n}]$. Now by the generating property of the $u_{1},\ldots ,u_{n}$, we may determine suitable coefficients $a_{i,j}\in S$ (where $i$ ranges in $\{1,\ldots ,n\}$ and $j$ in $\{1,\ldots ,m\}$) such that

$v_{j}=a_{1,j}u_{1}+\cdots +a_{n,j}u_{n}$, $j\in \{1,\ldots ,m\}~~~~~~~(*)$.

Furthermore, there exist suitable $b_{i,j,k}$ ($i,j,k\in \{1,\ldots ,n\}$) such that

$u_{j}u_{k}=b_{1,j,k}u_{1}+\cdots +b_{n,j,k}u_{n}~~~~~~~(**)$.

We define $S':=R[a_{i,j}(1\leq i\leq n,1\leq j\leq m),b_{i,j,k}(1\leq i,j,k\leq n)]\subseteq T$; this notation shall mean: $S'$ is the algebra generated by all the elements $a_{i,j},b_{i,j,k}$. Since the algebra operations of $T$ are the ones induced by its ring operations, $S'$, being a subalgebra, is a subring of $T$. Furthermore, $S'\subseteq S$ and $R\subseteq S'$. Since $R$ is a Noetherian ring, $S'$ is also Noetherian by theorem 16.?.

We claim that $T$ is even finitely generated as an $S'$-module. Indeed, if any element $t\in T$ is given, we may write it as a polynomial in the $v_{1},\ldots ,v_{m}$. Using $(*)$, multiplying everything out, and then using $(**)$ repeatedly, we can write this polynomial as a linear combination of the $u_{1},\ldots ,u_{n}$ with coefficients all in $S'$. This proves that indeed, $T$ is finitely generated as an $S'$-module. Hence, $T$ is Noetherian as an $S'$-module.

Therefore, $S$, being a submodule of $T$ as $S'$-module, is finitely generated as an $S'$-module. We claim that $S$ is finitely generated as an $R$-algebra. To this end, assume we are given a set of generators $m_{1},\ldots ,m_{l}\in S$ of $S$ as an $S'$-module. Any element $s\in S$ can be written

$s=c_{1}m_{1}+\cdots +c_{l}m_{l}$, $c_{1},\ldots ,c_{l}\in S'$.

Each of the $c_{i}$ is a polynomial in the generators of $S'$ (that is, the elements $a_{i,j},b_{i,j,k}$) with coefficients in $R$. Inserting this, we see that $s$ is a polynomial in the elements $a_{i,j},b_{i,j,k},m_{i}$ with coefficients in $R$. But this implies the claim.$\Box$

Theorem 24.3 (Zariski's lemma):

Let $L$ be a field extension of a field $K$. Assume that for some $\alpha _{1},\ldots ,\alpha _{n}$ in $L$, $R=K[\alpha _{1},\ldots ,\alpha _{n}]$ is a field. Then every $\alpha _{i}$ is algebraic over $K$.

Proof 1 (Azarang 2015):

Before giving the proof of the lemma, we recall the following two well-known facts.

Fact 1. If a field $F$ is integral over a subdomain $D$, then $D$ is a field.

Fact 2. If $D$ is any principal ideal domain (or just a UFD) with infinitely many (non-associate) prime elements, then its field of fractions is not a finitely generated $D$-algebra.

Proof of the Lemma: We use induction on $n$ for arbitrary fields $K$ and $L$. For $n=1$ the assertion is clear. Let us assume that $n>1$ and the lemma is true for less than $n$. Now to show it for $n$, one may assume that one of $\alpha _{i}$, say $\alpha _{1}$, is not algebraic over $K$ and since $K[\alpha _{1},\ldots ,\alpha _{n}]=K(\alpha _{1})[\alpha _{2},\ldots ,\alpha _{n}]$ is a field, by induction hypothesis, we infer $\alpha _{2},\ldots ,\alpha _{n}$ are all algebraic over $K(\alpha _{1})$. This implies that there are polynomials $f_{2}(\alpha _{1}),\ldots ,f_{n}(\alpha _{1})\in K[\alpha _{1}]$ such that all $\alpha _{i}$'s are integral over the domain $A=K[\alpha _{1}][1/f_{2}(\alpha _{1}),\ldots ,1/f_{n}(\alpha _{1})]$. Since $R$ is integral over $A$, by Fact 1, $A$ is a field. Consequently, $A=K(\alpha _{1})$, which contradicts Fact 2.

