Restriction/Extension of scalars and isomorphism

Question

Let $A,B$ be commutative rings (with unit) and $\varphi\colon A\to B$ a ring morphism. Denote by $\varphi_{*}$ the restriction of scalars functor and by $\varphi_{!}$ the extension of scalars functor. Since $\varphi_{*}$ and $\varphi_{!}$ are functors it is obvious that for any pair of $B$-modules $M,N$ such that $M\cong N$ we have $\varphi_{*}(M)\cong\varphi_{*}(N)$ (and similarly for $A$-modules and extension of scalas).

From this it follows that if $M$ is an $A$-module and $N$ a $B$-module, then $\varphi_{!}(M)\cong N$ implies $\varphi_{*}(\varphi_{!}(M))\cong\varphi_{*}(N)$.

What I'm looking for is a counter-example (if it exists) where $\varphi_{*}(\varphi_{!}(M))\cong\varphi_{*}(N)$ but $\varphi_{!}(M)$ is not isomorphic to $N$.

I came up with the following: if $\varphi$ is surjective, then $\varphi_{!}(M)\cong N$ whenever $\varphi_{*}(\varphi_{!}(M))\cong\varphi_{*}(N)$ since in this case any $A$-module homomorphism is also a $B$-module homomorphism (this is for example the case when $\varphi=\pi\colon A\to A/I$ is the projection) but I really can't find a counterexample. I've tried for example to take $A=M=\mathbb{Z}, B=N=\mathbb{Q}$ so that $Q\otimes_{\mathbb{Z}}\mathbb{Z}\cong \mathbb{Q}$ as $\mathbb{Z}$-module but I think in this case $Q\otimes_{\mathbb{Z}}\mathbb{Z}\cong \mathbb{Q}$ also as a $\mathbb{Q}$-module, so it does not work. The problem may be related to the fact the the functors are adjoint but I'm not very familiar with adjunctions so please avoid using this argument.

Any help is appreciated!

Answer 1

The restriction of scalars $\varphi_\ast:\mathsf{Mod}_B \rightarrow \mathsf{Mod}_A$ is a conservative functor. This is the case, since being an isomorphism of modules can be tested on the level of underlying sets. So if there exists a counterexample, the isomorphism $\varphi_\ast\varphi_!M \cong \varphi_\ast N$ mustn't be $\varphi_\ast$ applied to some morphism $\varphi_!M\rightarrow N$ in $\mathsf{Mod}_B$. In your examples you have that $\varphi:A\rightarrow B$ is an epimorphism. But then, as @AndrewHubery notes $\varphi_\ast:\mathsf{Mod}_B\rightarrow\mathsf{Mod}_A$ is full and faithful, so every morphism $\varphi_\ast\varphi_! M \rightarrow \varphi_\ast N$ comes from a morphism $\varphi_! M \rightarrow N$...

Let us take the inclusion $A=\Bbb Z\overset{\varphi}\longrightarrow \Bbb Z[x]=B$. Let $M=\Bbb Z$ so $\varphi_! M = \Bbb Z[x] \otimes_\Bbb Z \Bbb Z = \Bbb Z[x]$ with the ring structure as $\Bbb Z[x]$-module structure. Consider $N=\bigoplus\limits_{n=0}^\infty\Bbb Z$ with the trivial $\Bbb Z[x]$-module structure given by $p(x)\cdot(z_n)_n = (p(0)\cdot z_n)_n$. The modules $\varphi_!M$ and $N$ are not isomorphic. This is because for any morphism $f:\varphi_!M \rightarrow N$ and any polynomial $p(x) \in \Bbb Z[x]$ with $p(0)=0$ we have $$f(p(x)) = f(p(x)\cdot 1)= p(x) \cdot f(1) = p(x)\cdot(u_n)_n = (p(0)\cdot u_n)_n = 0$$ so it cannot be a monomorphism. However, the underlying abelian groups (ie. $\Bbb Z$-module structures in both cases are given by $\bigoplus\limits_{n\in\Bbb N} \Bbb Z$.

Answer 2

Consider, for a field $k$, the diagonal embedding $\varphi\colon k\to k\times k$. For $M=k$ we have $\varphi_!M=k\times k$, which restricts back down to $\varphi_\ast\varphi_!M=k^2$. Of course, $N=k^2\times0$ also restricts to $\varphi_\ast N=k^2$, but $\varphi_!M$ is not isomorphic to $N$.