Noether Normalisation and Nullstellensatz

Fix $\beta \in K$. If $\beta \in S$, then $\beta \in k(S)$ and hence $\beta$ is algebraic over $k(S)$ because it is a root of $T-\beta \in k(S)[T]$.

Assume $\beta \mathrm{\notin }S$. By maximality, $S\cup \{\beta \}$ is algebraically dependent over $k$, so there is a nonzero polynomial $P(T_1,\dots ,T_m,Y)\in k[T_1,\dots ,T_m,Y]$ and elements $s_1,\dots ,s_m\in S$ such that $P(s_1,\dots ,s_m,\beta )=0$. The polynomial $P$ must involve $Y$, for otherwise $s_1,\dots ,s_m$ would already satisfy a nontrivial polynomial relation over $k$, contradicting algebraic independence of $S$.

View $P(s_1,\dots ,s_m,Y)$ as a nonzero polynomial in $k[S][Y]$ and then in $k(S)[Y]$. After dividing by its leading coefficient in the field $k(S)$, we obtain a monic polynomial $Q(Y)\in k(S)[Y]$ with $Q(\beta )=0$. Therefore $\beta$ is algebraic over $k(S)$.

Steps 1.1 and 2.1 cover every $\beta \in K$, so $K\mathrm{/}k(S)$ is algebraic. Hence $S$ is a transcendence basis of $K$ over $k$.

Because $T$ is a transcendence basis, $K$ is algebraic over $k(T)$; in particular $s$ is algebraic over $k(T)$. Choose a finite subset $U=\{t_1,\dots ,t_m\}\subseteq T$ of minimal size such that $s$ is algebraic over $k(U)$.

Minimality forces $m\ge 1$. Choose a nonzero polynomial relation for $s$ over $k(U)$ and clear denominators to obtain $P(s,t_1,\dots ,t_m)=0$ with $P\in k[X,Y_1,\dots ,Y_m]$ nonzero. The variable $Y_m$ must occur in $P$; otherwise the same relation would show that $s$ is algebraic over $k(t_1,\dots ,t_{m-1})$, contradicting minimality of $U$. Therefore, viewing $P$ as a polynomial in $Y_m$ over $k(s,t_1,\dots ,t_{m-1})$, we see that $t_m$ is algebraic over $k(s,t_1,\dots ,t_{m-1})$.

Put $T^{\mathrm{\prime }}=(T\setminus \{t_m\})\cup \{s\}$. If $T^{\mathrm{\prime }}$ were algebraically dependent, then $s$ would be algebraic over $k(T\setminus \{t_m\})$. Together with step 2.1 this would make $t_m$ algebraic over $k(T\setminus \{t_m\})$, contradicting algebraic independence of $T$. Hence $T^{\mathrm{\prime }}$ is algebraically independent.

Step 2.1 shows that $t_m$ is algebraic over $k(T^{\mathrm{\prime }})$, so $k(T)$ is algebraic over $k(T^{\mathrm{\prime }})$. Since $K$ is algebraic over $k(T)$, it is also algebraic over $k(T^{\mathrm{\prime }})$. By [L1], $T^{\mathrm{\prime }}$ is a transcendence basis of $K$ over $k$.

Choose a transcendence basis $S=\{s_1,\dots ,s_r\}$ of $K$ over $k$ and a transcendence basis $T=\{t_1,\dots ,t_m\}$ of $L$ over $K$. Then $r$ and $m$ are the two given transcendence degrees.

The union $S\cup T$ is algebraically independent over $k$. Indeed, a polynomial relation over $k$ among $S\cup T$ would also be a relation over $K$, contradicting algebraic independence of $T$ over $K$. Moreover $L$ is algebraic over $K(T)$, and $K$ is algebraic over $k(S)$, so [L2] shows that $L$ is algebraic over $k(S,T)$. Therefore [L1] makes $S\cup T$ a transcendence basis of $L$ over $k$.

Since $S\cup T$ is a transcendence basis of $L$ over $k$ with $r+m$ elements, ${\mathrm{trdeg}}_{k}L=r+m={\mathrm{trdeg}}_{k}K+{\mathrm{trdeg}}_{K}L$.

Let $d$ be the total degree of $f$, and let $H$ be the homogeneous degree $d$ part of $f$. Then $H$ is nonzero.

The polynomial $H(X_1,\dots ,X_{n-1},1)$ in $k[X_1,\dots ,X_{n-1}]$ is nonzero, so [L1] yields $a_1,\dots ,a_{n-1}\in k$ with $H(a_1,\dots ,a_{n-1},1)\mathrm{\ne }0$.

