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It's well-known that over an infinite integral domain $R$, the ring of univariate polynomials $R\left[X_{1}\right]$ is isomorphic to a ring of one-argument "polynomial functions" (see, for example, Mac Lane and Birkhoff's Algebra).

It seems to me that this result should extend by induction to $R\left[X_{1},X_{2},\ldots,X_{n}\right]$ with the corresponding function ring of maps from $R^{n}$ to $R$, but I have been unable to find the more general result written down.

Is it true? If so, can someone provide a citation?

Edit: since no one has jumped at the chance to tell me that I'm blind, I'll add a bit more information for future searchers.

Over an infinite field, this result is well-known to algebraic geometers. For example, it's Corollary 6 on page 6 of Cox, Little, and O'Shea's Ideals, Varieties, and Algorithms. Fulton gives it as an exercise. Eisenbud mentions it in Commutative Algebra with a View Toward Algebraic Geometry. The proof, at least in the first case, is by a root-counting argument, which is why I suspect that it should work unmodified with an infinite integral domain.

Why bother? Suppose we rename one of the indeterminates, so that we're working in $R\left[X_{1},X_{2},\ldots,X_{n},\Lambda\right]$. Now we can consider (among other options) the maps from $R^{n}$ to $R\left[\Lambda\right]$ that take $n$ coordinates and return a univariate polynomial. The ring of these functions becomes relevant in linear algebra when we start to look at minimal and characteristic polynomials, and should also wind up being isomorphic to the polynomial ring from which it arises.

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  • $\begingroup$ Not an answer, but this post contains some interesting discussion on when the map $R[x] \to \operatorname{Map}(R,R)$ is injective (beyond the case of an integral domain). $\endgroup$
    – R. van Dobben de Bruyn
    Commented Feb 18, 2021 at 3:20
  • $\begingroup$ I'm not sure about a citation, but I think the same induction proof should work as in the field case, right? If some $p(x_1,\dotsc,x_n)$ vanishes identically on $R^n$, then expanding as $p_0 + p_1 x_n + \dotsb + p_d x_n^d$, each $p_i \in R[x_1,\dotsc,x_{n-1}]$, each $p_i$ must vanish identically on $R^{n-1}$ (or else for some choice of $(r_1,\dotsc,r_{n-1})$, $p(r_1,\dotsc,r_{n-1},x_n)$ wouldn't vanish identically on $R$, by the $n=1$ case). But then by induction each $p_i$ is actually the zero polynomial, hence so is $p$. I don't think this induction is different from the field case, is it? $\endgroup$
    – Zach Teitler
    Commented Feb 19, 2021 at 8:07
  • $\begingroup$ Yeah that's my thinking. If you take $S := R\left[\Lambda\right]$ and then proceed in $S\left[X_{1},X_{2},\ldots,X_{n}\right]$, you will actually wind up with a proof for the functions $S^{n} \to S$, and you can make a function $R^{n} \to S$ using the canonical embedding of $R$ into $S$. $\endgroup$
    – Michael Orlitzky
    Commented Feb 20, 2021 at 16:05
  • $\begingroup$ The main reason I'm looking for a citation is that the first (or last) $n$ indeterminates aren't really special. You could work with the functions that substitute any subset of the coordinates, and use the various canonical embeddings to get whatever domain you prefer. I think it should all work, and would be only a tedious bookkeeping exercise. But this isn't my area of expertise (or that of my readers), so the further I stray from the "obvious," the less my intuition should be trusted. $\endgroup$
    – Michael Orlitzky
    Commented Feb 20, 2021 at 16:11

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