I have proved a result that seems (to me) interesting, and I am wondering whether it is a known result, and if not whether it seems interesting to others. The result is as follows:
Let $R=\mathbb Z / n \mathbb Z$ and let $\mathcal A = R^R$ denote the algebra of functions from $R$ to itself. Then:
- If $n$ is prime, every function in $\mathcal A$ can be represented by a unique polynomial in $R[x]$ of degree at most $n-1$; more specifically, there exists a natural isomorphism $R[x]/(x^n-x) \cong \mathcal A$.
- If $n$ is composite, then most functions in $\mathcal A$ cannot be represented by any polynomial in $R[x]$; that is, the natural homomorphism $R[x] \to \mathcal A$ is not onto.
Is this a well-known result? A trivial one? It certainly seemed interesting to me when I came across it, and I have been unsuccessful in finding any statement of this online.
Just to clarify, I am not asking for a proof -- I have one already -- just for an assessment of the value of the result.
It's certainly well-known. It is an immediate corollary of the Chinese Remainder Theorem. Notice that you could also work with $R^S$ for any finite set $S$. (And you don't mean $R^R$, which doesn't exist; you mean $R^S$, where $S$ is the underlying set of the ring $R$.)