I don't know much about the proof of the Freyd–Mitchell embedding theorem and I could not find an answer to my question looking naïvely online, but at the same time I feel like this is the kind of question to which someone who knows some of the details of the proof might be able to answer immediately, so it's probably worth trying. Here it is:

Can the Freyd-Mitchell embedding theorem be made stronger for $k$-linear abelian categories (where $k$ is a field), saying that not only, if $\mathcal{A}$ is a small abelian $k$-linear category, there exists a ring $R$ and a full, faithful, exact functor $F: \mathcal{A} → \text{$R$-$\mathrm{Mod}$}$, but that, moreover, $R$ can be assumed to be a $k$-algebra and $F$ to be $k$-linear?

More in general (also for non-$k$-linear categories): can one say anything about $R$? Is there even a unique "minimal" $R$ (up to Morita equivalence)?


1 Answer 1


Well, if $\mathcal{A}$ is a small $k$-linear abelian category, then the embedding is given by the following:

First we put $\mathcal{A}$ inside $\mathcal{L}(\mathcal{A},\operatorname{Ab})$, the category of left exact additive functors from $\mathcal{A}$ to the category of abelian groups $\operatorname{Ab}$, by considering the contravariant Yoneda embedding $\mathcal{Y} : \mathcal{A} \longrightarrow \mathcal{L}(\mathcal{A},\operatorname{Ab})$ which sends $A$ to $\operatorname{Hom}_{\mathcal{A}}(A,{-})$. Since $\mathcal{A}$ is $k$-linear, we may show that $\mathcal{L}(\mathcal{A},\operatorname{Ab})$ is also $k$-linear and that $\mathcal{Y}$ is a $k$-linear functor. ($\mathcal{Y}$ is also exact.)

Now, $\mathcal{L}(\mathcal{A},\operatorname{Ab})$ is a complete abelian $k$-linear category possessing an injective cogenerator. Then we apply the duality functor $D$ in $\mathcal{L}(\mathcal{A},\operatorname{Ab})$ and we obtain a covariant (exact) $k$-linear embedding $D \mathcal{Y} :\mathcal{A} \longrightarrow \mathcal{L}(\mathcal{A},\operatorname{Ab})^{op}$.

Finally, we know that $\mathcal{L}(\mathcal{A},\operatorname{Ab})^{op}$ is a cocomplete abelian category possesing a projective generator $P$, and we take a certain coproduct of copies of $P$, obtaining an object $Q$. Then we take the ring $R = \operatorname{End}(Q)$, which is a $k$-algebra and we consider the exact embedding $T : \mathcal{L}(\mathcal{A},\operatorname{Ab})^{op} \longrightarrow {\operatorname{Mod}}R$ defined by $T(X) = \operatorname{Hom}(Q,X)$, which is also $k$-linear.

Therefore, the embedding of $\mathcal{A}$ into ${\operatorname{Mod}}R$ is given by $TD \mathcal{Y} : \mathcal{A} \longrightarrow {\operatorname{Mod}}R$ and it is a $k$-linear functor.

Remarks: I took Mitchell's book "Theory of Categories" (MSN) as a reference for this answer.

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    $\begingroup$ I agree with the conclusion that $R$ can be taken to be a $k$-algebra, with the embedding $k$-linear. But this is not how the Freyd-Mitchell embedding is constructed. Firstly, your construction embeds $\mathcal{A}$ into $\mathcal{L}(\mathcal{A}^{\text{op}},\operatorname{Ab})$, not $\mathcal{L}(\mathcal{A},\operatorname{Ab})$. And also, it is not true in general that $\mathcal{L}(\mathcal{A},\operatorname{Ab})$ has a projective generator. But it does have an injective cogenerator, which is what is used to construct the Freyd-Mitchell embedding. $\endgroup$ Apr 22, 2020 at 9:27
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    $\begingroup$ @JeremyRickard thank you! I edited the answer and I think that now it is right. $\endgroup$
    – user144185
    Apr 22, 2020 at 11:29

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