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Feb 20, 2021 at 16:11 comment added Michael Orlitzky 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.
Feb 20, 2021 at 16:05 comment added Michael Orlitzky 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 $S$.
Feb 19, 2021 at 8:07 comment added Zach Teitler 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?
Feb 18, 2021 at 3:20 comment added R. van Dobben de Bruyn 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).
Feb 18, 2021 at 2:52 history edited Michael Orlitzky CC BY-SA 4.0
added some background information and references
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Jan 23, 2021 at 14:25 history asked Michael Orlitzky CC BY-SA 4.0