Proof 2 (Artin–Tate):

If all of the generators of $A$ over $\mathbb {F}$ are algebraic over $\mathbb {F}$, the last paragraph of the preceding proof shows that $A$ is a finite field extension of $\mathbb {F}$. Hence, we only have to consider the case where at least one of the generators of $A$ over $\mathbb {F}$ is transcendental over $\mathbb {F}$.

Indeed, assume that $A=\mathbb {F} [a_{1},\ldots ,a_{n}]$. By reordering, we may assume that $a_{1},\ldots ,a_{r}$ are transcendental over $\mathbb {F}$ ($r\geq 1$) and $a_{r+1},\ldots ,a_{n}$ are algebraic over $\mathbb {F}$. We have $A=\mathbb {F} [a_{1},\ldots ,a_{n}]\subseteq \mathbb {F} (a_{1},\ldots ,a_{n})$, and furthermore $\mathbb {F} (a_{1},\ldots ,a_{n})\subseteq A$ since $A$ is a field extension of $\mathbb {F}$ containing all the elements $a_{1},\ldots ,a_{n}$. Hence, $A=\mathbb {F} (a_{1},\ldots ,a_{n})$.

Since all the $a_{r+1},\ldots ,a_{n}$ are algebraic over $\mathbb {F}$, they are also algebraic over $\mathbb {F} (a_{1},\ldots ,a_{r})$. Assume that there exists a polynomial $f\in \mathbb {F} [x_{1},\ldots ,x_{n}]\setminus \{0\}$ such that $f(a_{1},\ldots ,a_{r})=0$. Then $a_{r}$ is algebraic over $\mathbb {F} (a_{1},\ldots ,a_{r-1})$; for, the part of the monomials not being a power of $a_{r}$ may be seen as coefficients within that field. Hence, we may lower $r$ by one and still obtain that $a_{r+1},\ldots ,a_{n}$ are algebraic over $\mathbb {F} (a_{1},\ldots ,a_{r})$. Repetition of this process eventually terminates, or otherwise $a_{1}$ would be algebraic over $\mathbb {F}$, and $A$ would be a finite tower of algebraic extensions ($\mathbb {F} (a_{1})$, $\mathbb {F} (a_{1})(a_{2})$ and so on) and thus a finite field extension.

Therefore, we may assume that $a_{1},\ldots ,a_{r}$ are algebraically independent over $\mathbb {F}$. In this case, the map

$\mathbb {F} [x_{1},\ldots ,x_{r}]\to \mathbb {F} [a_{1},\ldots ,a_{r}],f(x_{1},\ldots ,x_{r})\mapsto f(a_{1},\ldots ,a_{r})$

is an isomorphism (it is a homomorphism, surjective and injective), and hence, $\mathbb {F} [a_{1},\ldots ,a_{r}]$ is a unique factorisation domain (since $\mathbb {F} [x_{1},\ldots ,x_{r}]$ is).

Now set $\mathbb {G} :=\mathbb {F} (a_{1},\ldots ,a_{r})$. Then $\mathbb {F} \subseteq \mathbb {G} \subseteq A$, and $A$ is finitely generated as an $\mathbb {F}$-algebra and finitely generated as a $\mathbb {G}$-module (since it is a finite field extension of $\mathbb {G}$). Therefore, by lemma 24.2, $\mathbb {G}$ is finitely generated as an $\mathbb {F}$-algebra. Let

${\frac {f_{1}(a_{1},\ldots ,a_{n})}{g_{1}(a_{1},\ldots ,a_{n})}},\ldots ,{\frac {f_{m}(a_{1},\ldots ,a_{n})}{g_{m}(a_{1},\ldots ,a_{n})}}$

be generators of $\mathbb {G}$ as $\mathbb {F}$-algebra. Let $p_{1},\ldots ,p_{l}$ be all the primes occurring in the (unique) prime factorisations of $g_{1},\ldots ,g_{m}$. Now $\mathbb {F} [a_{1},\ldots ,a_{r}]$ contains an infinite number of primes. This is seen as follows.

