The Stacks project

Lemma 15.65.3. Let $R$ be a ring. Let $L$, $M$, $N$ be $R$-modules.

  1. If $M$ is finitely presented and $L$ is flat, then the canonical map $\mathop{\mathrm{Hom}}\nolimits _ R(M, N) \otimes _ R L \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N \otimes _ R L)$ is an isomorphism.

  2. If $M$ is $(-m)$-pseudo-coherent and $L$ is flat, then the canonical map $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N) \otimes _ R L \to \mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N \otimes _ R L)$ is an isomorphism for $i < m$.

Proof. Choose a resolution $F_\bullet \to M$ whose terms are free $R$-modules, see Algebra, Lemma 10.71.1. The complex $\mathop{\mathrm{Hom}}\nolimits _ R(F_\bullet , N)$ computes $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N)$ and the complex $\mathop{\mathrm{Hom}}\nolimits _ R(F_\bullet , N \otimes _ R L)$ computes $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N \otimes _ R L)$. There always is a map of cochain complexes

\[ \mathop{\mathrm{Hom}}\nolimits _ R(F_\bullet , N) \otimes _ R L \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ R(F_\bullet , N \otimes _ R L) \]

which induces canonical maps $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N) \otimes _ R L \to \mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N \otimes _ R L)$ for all $i \geq 0$ (canonical for example in the sense that these maps do not depend on the choice of the resolution $F_\bullet $). If $L$ is flat, then the complex $\mathop{\mathrm{Hom}}\nolimits _ R(F_\bullet , N) \otimes _ R L$ computes $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N) \otimes _ R L$ since taking cohomology commutes with tensoring by $L$.

Having said all of the above, if $M$ is $(-m)$-pseudo-coherent, then we may choose $F_\bullet $ such that $F_ i$ is finite free for $i = 0, \ldots , m$. Then the map of cochain complexes displayed above is an isomorphism in degrees $\leq m$ and hence an isomorphism on cohomology groups in degrees $< m$. This proves (2). If $M$ is finitely presented, then $M$ is $(-1)$-pseudo-coherent by Lemma 15.64.4 and we get the result because $\mathop{\mathrm{Hom}}\nolimits = \mathop{\mathrm{Ext}}\nolimits ^0$. $\square$

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