Lemma 89.11.6. Let $K$ be a field. Let $L: \text{Mod}^{fg}_ K \to \text{Mod}_ K$ be a $K$-linear functor. Then $L$ is isomorphic to the functor $L(K) \otimes _ K - : \text{Mod}^{fg}_ K \to \text{Mod}_ K$.

**Proof.**
For $V \in \mathop{\mathrm{Ob}}\nolimits (\text{Mod}^{fg}_ K)$, the isomorphism $L(K) \otimes _ K V \to L(V)$ is given on pure tensors by $x \otimes v \mapsto L(f_ v)(x)$, where $f_ v: K \to V$ is the $K$-linear map sending $1 \mapsto v$. When $V = K$, this is the isomorphism $L(K) \otimes _ K K \to L(K)$ given by multiplication by $K$. For general $V$, it is an isomorphism by the case $V = K$ and the fact that $L$ commutes with finite products (Remark 89.11.3).
$\square$

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