Assume $q_{1},\ldots ,q_{k}$ were the only primes of $\mathbb {F} [a_{1},\ldots ,a_{r}]$. Since we have prime factorisation, the element $q_{1}\cdot q_{2}\cdots q_{n}+1\in \mathbb {F} [a_{1},\ldots ,a_{r}]$ is divisible by at least one of $q_{1},\ldots ,q_{k}$, say $q_{j}$. This means

$1=q_{j}(q_{1}\cdots q_{j-1}q_{j+1}\cdots q_{k}+s)$

for a certain $s\in \mathbb {F} [a_{1},\ldots ,a_{r}]$, which is absurd, since applying the inverse of the above isomorphism to $\mathbb {F} [x_{1},\ldots ,x_{r}]$, we find that $1$ is mapped to $1$, but the right hand side has strictly positive degree.

Hence, we may pick $p\notin \{p_{1},\ldots ,p_{l}\}$ prime. Then $1/p$ can not be written as a polynomial in terms of the generators, but is nonetheless contained within $\mathbb {G}$. This is a contradiction.$\Box$

Proof 3 (using Noether normalisation):

According to Noether's normalisation lemma for fields, we may pick $c_{1},\ldots ,c_{k}\in A$ algebraically independent over $\mathbb {F}$ such that $A$ is a finitely generated $\mathbb {F} [c_{1},\ldots ,c_{k}]$-module. Let $m_{1},\ldots ,m_{l}$ be elements of $A$ that generate $A$ as an $\mathbb {F} [c_{1},\ldots ,c_{k}]$-module. Then according to theorem 21.10 3. $\Rightarrow$ 1., the generators are all integral over $\mathbb {F} [c_{1},\ldots ,c_{k}]$, and since the integral elements form a ring, $A$ is integral over $\mathbb {F} [c_{1},\ldots ,c_{k}]$. Hence, $\mathbb {F} [c_{1},\ldots ,c_{k}]$ is a field by theorem 21.11. But if $k\geq 1$, then the $c_{1},\ldots ,c_{k}$ being algebraically independent means that the homomorphism

$\mathbb {F} [x_{1},\ldots ,x_{k}]\to \mathbb {F} [c_{1},\ldots ,c_{k}],f(x_{1},\ldots ,x_{k})\mapsto f(c_{1},\ldots ,c_{k})$

is in fact an isomorphism, whence $\mathbb {F} [c_{1},\ldots ,c_{k}]$ is not a field, contradiction. Thus, $k=0$, and hence $A$ is finitely generated as an $\mathbb {F}$-module. This implies that we have a finite field extension; all elements of $A$ are finite $\mathbb {F}$-linear combinations of certain generators.$\Box$

Hilbert's Nullstellensatz

There are several closely related results bearing the name Hilbert's Nullstellensatz. We shall state and prove the ones commonly found in the literature. These are the "weak form", the "common roots form" and the "strong form". The result that Hilbert originally proved was the strong form.

Weak form

The formulation and proof of the weak form of Hilbert's Nullstellensatz are naturally preceded by the following lemma.

Lemma 24.5:

Let $\mathbb {F}$ be any field. For any maximal ideal $m\leq \mathbb {F} [x_{1},\ldots ,x_{n}]$, the field $\mathbb {F} [x_{1},\ldots ,x_{n}]/m$ is a finite field extension of the field $\{c+m|c\in \mathbb {F} \}\subseteq \mathbb {F} [x_{1},\ldots ,x_{n}]/m$. In particular, if $\mathbb {F}$ is algebraically closed (and thus has no proper finite field extensions), then $\mathbb {F} [x_{1},\ldots ,x_{n}]/m=\{c+m|c\in \mathbb {F} \}$.

Proof 1 (using Zariski's lemma):

$\mathbb {F} [x_{1},\ldots ,x_{n}]/m$ is a finitely generated $\{c+m|c\in \mathbb {F} \}$-algebra, where all the operations are induced by the ring structure of $\mathbb {F} [x_{1},\ldots ,x_{n}]$; this is because the set $\{x_{1}+m,\ldots ,x_{n}+m\}$ constitutes a set of generators, since every element in $\mathbb {F} [x_{1},\ldots ,x_{n}]/m$ can be written as polynomials in those elements over $\{c+m|c\in \mathbb {F} \}$. Therefore, Zariski's lemma implies that $\mathbb {F} [x_{1},\ldots ,x_{n}]/m$ is a finite field extension of the field $\{c+m|c\in \mathbb {F} \}$.$\Box$

Proof 2 (using Jacobson rings):

We proceed by induction on $n$.