Substitute $x_i+a_i{x}_n$ for $x_i$ when $i<n$. Every degree-$d$ monomial of $f$ contributes to the coefficient of $x_n^d$, and the lower-degree part of $f$ contributes only lower powers of $x_n$. Therefore the coefficient of $x_n^d$ in the transformed polynomial is exactly $H(a_1,\dots ,a_{n-1},1)\in k^{\times }$.

Multiplying by the inverse scalar $c=H(a_1,\dots ,a_{n-1},1{)}^{-1}$ makes the transformed polynomial monic in $x_n$.

Write $f={\sum }_{\alpha =({\alpha }_1,\dots ,{\alpha }_n)}{c}_{\alpha }{x}_1^{{\alpha }_1}\cdots x_n^{{\alpha }_n}$ with only finitely many nonzero coefficients. Choose $N$ larger than every exponent ${\alpha }_i$ occurring with $c_{\alpha }\mathrm{\ne }0$.

For each exponent vector $\alpha$, define the weight $w(\alpha )={\alpha }_n+{\alpha }_{1}N+{\alpha }_2{N}^2+\cdots +{\alpha }_{n-1}{N}^{n-1}\mathrm{.}$ Because every ${\alpha }_i<N$, base-$N$ expansion is unique, so distinct exponent vectors have distinct weights.

After substituting $x_i+x_n^{N^i}$ for $x_i$, the monomial $x_1^{{\alpha }_1}\cdots x_n^{{\alpha }_n}$ contributes the term $c_{\alpha }{x}_n^{w(\alpha )}$ to the highest $x_n$-power coming from that monomial; all other terms from its binomial expansion have smaller $x_n$-power. Let ${\alpha }^{\ast }$ be the exponent vector with maximal weight among those with $c_{\alpha }\mathrm{\ne }0$. By step 2.1 this ${\alpha }^{\ast }$ is unique, so $c_{{\alpha }^{\ast }}{x}_n^{w({\alpha }^{\ast })}$ is the unique highest $x_n$-term of the transformed polynomial.

The transformed polynomial is therefore nonzero and has a unique highest $x_n$-term with nonzero coefficient. Multiplying by the inverse of that coefficient makes it monic in $x_n$.

Let $B=k[y_1,\dots ,y_{n-1}]$. By [L1], $B$ is a subalgebra of $A$. The displayed relation is a monic polynomial in $B[Y]$ with value zero at $y_n$, so [L2] shows that $y_n$ is integral over $B$.

Since $A$ is generated by $y_1,\dots ,y_n$ over $k$, it is generated by $y_n$ as a $B$-algebra: $A=B[y_n]$. Every element of $B$ is integral over $B$, and step 1.1 gives integrality of $y_n$ over $B$, so every element of $A=B[y_n]$ is integral over $B$.

Therefore $A$ is integral over the subalgebra $k[y_1,\dots ,y_{n-1}]$.

By [L1], choose generators $x_1,\dots ,x_n$ of $A$ as a $k$-algebra. We prove the theorem by induction on $n$.

Inductive hypothesis: assume the statement for every $k$-algebra generated by at most $n-1$ elements.

Base case $n=0$: then $A$ is the image of $k$. Because $A\mathrm{\ne }0$ and $k$ is a field, the structure map $k\to A$ is injective and hence identifies $A$ with $k=k[]$. Thus $A$ is integral over the empty polynomial algebra.

If $x_1,\dots ,x_n$ are algebraically independent over $k$, then $A=k[x_1,\dots ,x_n]$ already has the required form.

Otherwise there is a nonzero polynomial relation among $x_1,\dots ,x_n$. If $k$ is infinite, apply [L3] to make such a relation monic in the last variable after a triangular change. If $k$ is finite, apply [L4] to make such a relation monic in the last variable after the exponent substitution. In either case we obtain new generators $y_1,\dots ,y_n$ of $A$ such that $A$ is integral over $B:=k[y_1,\dots ,y_{n-1}]$ by [L5].

The algebra $B$ is generated over $k$ by $n-1$ elements, so the induction hypothesis yields algebraically independent elements $z_1,\dots ,z_d\in B$ such that $B$ is integral over $k[z_1,\dots ,z_d]$. Then [L2] and step 2.3 show that $A$ is integral over $k[z_1,\dots ,z_d]$.

Step 2.2 handles the algebraically independent case, while steps 2.3 and 3.1 handle the dependent case. Therefore the theorem holds for $n$ generators, and hence for every finite-type $k$-algebra.

By [L1], choose algebraically independent elements $z_1,\dots ,z_d\in A$ such that $A$ is integral over $R:=k[z_1,\dots ,z_d]$.

Because $A$ is of finite type over $k$, choose generators $a_1,\dots ,a_m\in A$ with $A=k[a_1,\dots ,a_m]$. Since $R$ contains the image of $k$ and lies in $A$, we also have $A=R[a_1,\dots ,a_m]$. Each $a_i$ is integral over $R$ by step 1.1, so [L2] implies that $A$ is module-finite over $R$.

