Corollary 2

— Suppose that A is an infinite integral domain or that A is an algebra over an infinite field. For every $f \in A[(X_i)_{i \in I}]$, let $\tilde{f}: A^I \to A$ be the polynomial function defined by f (IV, p. 4). Then the mapping $f \mapsto \tilde{f}$ is injective.

When A is an infinite integral domain, the corollary follows at once from Prop. 9. Suppose that A is an algebra over an infinite field k. Let $f = \sum_{\nu \in \mathbf{N}^{(I)}} \alpha_\nu X^\nu$ be a non-zero element of $A[(X_i)_{i \in I}]$; then there exists $\nu_0 \in \mathbf{N}^{(I)}$ such that $\alpha_{\nu_0} \neq 0$, and a k-linear form $\varphi$ on A such that $\varphi(\alpha_{\nu_0}) \neq 0$. Let $g = \sum_{\nu \in \mathbf{N}^{(I)}} \varphi(a_\nu) X^\nu \in k[(X_i)_{i \in I}]$; we have $g \neq 0$, hence there exists $x \in k^I$ such that $g(x) \neq 0$. Then $\varphi(f(x)) = g(x) \neq 0$, and so $f(x) \neq 0$.

When A is an infinite integral domain or when A is an algebra over an infinite field, we shall usually identify f with $\tilde{f}$.

Suppose that A is finite and let $f(X) = \prod_{a \in A} (X - a)$, then $f \neq 0$, but $\tilde{f} = 0$. For other examples see IV, p. 88, exercises 7 and 8.