The case $n=1$ follows by noting that $\mathbb {F} [x_{1}]$ is a principal ideal domain (as an Euclidean domain) and hence, if $m\leq \mathbb {F} [x_{1}]$ is a (maximal) ideal, then $m=\langle f\rangle$ for a suitable $f\in \mathbb {F} [x_{1}]$. Now $\mathbb {F} [x_{1}]/m$ is a field if $m$ is maximal; we claim that it is a finite field extension of the field $\{c+m|c\in \mathbb {F} \}$. Indeed, as basis elements we may take $1+m,x_{1}+m,x_{1}^{2}+m,\ldots ,x_{1}^{d-1}+m$, where $d:=\deg f$ is the degree of the generating polynomial of the maximal ideal $m=\langle f\rangle$. Any element of $\mathbb {F} [x_{1}]/m$ can thus be expressed as linear combination of these basis elements, since the relation

$a_{d}x^{d}+m=-(a_{d-1}x^{d-1}+\cdots +a_{1}x+a_{0})+m$ (where $f(x)=a_{d}x^{d}+\cdots +a_{1}x+a_{0}$)

allows us to express monomials of degree $\geq d$ in terms of smaller ones.

Assume now the case $n-1$ is proven. Let $m\leq \mathbb {F} [x_{1},\ldots ,x_{n}]$ be a maximal ideal. According to Jacobson's first criterion, $\mathbb {F} [x_{1},\ldots ,x_{n-1}]$ is a Jacobson ring (since $\mathbb {F}$ is, being a field). Now $\mathbb {F} [x_{1},\ldots ,x_{n}]=\mathbb {F} [x_{1},\ldots ,x_{n-1}][x_{n}]$ and hence $m$ is a maximal ideal of $\mathbb {F} [x_{1},\ldots ,x_{n-1}][x_{n}]$. Thus, Goldman's second criterion asserts that $m_{0}:=\mathbb {F} [x_{1},\ldots ,x_{n-1}]\cap m$ is a maximal ideal of $\mathbb {F} [x_{1},\ldots ,x_{n-1}]$. Thus, $\mathbb {F} [x_{1},\ldots ,x_{n-1}]/m_{0}$ is a field, and, by the induction hypothesis, a finite field extension of $\{c+m_{0}|c\in \mathbb {F} \}$.

We define the ideal $p:=m_{0}\mathbb {F} [x_{1},\ldots ,x_{n}]$. The following map is manifestly an isomorphism:

${\begin{aligned}\varphi :\mathbb {F} [x_{1},\ldots ,x_{n-1}][x_{n}]/p&\mapsto (\mathbb {F} [x_{1},\ldots ,x_{n-1}]/m_{0})[x_{n}]\\a_{k}x_{n}^{k}+\cdots +a_{1}x_{n}+a_{0}+p&\mapsto (a_{k}+m_{0})x_{n}^{k}+\cdots +(a_{1}+m_{0})x_{n}+(a_{0}+m_{0})\end{aligned}}$

This map sends $\{c+p|c\in \mathbb {F} \}$ to $\{(c+m_{0})|c\in \mathbb {F} \}$ (and, being an isomorphism, vice versa).

Furthermore, since $m\subsetneq p$, the ideal $\pi _{p}(m)$ is maximal in $\mathbb {F} [x_{1},\ldots ,x_{n-1}][x_{n}]/p$. Hence, $\varphi (p/m)$ is maximal in $(\mathbb {F} [x_{1},\ldots ,x_{n-1}]/m_{0})[x_{n}]$ and thus $\left((\mathbb {F} [x_{1},\ldots ,x_{n-1}]/m_{0})[x_{n}]\right){\big /}\varphi (\pi _{p}(m))$ is a field. By the case $n=1$ it is a finite field extension of the field $\{d+\varphi (m/p)|d\in \mathbb {F} [x_{1},\ldots ,x_{n-1}]/m_{0}\}$.

In general, any proper ideal of $\mathbb {F} [x_{1},\ldots ,x_{n}]$, where $\mathbb {F}$ is a field, does not contain any constants (apart from zero), for else it would contain a unit and thus be equal to the whole of $\mathbb {F} [x_{1},\ldots ,x_{n}]$. This applies, in particular, to all maximal ideals of $\mathbb {F} [x_{1},\ldots ,x_{n}]$. Thus, elements of $\mathbb {F} [x_{1},\ldots ,x_{n}]/m$ of the form $c+m$ are distinct for pairwise distinct $c$. By definition of addition and multiplication of residue class rings, this implies that we have an isomorphism of rings (and thus, of fields)

$\mathbb {F} [x_{1},\ldots ,x_{n}]/m\supseteq \{c+m|c\in \mathbb {F} \}\cong \mathbb {F} ,c+m\mapsto c$.

