Let B —> A be a surjection of rings. Let M, X, Y be a A-modules.<\/p>\n
If φ : X —> Y is a B-module map, then φ is an A-module map. We obtain X ⊗A<\/sub> M —> Y ⊗A<\/sub> M and X ⊗L<\/sup>A<\/sub> M —> Y ⊗L<\/sup>A<\/sub> M by functoriality.<\/p>\n Let ξ ∈ Ext^1_B(X, Y). I claim there is an element in Ext^2_A(X ⊗L<\/sup>A<\/sub> M, Y ⊗L<\/sup>A<\/sub> M) associated to ξ. Here is my construction. Choose a complex of free A-modules F_* resolving M. Choose a sequence (not<\/strong> a complex) of free B-modules F’_* such that F’_* ⊗B<\/sub> A is isomorphic to F_*. Let 0 —> Y —> E —> X —> 0 be the short exact sequence representing ξ. Then consider the composition<\/p>\n F’_{n + 2} ⊗B<\/sub> E —> F’_{n + 1} ⊗B<\/sub> E —> F’_n ⊗B<\/sub> E<\/p>\n Clearly this factors through a map<\/p>\n F_{n + 2} ⊗A<\/sub> X = F’_{n + 2} ⊗B<\/sub> X —> F’_n ⊗B<\/sub> Y = F_n ⊗A<\/sub> Y<\/p>\n The collection of these map gives X ⊗L<\/sup>A<\/sub> M —> Y ⊗L<\/sup>A<\/sub> M[2] as desired.<\/p>\n Questions: Thanks!<\/p>\n","protected":false},"excerpt":{"rendered":" Let B —> A be a surjection of rings. Let M, X, Y be a A-modules. If φ : X —> Y is a B-module map, then φ is an A-module map. We obtain X ⊗A M —> Y ⊗A … Continue reading
\n(a) Does this actually work?
\n(b) What is a “better” description of this construction?
\n(c) Is there a similar map Ext^2_B(X, Y) —> Ext^3_A(X ⊗L<\/sup>A<\/sub> M, Y ⊗L<\/sup>A<\/sub> M)?
\n(d) If you have a reference, could you please let us know?.<\/p>\n