Lemma 10.73.1. Given a flat ring map $R \to R'$, an $R$-module $M$, and an $R'$-module $N'$ the natural map
\[ \mathop{\mathrm{Ext}}\nolimits ^ i_{R'}(M \otimes _ R R', N') \to \text{Ext}^ i_ R(M, N') \]
is an isomorphism for $i \geq 0$.
10.73 Functorialities for Ext
In this section we briefly discuss the functoriality of $\mathop{\mathrm{Ext}}\nolimits $ with respect to change of ring, etc. Here is a list of items to work out.
Given $R \to R'$, an $R$-module $M$ and an $R'$-module $N'$ the $R$-module $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N')$ has a natural $R'$-module structure. Moreover, there is a canonical $R'$-linear map $\mathop{\mathrm{Ext}}\nolimits ^ i_{R'}(M \otimes _ R R', N') \to \mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N')$.
Given $R \to R'$ and $R$-modules $M$, $N$ there is a natural $R$-module map $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N) \to \text{Ext}^ i_ R(M, N \otimes _ R R')$.
Proof. Choose a free resolution $F_\bullet $ of $M$. Since $R \to R'$ is flat we see that $F_\bullet \otimes _ R R'$ is a free resolution of $M \otimes _ R R'$ over $R'$. The statement is that the map
\[ \mathop{\mathrm{Hom}}\nolimits _{R'}(F_\bullet \otimes _ R R', N') \to \mathop{\mathrm{Hom}}\nolimits _ R(F_\bullet , N') \]
induces an isomorphism on homology groups, which is true because it is an isomorphism of complexes by Lemma 10.14.3. $\square$
Comments (1)
Comment #683 by Keenan Kidwell on