Hence, in the case that $\mathbb {F}$ is algebraically closed, the above lemma implies $\mathbb {F} [x_{1},\ldots ,x_{n}]/m\cong \mathbb {F}$ via that isomorphism.

Theorem 24.6 (Hilbert's Nullstellensatz, weak form):

Let $\mathbb {F} ={\overline {\mathbb {F} }}$ be an algebraically closed field. For any $\xi =(\xi _{1},\ldots ,\xi _{n})\in \mathbb {F} ^{n}$, set

$m_{\xi }:=\langle x_{1}-\xi _{1},\ldots ,x_{n}-\xi _{n}\rangle \leq \mathbb {F} [x_{1},\ldots ,x_{n}]$;

according to lemma 21.12, $m_{\xi }$ is a maximal ideal.

The claim of the weak Hilbert's Nullstellensatz is this: Every maximal ideal $m\leq \mathbb {F} [x_{1},\ldots ,x_{n}]$ has the form $m_{\xi }$ for a suitable $\xi \in \mathbb {F} ^{n}$.

Proof:

Let $m\leq \mathbb {F} [x_{1},\ldots ,x_{n}]$ be any maximal ideal of $\mathbb {F} [x_{1},\ldots ,x_{n}]$. According to the preceding lemma, and since $\mathbb {F}$ is algebraically closed, we have $\mathbb {F} [x_{1},\ldots ,x_{n}]/m\cong \mathbb {F}$ via an isomorphism that sends elements of the type $c+m$ to $c$. Now this isomorphism must send any element of the type $x_{j}+m$ to some element $\alpha _{j}$ of $\mathbb {F}$. But further, the element $\alpha _{j}+m$ is sent to $\alpha _{j}\in \mathbb {F}$. Since we have an isomorphism (in particular injectivity), we have $\alpha _{j}+m=x_{j}+m\Leftrightarrow x_{j}-\alpha _{j}\in m$. Thus $\langle x_{1}-\alpha _{1},\ldots ,x_{n}-\alpha _{n}\rangle \subseteq m$ for suitable $\alpha _{1},\ldots ,\alpha _{n}$. Since the ideal $\langle x_{1}-\alpha _{1},\ldots ,x_{n}-\alpha _{n}\rangle$ is maximal (lemma 21.12), we have equality: $\langle x_{1}-\alpha _{1},\ldots ,x_{n}-\alpha _{n}\rangle =m$.$\Box$

Common roots form

Theorem 24.7 (Hilbert's Nullstellensatz, common root form):

Let $\mathbb {F} ={\overline {\mathbb {F} }}$ be an algebraically closed field and let $f_{1},\ldots ,f_{k}\in \mathbb {F} [x_{1},\ldots ,x_{n}]$. If

$\langle f_{1},\ldots ,f_{k}\rangle \subsetneq \mathbb {F} [x_{1},\ldots ,x_{n}]$,

then there exists $\xi =(\xi _{1},\ldots ,\xi _{n})\in \mathbb {F} ^{n}$ such that $f_{1}(\xi )=f_{2}(\xi )=\ldots =f_{k}(\xi )=0$.

Proof:

This follows from the weak form, since $\langle f_{1},\ldots ,f_{k}\rangle$ is contained within some maximal ideal $m\leq \mathbb {F} [x_{1},\ldots ,x_{n}]$, which by the weak form has the form $m=\langle x_{1}-\alpha _{1},\ldots ,x_{n}-\alpha _{n}\rangle$ for suitable $\alpha _{1},\ldots ,\alpha _{n}\in \mathbb {F}$ and hence $\{(\alpha _{1},\ldots ,\alpha _{n})\}=V(m)\subseteq V(\langle f_{1},\ldots ,f_{k}\rangle )$; in particular, $(\alpha _{1},\ldots ,\alpha _{n})\in V(\langle f_{1},\ldots ,f_{k}\rangle )$, that is, $\xi :=(\alpha _{1},\ldots ,\alpha _{n})$ is a common root of $f_{1},\ldots ,f_{k}$.$\Box$

Strong form

Theorem 24.8 (Hilbert's Nullstellensatz, strong form):

Let $\mathbb {F} ={\overline {\mathbb {F} }}$ be an algebraically closed field. If $I\leq \mathbb {F} [x_{1},\ldots ,x_{n}]$ is an arbitrary ideal, then

$I(V(I))=r(I)$;

recall: $r(I)$ is the radical of $I$.

