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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:

  1. 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 $R[x]/(x^n-x) \cong \mathcal A$.
  2. 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.

user26857
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mweiss
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    A related question. See the answers by Bill Dubuque and yours truly. Not an exact duplicate, but close enough to prevent me from answering this with a clear conscience. – Jyrki Lahtonen Apr 20 '15 at 15:07
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    IMO the simplest argument showing that you cannot get all the functions from polynomials is the following. If $p$ is a prime divisor of $n$, $n$ composite, then all the polynomials $f$ with coefficients in $R$ will satisfy $$a\equiv b\pmod p\implies f(a)\equiv f(b)\pmod p.$$ IIRC in a paper by Carlitz there is a description of the set of functions that you get by evaluating polynomials. – Jyrki Lahtonen Apr 20 '15 at 15:10
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    The fact that every function from $\mathbb Z/n\mathbb Z$ to itself is a polynomial when $n$ is prime is something that I think of as well known because I know it and because the proof is obvious, but I haven't taken any poll of mathematicians to see how many are aware of it. The other point I don't recall having thought of before. However, if you call a function algebraic, I am at first inclined to think it means the function somehow is a zero of a polynomial, rather than that the function itself is a polynomial function. ${}\qquad{}$ – Michael Hardy Apr 20 '15 at 15:47

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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 $R$.)

Martin Brandenburg
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