Theorem 2

— Let K be an infinite field, L an extension of K and A an algebra over K. Let $u_1, \ldots, u_n$ be distinct K-algebra homomorphisms of A into L and f a polynomial in $L[X_1, \ldots, X_j]$. If we have $f(u_1(x), \ldots, u_n(x)) = 0$ for all $x \in A$, then $f = 0$.

Let B be the set of elements of $L^n$ of the form $(u_1(x), \ldots, u_n(x))$ with $x \in A$. By Th. 1, there is no sequence $(\alpha_1, \ldots, \alpha_n)$ of elements not all zero in L such that $\sum_{i=1}^n \alpha_i u_i(x) = 0$ for all $x \in A$; therefore (II, p. 301, Th. 7) B generates the vector space $L^n$ over L. So there exist elements $a_1, \ldots, a_n$ of $A$ such that the matrix $(u_i(a_j))_{1 \leq i,j \leq n}$ is invertible.

Let us define the polynomial $g \in L[Y_1, \ldots, Y_n]$ by

Let $y_1, \ldots, y,$ be in $K$; writing $x = \sum_{i=1}^n y_i a_i$, we have whence $g(y_1, \ldots, y_n) = 0$ by the hypothesis on $f$. Since the field $K$ is infinite, we have $g = 0$ (IV, p. 18, Cor. 2); now the matrix $(u_i(a_j))$ has an inverse $(b_{ij})$ and we have whence $f = 0$.

Th. 2 has no analogue for finite fields. Thus let $K$ be a finite field with q elements, $A = L = K$ and $f(X) = X^q - X$. We have $x^q = x$ for all $x \in K$ (V, p. 93, Prop. 2); therefore if $u$ is the identity automorphism of $K$, we have $f(u(x)) = 0$ for all $x \in K$, even though $f$ is not zero.