In particular, if $I$ is a radical ideal (that is, $r(I)=I$), then

$I(V(I))=I$.

Note that together with the rule

$V(I(V(S)))=V(S)$

for any algebraic set $V(S)$ (that was established in chapter 22), this establishes a bijective correspondence between radical ideals of $\mathbb {F} [x_{1},\ldots ,x_{n}]$ and algebraic sets in $\mathbb {F} ^{n}$, given by the function

$V(\cdot ):\{{\text{radical ideals of }}\mathbb {F} [x_{1},\ldots ,x_{n}]\}\to \{{\text{algebraic sets in }}\mathbb {F} ^{n}\}$

and inverse

$I(\cdot ):\{{\text{algebraic sets in }}\mathbb {F} ^{n}\}\to \{{\text{radical ideals of }}\mathbb {F} [x_{1},\ldots ,x_{n}]\}$.

Proof 1 (using Jacobson rings):

Certainly, a field is a Jacobson ring. Furthermore, from Goldman's first criterion (theorem 14.4) we may infer that $\mathbb {F} [x_{1},\ldots ,x_{n}]$ is a Jacobson ring as well. Let now $f\in \mathbb {F} [x_{1},\ldots ,x_{n}]$ be a polynomial vanishing at all of $V(I)$, and let $m\leq \mathbb {F} [x_{1},\ldots ,x_{n}]$ be any maximal ideal of $\mathbb {F} [x_{1},\ldots ,x_{n}]$ that contains $I$. By the weak Nullstellensatz, $m$ has the form $m_{\xi }=\langle x_{1}-\xi _{1},\ldots ,x_{n}-\xi _{n}\rangle$ for a suitable $\xi =(\xi _{1},\ldots ,\xi _{n})\in \mathbb {F} ^{n}$.

Now we have $\xi \in V(m_{\xi })$, since any polynomial in $m_{\xi }$ can be written as a $\mathbb {F} [x_{1},\ldots ,x_{n}]$-linear combination of the generators $x_{1}-\xi _{1},\ldots ,x_{n}-\xi _{n}$. Hence, $I(V(m_{\xi }))$ is not all of $\mathbb {F} [x_{1},\ldots ,x_{n}]$; due to the constant functions, only the empty set has this ideal of vanishing. This, in combination with the fact that $m_{\xi }\subseteq I(V(m_{\xi }))$ and the maximality of $m_{\xi }$ implies $I(V(m_{\xi }))=m_{\xi }$.

Furthermore, $V(m_{\xi })\subseteq V(I)$, and hence $I(V(I))\subseteq I(V(m_{\xi }))$. Therefore, $f\in I(V(m_{\xi }))=m_{\xi }$.

Since $f\in I(V(I))$ was arbitrary, $I(V(I))$ is thus contained in all maximal ideals containing $I$ and hence, since $\mathbb {F} [x_{1},\ldots ,x_{n}]$ is Jacobson, $I(V(I))\subseteq r(I)$. However, the other direction $r(I)\subseteq I(V(I))$ is easy to see (we will prove this in the first paragraph of the next proof; there is no need to repeat the same argument in two proofs). Thus, $I(V(I))=r(I)$.$\Box$

Proof 2 (Rabinowitsch trick):

First we note $\supseteq$: Indeed, if $g^{n}\in I$, then $g(x)^{n}=0$ for all $x\in V(I)$. Hence also $g(x)=0$ for all $x\in V(I)$ since a field does not have nilpotent elements except zero (in fact, not even zero divisors). This implies $g\in I(V(I))$.

$\subseteq$ is the longer direction. Note that any field is Noetherian, and thus, by Hilbert's basis theorem, so is $\mathbb {F} [x_{1},\ldots ,x_{n}]$. Hence, $I$, being an ideal of $\mathbb {F} [x_{1},\ldots ,x_{n}]$, is finitely generated. Write

$I=\langle f_{1}(x),\ldots ,f_{k}(x)\rangle$.

