= Solution
The <Noether normalization lemma> is valid over every <field> $k$, including finite fields: a nonzero <finitely generated algebra> $C$ contains elements $y_1,\ldots,y_d$ with <algebraic independence> over $k$ such that $C$ is finite as a <module> over the <polynomial ring> $k[y_1,\ldots,y_d]$. We prove it without an infinite-field hypothesis.
Write $C=k[a_1,\ldots,a_n]$ and induct on $n$. If the generators have <algebraic independence>, they already give a <polynomial ring> and the assertion is immediate. Otherwise choose a nonzero relation $F(a_1,\ldots,a_n)=0$. Choose an integer $M$ larger than every exponent in the monomials of $F$, and set
$$
b_i=a_i-a_n^{M^i}\quad(1\leq i<n).
$$
In $F(Y_1+T^{M},Y_2+T^{M^2},\ldots,Y_{n-1}+T^{M^{n-1}},T)$, the largest power of $T$ has a nonzero coefficient in $k$. Indeed, a monomial with exponents $(e_1,\ldots,e_n)$ contributes top weight $e_n+\sum_{i<n}e_iM^i$. These weights are distinct by uniqueness of base-$M$ expansion, so only one monomial contributes the highest power. After rescaling, the substituted relation is monic in $T$.
Consequently $a_n$ is an <integral element> over $C'=k[b_1,\ldots,b_{n-1}]$, and $C=C'[a_n]$ is finite over $C'$. Apply the induction hypothesis to $C'$ and compose the finite <module> extensions. This proves <Noether normalization by weighted substitutions>. The induction reaches $n=0$, where $C=k$. For an <integral domain> $C$, taking <fraction fields> makes the resulting extension finite algebraic, so $d$ is the <transcendence degree> of $\operatorname{Frac}(C)/k$. More generally $d=\dim C$: <integral extensions preserve Krull dimension>, and a polynomial algebra in $d$ variables has <Krull dimension> $d$.
A useful bridge to the <Hilbert Nullstellensatz> is the <Zariski lemma>. If a <field> $E$ is a <finitely generated algebra> over $k$, normalization makes it finite and integral over a <polynomial ring> $P=k[y_1,\ldots,y_d]$. A subring over which a field is integral is a field: for nonzero $a\in P$, an integral equation for $a^{-1}\in E$, multiplied by $a^{r-1}$, expresses $a^{-1}$ as an element of $P$. Therefore $P$ must be a field. A <polynomial ring> in a positive number of variables is not a field, since a variable has no polynomial inverse. Hence $d=0$ and $E/k$ is a <finite field extension>. This proves the <Zariski lemma>.
Now let $k$ be an <algebraically closed field>. The <Weak Hilbert Nullstellensatz> says that every <maximal ideal> of $k[X_1,\ldots,X_n]$ is uniquely of the form
$$
\boxed{\mathfrak m=(X_1-a_1,\ldots,X_n-a_n),\qquad a\in k^n.}
$$
To prove it, the residue field $k[X]/\mathfrak m$ is a field generated as a $k$-algebra by the images of the variables. The <Zariski lemma> makes it finite algebraic over $k$, and algebraic closedness makes it $k$. Thus each $X_i$ has an image $a_i\in k$. The evaluation map has kernel the displayed ideal: subtracting the constant value of a polynomial expresses its difference as a combination of $X_i-a_i$. That kernel is maximal and contained in $\mathfrak m$, so equality holds. Conversely every evaluation kernel is maximal because its quotient is $k$. Uniqueness follows from the variable images. Every proper <ideal> is contained in a <maximal ideal>, so it has a common zero; equivalently, an ideal with no common zero is the whole ring.
For an <ideal> $I\subseteq k[X_1,\ldots,X_n]$, let $V(I)$ be its common-zero set and let $I(V(I))$ be all polynomials vanishing on that set. The <Strong Hilbert Nullstellensatz> states
$$
\boxed{I(V(I))=\sqrt I.}
$$
The inclusion $\sqrt I\subseteq I(V(I))$ follows because a <field> has no nonzero <nilpotent elements>. For the other inclusion, take $f$ vanishing on $V(I)$, with $f\ne0$, and form the <Rabinowitsch trick> ideal
$$
J=I\,k[X_1,\ldots,X_n,T]+(1-Tf).
$$
It has no common zero: at a zero of $I$ the second generator has value one. The <Weak Hilbert Nullstellensatz> in $n+1$ variables gives $J=(1)$. Hence a finite identity has the form $1=\sum_j h_j(X,T)g_j(X)+h(X,T)(1-Tf)$, with $g_j\in I$. Substitute $T=f^{-1}$ in the <localization of a ring> $k[X]_f$. Clearing the finitely many powers of $f$ occurring in denominators gives $f^r\in I$ for some $r$, so $f\in\sqrt I$. The case $f=0$ is immediate. This completes all three proofs. The algebraically closed hypothesis belongs to the two forms of the <Hilbert Nullstellensatz>; it was not needed for the <Noether normalization lemma> or the <Zariski lemma>.
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