Timeline for Action on group $\operatorname{Ext}^i(\mathcal{L}, \mathcal{M})$ by scalar multiplication
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 23, 2020 at 1:21 | comment | added | user267839 | Could you take couple of minites to look through if I have implemented your hint on my QUESTION 1 regarding the scalar action of an arbitrary $a \in k^*$ on $Ext(L,M)$ correctly? Since I can't draw commutative diagrams in comments I wrote my approach as an answer below: | |
| May 19, 2020 at 10:08 | comment | added | Sasha | The functor $Hom(-,-)$ has two arguments, you can derived with respect to any of these, and the results $Ext^i(-,-)$ are known to be isomorphic. Therefore, if you are interested in $Ext^1(L,M)$ you start with an exact sequence $$0 \to M \to E \to L \to 0$$ and either apply $Hom(L,-)$ to get a morphism $Hom(L,L) \to Ext^1(L,M)$, or apply $Hom(-,M)$ to get $Hom(M,M) \to Ext^1(L,M)$. | |
| May 19, 2020 at 9:10 | vote | accept | user267839 | ||
| May 19, 2020 at 9:09 | comment | added | user267839 | more concretly we write an arbitrary sequence $0 \to L \to E \to M \to 0$ draw a vertical arrow $a: L \to L$ (=the multiplication by $a \in k$) and obtain pullback $a^*E$. Thus we obtain a commutative diagram between the sequence $0 \to L \to E \to M \to 0$ and $0 \to L \to a^*E \to M \to 0$ above conncted by the vertical maps. Then we apply $Hom(-,L)$ and compare the classes in $Ext^1(M,L)$ in the related commutative square. I think that is it? | |
| May 19, 2020 at 9:02 | comment | added | user267839 | ...and a question on your hint on the verification about what kind of action is induced by $k$: Essentially this follows simply from the fact that the map $Hom(L,L) \to Ext^1(L,M)$ in compatible the $k$-multiplication; ie it's a $k$-morphism. That was the spice in your hint, right? More precisely, in addition we have to exploit the naturality of the boundary map... | |
| May 19, 2020 at 8:32 | comment | added | user267839 | Another remark: probably your hint on second way to show that that $L \oplus M$ corresponds to $0$ shold be beginn with applying $Hom(-,L)$ instead of $Hom(L,-)$ to the sequence, right? I learned this topic on that way that in general if $0 \to L \to X \to M \to 0$ is a ex sequence then it's corresponding class arrise as image of $id_L$ of the boundary map after application of $Hom(-,L)$ and forming the long exact sequence | |
| May 18, 2020 at 16:22 | comment | added | Sasha | This is a good exercise. For instance, you can easily deduce this from the above exact sequence. | |
| May 18, 2020 at 15:40 | comment | added | user267839 | Do you know a recommendable reference where it is explaned why this inherited action by scalar multiplication induce exactly the pullback resp pushout. Up to now I nowhere found a homological algebra book treating exactly this point. | |
| May 18, 2020 at 15:28 | history | answered | Sasha | CC BY-SA 4.0 |