Scholium

— Let A be an integral domain and $f \in A[(X_i)_{i \in I}]$. Th. 2 provides a convenient means of proving that $f = 0$. It suffices to consider an infinite integral domain E containing A as subring; if we can show that $f((x_i)) = 0$ for all $(x_i) \in E^I$ (or even for those $(x_i) \in E^I$ at which a finite number of given non-zero polynomials do not vanish) then it follows that $f = 0$. If A itself is not infinite, we can for example take E to be the ring $A[X]$ or its field of fractions.

Once we have proved the relation $f = 0$, we can clearly deduce that $f((y_i)) = 0$ for all $(y_i) \in F^I$ where F is any unital associative and commutative A-algebra whatsoever; in particular F may be finite or non-integral.

In other words, the proof of the identity $f((x_i)) = 0$ when the $x_i$ run over an infinite integral domain containing A as subring (with the possible restriction that $g_k((x_i)) \neq 0$ for $1 \leq k \leq m$, where the $g_k$ are non-zero polynomials) implies the same identity when the $x_i$ run over any unital associative and commutative A-algebra.

In particular let $f \in Z[(X_i)_{i \in I}]$. If $f((x_i)) = 0$ when the $x_i$ run over $\mathbf{Z}$ (with the possible restriction that $g_k((x_i)) \neq 0$ for $1 \leq k \leq m$, where the $g_k$ are non-zero elements of $\mathbf{Z}[(X_i)]$), then we have the same identity when the $x_i$ run over an arbitrary commutative ring.