Proposition 1

Let $A$ be a finitely generated algebra over a field $k$ and $L$ the algebraic closure of $k$.

(i) If $A \neq \{0\}$, there exists a $k$-homomorphism from $A$ to $L$.

(ii) Let $f_1, f_2$ be two $k$-homomorphisms from $A$ to $L$. For $f_1$ and $f_2$ to have the same kernel, it is necessary and sufficient that there exist a $k$-automorphism $s$ of $L$ such that $f_2 = s \circ f_1$.

(iii) Let $a$ be an ideal of $A$. For $a$ to be maximal, it is necessary and sufficient that it be the kernel of a k-homomorphism from $A$ to $L$.

(iv) For an element $x$ of $A$ to be such that $f(x) = 0$ for every k-homomorphism $f$ from $A$ to $L$, it is necessary and sufficient that $x$ be nilpotent.

Assertion (i) follows from no. 1, Corollary 3 to Theorem 1 applied replacing $A$ by $k$, $B$ by $A$, $b$ by the unit element of $B$ and $f$ by the canonical injection of $k$ into $L$.

If $f$ is a k-homomorphism from $A$ to $L$, $f(A)$ is a subring of $L$ containing $k$; as $L$ is an algebraic extension of $k$, $f(A)$ is a field (Algebra, Chapter V, § 3, no. 2, Proposition 3) and, if $a$ is the kernel off, $A/a$, isomorphic to $f(A)$, is therefore a field, which proves that $a$ is maximal. Conversely, if $a$ is a maximal ideal of $A$, it follows from (i) that there exists a k-homomorphism from $A/a$ to $L$ and hence a k-homomorphism of $A$ to $L$ whose kernel $b$ contains $a$; but as $a$ is maximal, $b = a$; this proves (iii).

We now prove (ii). If $s$ is a k-automorphism of $L$ such that $f_2 = s \circ f_1$, clearly $f_1$ and $f_2$ have the same kernel. Conversely, suppose that $f_1$ and $f_2$ have the same kernel; then there exists a k-isomorphism $s_0$ of the field $f_1(A)$ onto the field $f_2(A)$ such that $f_2 = s_0 \circ f_1$; but by Algebra, Chapter V, § 6, no. 3, Proposition 7, $s_0$ extends to a k-automorphism of $L$ and hence $f_2 = s \circ f_1$.

Finally, if $x \in A$ is such that $x^n = 0$, for every k-homomorphism $f$ from $A$ to $L$, $(f(x))^n = f(x^n) = 0$ and hence $f(x) = 0$ since $L$ is a field. Conversely, suppose that $x \in A$ is not nilpotent; then $A[x^{-1}]$ is a finitely generated $A$-algebra (and therefore a finitely generated k-algebra) not reduced to 0 (Chapter 11, § 2, no. 1, Remark 3) and hence there exists a k-homomorphism $g$ from $A[x^{-1}]$ to $L$ by (i). If $j : A \to A[x^{-1}]$ is the canonical homomorphism, $f = g \circ j$ is a k-homomorphism from $A$ to $L$ and $f(x)g(1/x) = g(x/1)g(1/x) = g(1) = 1$, whence $f(x) \neq 0$.

Let $k$ be a field and $L$ an extension field of $k$; an element $x = (x_1, \ldots, x_n)$ of $L^n$ is called a zero in $L^n$ of an ideal $c$ of the polynomial ring $k[X_1, \ldots, X_n]$ if

for all $P \in r$.