Hence $A$ is module-finite over the polynomial ring $k[z_1,\dots ,z_d]$.

Because $A$ is finitely generated as a module over $R:=k[z_1,\dots ,z_d]$, multiplication by any $a\in A$ is an $R$-linear endomorphism of a finite $R$-module. Cayley-Hamilton therefore gives a monic polynomial over $R$ satisfied by $a$, so $A$ is integral over $R$.

The polynomial ring $R$ has the prime chain $(0)\subset (z_1)\subset (z_1,z_2)\subset \cdots \subset (z_1,\dots ,z_d),$ of length $d$: each quotient by $(z_1,\dots ,z_i)$ is again a polynomial ring over $k$ and hence an integral domain, so those ideals are prime.

Apply [L2] to the chain from step 1.2 and the integral inclusion $R\subseteq A$ from step 1.1. This gives a chain of $d+1$ prime ideals in $A$, so by [L1] the dimension of $A$ is at least $d$.

Suppose that $k(t)$ is generated as a $k[t]$-module by finitely many fractions $f_1\mathrm{/}{g}_1,\dots ,f_m\mathrm{/}{g}_m$ with $f_i,g_i\in k[t]$ and $g_i\mathrm{\ne }0$. Let $g=g_1\cdots g_m$. Then every $k[t]$-linear combination of the generators has denominator dividing $g$, so it can be written as $h\mathrm{/}g$ for some $h\in k[t]$.

If $g$ is constant, then $k[t]$ itself would equal $k(t)$, which is false because $1\mathrm{/}t\mathrm{\notin }k[t]$. So $g$ is nonconstant. By [L2], the nonunit $g+1\in k[t]$ has an irreducible factor $q$. Since $q$ divides $g+1$, it does not divide $g$.

The fraction $1\mathrm{/}q$ lies in $k(t)$ by [L1]. If it belonged to the $k[t]$-module generated by the chosen fractions, step 1.1 would give $1\mathrm{/}q=h\mathrm{/}g$ for some $h\in k[t]$, hence $g=hq$. But then $q$ would divide $g$, contrary to step 2.1. Therefore the assumed finite generating set cannot exist, so $k(t)$ is not finite over $k[t]$.

Choose an integer $N>1$ larger than every exponent occurring in $s$, and define a $k$-algebra map $\phi :k[t_1,\dots ,t_r]\to k[u],t_i\mapsto u^{N^{i-1}}\mathrm{.}$ By uniqueness of base-$N$ expansion, distinct monomials of $s$ acquire distinct $u$-degrees under $\phi$, so $\phi (s)\mathrm{\ne }0$.

Because $\phi (s)\mathrm{\ne }0$, the universal property of localization gives a ring homomorphism $\mathrm{\Phi }:k[t_1,\dots ,t_r]\left[\frac{1}{s}\right]\to k[u]\left[\frac{1}{\phi (s)}\right]\mathrm{.}$ If the source were a field, its image would also be a field.

The target is not a field. If $\phi (s)$ is constant, then the target is just $k[u]$, and $u$ is not invertible in $k[u]$. If $\phi (s)$ is nonconstant and $1\mathrm{/}(\phi (s)+1)=h\mathrm{/}\phi (s{)}^m$ in the localization, then $\phi (s{)}^m=h(\phi (s)+1)$ in $k[u]$. Dividing the left-hand side by $\phi (s)+1$ leaves remainder $(-1{)}^m\mathrm{\ne }0$, contradiction.

Step 3.1 contradicts the conclusion of step 2.1. Therefore $k[t_1,\dots ,t_r][1\mathrm{/}s]$ is not a field.

Choose generators $a_1,\dots ,a_m\in K$ of $K$ as a $k$-algebra. Since $t_1,\dots ,t_r$ is a transcendence basis, [L2] shows that each $a_i$ is algebraic over $F:=k(t_1,\dots ,t_r)$.

For each $i$, choose a nonzero polynomial $c_{i,d_i}{X}^{d_i}+\cdots +c_{i,0}\in k[t_1,\dots ,t_r][X]$ with $c_{i,d_i}\mathrm{\ne }0$ and value zero at $a_i$. Let $s$ be the product of all leading coefficients $c_{i,d_i}$. After localizing at $s$, each $c_{i,d_i}$ becomes invertible, so each $a_i$ satisfies a monic polynomial over $k[t_1,\dots ,t_r][1\mathrm{/}s]$. Hence every $a_i$ is integral over that localization.

