Proposition 9

— Assume that $A$ is an integral domain. Let $I$ be a set, $(\mathbf{H}_i)_{i \in I}$ a family of infinite subsets of $\mathbf{A}$ and $H = \prod_{i \in I} \mathbf{H}_i \subset A'$. Iff is a non-zero element of $\mathbf{A}[(X_i)_{i \in I}]$, and $H_f$ the set of all $x \in H$ such that $f(x) \neq 0$, then $H$ and $H_f$ are equipotent.

a) First suppose that $I$ is finite and put $n = \mathrm{Card}\ I$. The proposition is clear for $n = 0$; we shall prove it by induction on $n$. Choose an element $i_0$ of $I$ and put $J = I - \{i_0\}$, $B = \mathbf{A}[(X_i)_{i \in J}]$. Since $f \neq 0$, we can write $f = \sum_{k=0}^m g_k X_{i_0}^k$, where $g_0, \ldots, g_m \in B$ and $g_m \neq 0$. By the induction hypothesis the set K of all $x \in \prod_{i \in s} H_i$ such that $g_m(x) \neq 0$ is equipotent with $\prod_{i \in J} H_i$. For $x \in K$ the polynomial

is non-zero. By Th. 1 (IV, p. 16) the set of $a \in H_{i_0}$ such that $h(\alpha) \neq 0$ is equipotent with $H_{i_0}$ whence and so $\operatorname{Card} H = \operatorname{Card} H_f$.

b) In the general case there is a finite subset $I'$ of I such that $f \in A[(X_i)_{i \in I'}]$. Let $H'_f$ be the set of all $x \in \prod_{i \in I'} H_i$ such that $f(x) \neq 0$. Then

and it suffices to apply the first part of the proof to $H'_f$.