When is the restriction of scalars functor essentially surjective?

Question

For an algebra map $f: A \to B$, under what conditions is the restrictions of scalars functor $\mathbf{Mod}_B \to \mathbf{Mod}_A$ essentially surjective?

Answer 1

I don't have a general classification, but some interesting results to offer.

Let us start with an instructive example first. The forgetful functor $$\mathbf{Mod}_{\mathbb{C}} \to \mathbf{Mod}_{\mathbb{R}},\quad V \mapsto V|_{\mathbb{R}}$$ is not essentially surjective: its essential image (which coincides with the image, see below) consists of all infinite-dimensional real vector spaces and the finite-dimensional real vector spaces whose dimension is even.

Let $f : A \to B$ be a homomorphism of $k$-algebras, where $k$ is any commutative ring. Let us denote the restriction of scalars functor by $$f^* : \mathbf{Mod}_B \to \mathbf{Mod}_A$$ between the categories of left modules.

Lemma 1. If $f : A \to B$ is left invertible, i.e. there is an algebra homomorphism $g : B \to A$ with $g \circ f = \mathrm{id}_A$, then $f^*$ is surjective.

This follows directly from $f^* \circ g^* = (g \circ f)^* = (\mathrm{id}_A)^* = \mathrm{id}_{\mathbf{Mod}_A}$.

Lemma 1 gives lots of examples where $f^*$ is surjective, but we will see later that being left invertible is not a necessary condition. But we will see below that there are other examples as well.

Lemma 2. The image of $f^*$ is equal to the essential image of $f^*$. In particular, $f^*$ is surjective iff $f^*$ is essentially surjective.

This follows from the structure transport principle. Let $M$ be some $A$-module in the essential image, i.e. it is isomorphic to $f^*(N) = N|_A$ for some $B$-module $N$, say via $\varphi : M \to N|_A$. We can define a $B$-module structure on $M$ via $b \cdot m := \varphi^{-1}(b \cdot\varphi(m))$, and it extends the given $A$-module structure. This defines a $B$-module $M'$ such that $M$ is equal to $M'|_A$, which means that $M$ is in the image of $f^*$. $\square$

Because of Lemma 2, we will only deal with surjectivity in the following.

Lemma 3. The functor $f^*$ is surjective iff for every $k$-module $M$ every $k$-algebra homomorphism $A \to \mathrm{End}_k(M)$ lifts to a $k$-algebra homomorphism $B \to \mathrm{End}_k(M)$ along $f$.

This follows immediately from the equivalent description of an $A$-module as a $k$-module $M$ equipped with an $k$-algebra homomorphism $A \to \mathrm{End}_k(M)$. The functor $f^*$ just precomposes this map with $f$, the underlying $k$-module stays the same. $\square$

Of course, a stronger property would be that every algebra homomorphism $A \to C$ factors through $f : A \to B$, but this again would imply that $f$ is left invertible, which we covered in Lemma 1. This stronger property does not have to be satisfied!

Corollary 4. If $A,B$ are commutative and $f^*$ is surjective, then the morphism of affine schemes $\mathrm{Spec}(B) \to \mathrm{Spec}(A)$ is surjective on $k$-rational points.

This follows from Lemma 3 applied to $M=k$. $\square$

Another interesting special case is $A$ considered as an $A$-module. So if $f^*$ is surjective, $A$ actually has a $B$-module structure, extending the $A$-module structure. This is quite strong. In fact, we can prove (I will add the proof later):

Lemma 5. The $A$-module $A$ is in the image of $f^*$ iff there is a $k$-linear map $g : B \to A$ with $g \circ f = \mathrm{id}_A$ and $g(b \cdot b') = g(b \cdot f(g(b')))$ for all $b,b' \in B$.

As you can see, this is weaker than the multiplicativity of $g$.

Corollary 6. If $f^*$ is surjective, then $f$ is injective.

... To be continued ...

Answer 2

Not a complete answer. I don't see any reason to restrict attention to algebras over a field so let's assume that $f : A \to B$ is an arbitrary ring homomorphism. As others have noted, a necessary condition is that $f$ is injective, or else restriction of scalars only produces $A$-modules on which $\text{ker}(f)$ acts trivially. So from now on we'll assume this.

Next, restriction of scalars preserves colimits, and $\text{Mod}(A)$ is generated by colimits of copies of $A$. It follows that a sufficient condition for restriction of scalars to be essentially surjective is that $A$ lies in the essential image (this condition is also necessary) and so does every $A$-linear map $A \to A$, since we can construct an arbitrary $A$-module as a cokernel of a homomorphism between free $A$-modules and the above hypotheses imply that this whole diagram lifts to $\text{Mod}(B)$ and its colimit is preserved by restriction of scalars. Since this is about to matter, by "modules" here we mean right modules, so that the endomorphism ring of $A$ as a right $A$-module is $A$.

$A$ lies in the essential image iff the map $A \to \text{End}(A)$ given by right multiplication factors through $B$, or equivalently iff $A$ has a right $B$-module structure compatible with its existing $A$-module structure. The condition that every $A$-linear map also lifts is the condition that this right $B$-module structure also commutes with the natural left $A$-module structure on $A$. If $A, B$ are commutative this is automatic.

So, if I haven't missed anything, this argument appears to show that in the commutative case it is necessary and sufficient that $A$ is in the essential image, or equivalently that $A$ has a $B$-module structure compatible with its natural $A$-module structure. I will hazard a guess that in the general case it's going to turn out that $f$ must have a left inverse since nobody seems to have produced any examples in which this isn't true yet.