# A simple example of a tame tensor functor

Let $C$ be a $R$-linear cocomplete tame abelian tensor category (see here for definitions) with unit $1$, then there is a unique $R$-linear cocontinuous tensor functor $F : \text{Mod}(R) \to C$, sending an $R$-module $M$ to the external tensor product $M \otimes_R 1$. Does it follow that $F$ is a tame tensor functor? So there are two things to check:

a) If $M$ is a flat $R$-module, is then $M \otimes_R 1$ a flat object of $C$? This is indeed true if $C$ has the property that directed colimits of monos are monos: Write $M$ as a directed colimit of free modules (Lazard's Theorem). Then $M \otimes_R 1$ is a directed colimit of objects which are directs sums of copies of $1$ and thus flat. The assumption now implies that directed colimits of flat objects are flat. But what about general $C$?

b) If $0 \to M' \to M \to M'' \to 0$ is an exact sequence of $R$-modules with $M''$ flat, is then the induced sequence $0 \to M' \otimes_R 1 \to M \otimes_R 1 \to M'' \otimes_R 1 \to 0$ exact? In the case that $M''$ is free this is clear (the sequence splits), and in general we may write again $M''$ as a directed colimit of free modules $M''_i$. It now clear how to define compatible exact sequences $0 \to M'_i \to M_i \to M''_i \to 0$ mapping to the given sequence, but it is not clear to me if we can arrange this so that this is a colimit of sequences. If it was, and $C$ satisfies the axiom (AB5), then we would be done.

Remark that both a),b) are true if $C = \text{Mod}(S)$ for some ring $S$, and then more generally when $C = \text{Qcoh}(X)$ for some scheme $X$.

-