The field $K$ is generated over $k[t_1,\dots ,t_r][1\mathrm{/}s]$ by the integral elements $a_1,\dots ,a_m$. Therefore every element of $k[t_1,\dots ,t_r][1\mathrm{/}s][a_1,\dots ,a_m]=K$ is integral over $k[t_1,\dots ,t_r][1\mathrm{/}s]$.

Thus $K$ is integral over $k[t_1,\dots ,t_r][1\mathrm{/}s]$.

Let $t_1,\dots ,t_r$ be a transcendence basis of $K$ over $k$. By [L1] there exists a nonzero $s\in k[t_1,\dots ,t_r]$ such that $K$ is integral over $R:=k[t_1,\dots ,t_r][1\mathrm{/}s]$.

Assume $r>0$. Because $K$ is a field and integral over $R$, every nonzero element $a\in R$ is a unit of $R$: the inverse $a^{-1}\in K$ satisfies a monic equation over $R$, and multiplying by a large power of $a$ rewrites that equation as $a^{-1}=-(c_{n-1}+c_{n-2}a+\cdots +c_0{a}^{n-1})\in R\mathrm{.}$ Thus $R$ would be a field, contradicting [L2].

Therefore $r=0$, so the transcendence basis is empty and $K\mathrm{/}k$ is algebraic. Since $K$ is finitely generated as a $k$-algebra, choose algebra generators $a_1,\dots ,a_m$; they are algebraic over $k$, and [L3] makes $K=k(a_1,\dots ,a_m)$ finite over $k$.

By [L1], choose generators $a_1,\dots ,a_n$ of $A$ as a $k$-algebra. Their images in $A\mathrm{/}\mathfrak{m}$ generate the quotient as a $k$-algebra, so $A\mathrm{/}\mathfrak{m}$ is again a finite-type $k$-algebra.

By [L2], the quotient $A\mathrm{/}\mathfrak{m}$ is a field. Applying [L3] to this finite-type field over $k$ shows that $A\mathrm{/}\mathfrak{m}$ is finite over $k$.

Base case $n=1$: divide $f(x_1)$ by the monic polynomial $x_1-a_1$. By [L2], $f(x_1)=(x_1-a_1)q(x_1)+r$ with $r\in k$. Evaluating at $x_1=a_1$ gives $r=f(a_1)$ by [L1], so $f-f(a_1)\in (x_1-a_1)$. Therefore the kernel of ${\mathrm{ev}}_{a_1}$ is $(x_1-a_1)$.

Inductive hypothesis: assume the statement for $n-1$ variables.

Write $f$ as a polynomial in $x_n$ with coefficients in $k[x_1,\dots ,x_{n-1}]$ and divide by the monic polynomial $x_n-a_n$: $f=(x_n-a_n)q+r,$ where $r\in k[x_1,\dots ,x_{n-1}]$. Evaluating at $a$ gives $f(a)=r(a_1,\dots ,a_{n-1})$. By the induction hypothesis, $r-r(a_1,\dots ,a_{n-1})\in (x_1-a_1,\dots ,x_{n-1}-a_{n-1})$, and hence $f-f(a)\in (x_1-a_1,\dots ,x_n-a_n)\mathrm{.}$

Step 2.1 shows that every polynomial differs from its value at $a$ by an element of $(x_1-a_1,\dots ,x_n-a_n)$, so the quotient by that ideal is isomorphic to $k$. By [L3] the ideal is maximal, and its kernel description is the one established above.

By [L1], the residue field $K:=k[x_1,\dots ,x_n]\mathrm{/}\mathfrak{m}$ is a finite extension of $k$. Let ${\bar{x}}_i$ be the image of $x_i$ in $K$.

Each ${\bar{x}}_i$ is algebraic over $k$ because $K\mathrm{/}k$ is finite. Let $m_i(T)\in k[T]$ be its minimal polynomial. Since $k$ is algebraically closed, [L3] gives a root $a_i\in k$ of $m_i$. Minimality forces $m_i(T)=T-a_i$, so ${\bar{x}}_i=a_i$ in $K$.

The quotient map $k[x_1,\dots ,x_n]\to K$ is therefore evaluation at the point $a=(a_1,\dots ,a_n)$. Its kernel is $\mathfrak{m}$, while [L2] says the evaluation kernel is $(x_1-a_1,\dots ,x_n-a_n)$. Hence these ideals are equal. Uniqueness of $a$ follows because the quotient remembers each coordinate class ${\bar{x}}_i$.

Suppose $(a,b)\in k^{n+1}$ were a common zero of $J$. Then every element of $I$ vanishes at $a$, so $a\in V(I)$. By hypothesis, $f(a)=0$.

But $1-yf\in J$, so evaluating at $(a,b)$ gives $1-bf(a)=0$. Using step 1.1 this becomes $1=0$, which is impossible in a field. Therefore $J$ has no common zero.