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I'm currently trying to understand a line in a paper that would follow easily if the answer to the following question was yes:

Let $R, S$ be (commutative, unital) rings and let $\phi\colon R\hookrightarrow S$. Suppose that, as $\mathbb Z$-algebras, $\phi\otimes\mathrm{Id}_\mathbb Q\colon R\otimes \mathbb Q\to S\otimes\mathbb Q$ and $\phi\otimes\mathrm{Id}_{\mathbb F_p}\colon R\otimes \mathbb F_p\to S\otimes\mathbb F_p$ are isomorphisms for all primes $p$.

Is $\phi$ an isomorphism?

In the setting I care about, I know that $R, S$ are torsion free integral domains. I also know that $S$ is a subring of $\mathbb Z[X_1, \ldots, X_n]$ for some $n$ and that $S\otimes \mathbb Q$ is finitely generated as a $\mathbb Q$-algebra.

I know that the general statement is false when $R, S$ are just $\mathbb Z$-modules: a counterexample is $R = 0$ and $S = \mathbb {Q/Z}$.

Mathmo123
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    At least some assumption (for instance torsion-free) seems necessary, since otherwise you can adapt your example taking $R=\mathbb{Z}$ and $S=\mathbb{Z}\oplus \mathbb{Q}/\mathbb{Z}$ (where $\mathbb{Q}/\mathbb{Z}$ has a trivial product). – Captain Lama Dec 30 '20 at 14:23
  • @CaptainLama: Nice catch. In fact, I believe that this way, you can lift any module counterexample. So unless you rule out zero divisors, any module counterexample can be lifted to a unital ring example which is just as well-behaved. So you should either assume that the rings are domains or look for stronger hypotheses that make the statement true for modules. – tomasz Dec 30 '20 at 14:26
  • For abelian groups, I asked it a while ago: https://math.stackexchange.com/questions/1869690/if-f-otimes-textid-bbb-q-and-f-otimes-textid-bbbf-p-are-isom (you mentioned that it is wrong at the end of your post, I am just writing this for reference) – Watson Dec 20 '21 at 07:24

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For $R\subset S$ torsion-free abelian groups such that $S\otimes_\Bbb{Z} \Bbb{Q}=R\otimes_\Bbb{Z} \Bbb{Q}$.

For $a\in S$, if $a\not \in R$ then $a\Bbb{Z}\cap R=an \Bbb{Z}$ for some $n\ge 2$.

For $p|n$ then $an\ne 0 $ in $R/pR$ whereas $an=0\in S/pS$ so the map $R/pR\to R/(pS\cap R)\subset S/pS$ is not injective.

reuns
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  • What additional facts are you using? I'm a bit confused by the last line. For example, if $R = \mathbb Q$, then $R/pR$ is $0$ for all $p$. – Mathmo123 Dec 30 '20 at 14:31
  • $S$ torsion-free gives that $an \in p R \implies a\frac{n}p\in R$ thus $a\frac{n}p\not \in R$ gives that $an\not\in pR$ so $an\in ker(R/pR\to S/pS)$ – reuns Dec 30 '20 at 14:34
  • OK. Interestingly, you don't seem to need the fact that $\phi\otimes\mathrm{Id}_{\mathbb F_p}$ is surjective! – Mathmo123 Dec 30 '20 at 14:50
  • If $R\to S$ is not surjective but $R\otimes Q\to S\otimes Q$ is then $R/pR\to S/pS$ is not injective for some $p$, I admit it is a bit strange. – reuns Dec 30 '20 at 15:24
  • What's worrying me is that, in the paper I'm trying to understand, the statement is something along the lines of: if $\mathbb Z[\text{gens}]\subsetneq \mathbb Z[X_1,\ldots X_n]^G$, then for some $p$, $\mathbb F_p[\text{gens}]\subsetneq \mathbb F_p[X_1,\ldots X_n]^G$, then for some $p$. – Mathmo123 Dec 30 '20 at 15:27
  • If $R=\Bbb{Z},S=\Bbb{Z}[p^{-1}]=\Bbb{Z}[x]/(px-1)$ then $S/pS=0$ so $R/pR \to S/pS$ is surjective, only the injectivity is lacking. – reuns Dec 30 '20 at 15:35
  • The ambiguity in your last comment is if you meant $R/pR$ or $R/(pS\cap S)$, not the same. – reuns Dec 30 '20 at 15:38