Let $g\in I(V(I))$. Consider the polynomial ring $\mathbb {F} [x_{1},\ldots ,x_{n},z]$, which is augmented by an additional variable. In that ring, consider the polynomial $h(x_{1},\ldots ,x_{n},z):=1-zg(x_{1},\ldots ,x_{n})$. The polynomials $f_{1},\ldots ,f_{n},h$ have no common zero (where the polynomials $f_{1},\ldots ,f_{n}$ are seen as polynomials in the variables $x_{1},\ldots ,x_{n},z$ by the way of $f_{j}(x_{1},\ldots ,x_{n},z):=f_{j}(x_{1},\ldots ,x_{n})$), since if all the polynomials $f_{1},\ldots ,f_{n}$ are zero at $(\alpha _{1},\ldots ,\alpha _{n},\beta )$ (where the variable $\beta$ does not matter for the evaluation of $f_{1},\ldots ,f_{n}$), then so is $g$. Hence, in this case, $h(\alpha _{1},\ldots ,\alpha _{n},\beta )=1-\beta \cdot 0=1\neq 0$.

Now we may apply the common roots form of the Nullstellensatz for the case of $n+1$ variables. The polynomials $f_{1},\ldots ,f_{n},h$ have no common zero, and therefore, the common roots form Nullstellensatz implies that the ideal $\langle f_{1},\ldots ,f_{n},h\rangle$ must be all of $\mathbb {F} [x_{1},\ldots ,x_{n},z]$. In particular, we can find $\eta _{1},\ldots ,\eta _{n},\mu \in \mathbb {F} [x_{1},\ldots ,x_{n},z]$ such that

$1=\eta _{1}(x_{1},\ldots ,x_{n},z)f_{1}(x_{1},\ldots ,x_{n},z)+\cdots +\eta _{n}(x_{1},\ldots ,x_{n},z)f_{n}(x_{1},\ldots ,x_{n},z)+\mu (x_{1},\ldots ,x_{n},z)h(x_{1},\ldots ,x_{n},z)$.

Passing to the field of rational functions $\mathbb {F} (x_{1},\ldots ,x_{n})$, we may insert ${\frac {1}{g(x_{1},\ldots ,x_{n})}}$ for $z$ (recall that we assumed $g\not \equiv 0$) to obtain

$1=\eta _{1}(x_{1},\ldots ,x_{n},1/g)f_{1}(x_{1},\ldots ,x_{n},1/g)+\cdots +\eta _{n}(x_{1},\ldots ,x_{n},1/g)f_{n}(x_{1},\ldots ,x_{n},1/g)+\mu (x_{1},\ldots ,x_{n},1/g)h(x_{1},\ldots ,x_{n},1/g)$,

where we left out the variables of $g$ so that it still fits on the screen. Now $h(x_{1},\ldots ,x_{n},1/g)=1-{\frac {g(x_{1},\ldots ,x_{n})}{g(x_{1},\ldots ,x_{n})}}=0$, whence

$1=\eta _{1}(x_{1},\ldots ,x_{n},1/g)f_{1}(x_{1},\ldots ,x_{n},1/g)+\cdots +\eta _{n}(x_{1},\ldots ,x_{n},1/g)f_{n}(x_{1},\ldots ,x_{n},1/g)$.

Multiplying this equation by an appropriate power of $g$, call it $N\in \mathbb {N}$, sufficiently large such that we clear out all denominators, and noting that the last variable does not matter for $f_{1},\ldots ,f_{n}$, yields that $g^{N}$ equals an $\mathbb {F} [x_{1},\ldots ,x_{n}]$-linear combination of $f_{1},\ldots ,f_{n}$ and is thus contained within $I$. Hence, $g\in r(I)$.$\Box$

Note how Yuri Rainich ("Rabinowitsch") may have found this trick. Perhaps he realized that the weak Nullstellensatz is a claim for arbitrary $n$, and for the proof of the strong Nullstellensatz, we can do one $n$ at a time, using the infinitude of cases of the common roots form Nullstellensatz. That is, compared to a particular dimensional case in the strong Nullstellensatz, the infinitude of cases for the common roots form Nullstellensatz are not so weak at all, despite the common roots form being a consequence of the weak Nullstellensatz. This could have given Rainich evidence that using more cases, one obtains a stronger tool. And indeed, it worked out.

A diagram depicting the different paths to Hilbert's Nullstellensatz covered in this wikibook.