Let $K$ be a commutative ring with unity and define $\varphi$ by $$\varphi:K[x] \to \{f:K\to K\} \\ F\quad\mapsto \quad(a \mapsto F(a))$$ We know that $\varphi$ is injective if $K$ is an infinite (i.e. has infinitely many elements) integral domain. Prove or disprove: This is also true if you drop the "integral domain" condition.

I am pretty sure that this does no longer hold when we don't have an integral domain, however I could not come up with any counter example. I know a lot of rings, but none that are commutative and have infinitely many elements AND induce the same function. Can you help?


1 Answer 1


Let $K=\Bbb F_p^{\Bbb N}$, the direct product of countably many copies of the finite field $\Bbb F_p$. Then, $\ker \varphi\supseteq (x^p-x)$, just like it happens for $K=\Bbb F_p$.

Note: For the sake of clarity, if $\underline 1$ is the sequence $(1,1,\cdots)$, notice that $x^p-x:=\underline 1x^p-\underline 1x$, and that $\varphi(x^k)$ is the map $$(a_0,a_1,a_2,\cdots)\mapsto \left(a_0^k,a_1^k,a_2^k,\cdots\right)$$


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