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