Question about Yoneda product

Asked May 12 '15 at 15:01

Active May 12 '15 at 16:12

Viewed 189 times

Let $A$ be a ring and $M,N,K$ are modules over $A$. Let $\xi\in\text{Ext}_A^1(N,M)$ and $\eta\in\text{Ext}_A^1(K,N)$ are given by

$$\xi:\,\,\,0\to M\to X\to N\to0,$$ $$\eta:\,\,\,0\to N\to Y\to K\to0.$$

Then $$\eta\xi:\,\,\,0\to M\to X\to Y\to K\to0$$ is an element in $\text{Ext}_A^2(K,M)$.

My questions is, what are the necessary and sufficient conditions for $\xi$ and $\eta$ to have a zero product in $\text{Ext}^2$?

What is the definition of element being zero in $\text{Ext}^2$?

edited May 12 '15 at 16:12

asked May 12 '15 at 15:01

guest31

I don't understand what you want. – Mariano Suárez-Álvarez May 12 '15 at 15:06
I made an edit. – guest31 May 12 '15 at 15:34
There is not much more that can be said than the definition of what being zero in $Ext^2$, I am afraid. – Mariano Suárez-Álvarez May 12 '15 at 15:36
I don't know what is the definition of element being zero in $Ext^2$. I know only the definition of zero for $Ext^1$, which is precisely the split exact sequence. – guest31 May 12 '15 at 15:43
Look in the book by Hilton-Stammbach, for example. – Mariano Suárez-Álvarez May 12 '15 at 16:13
There is no precise statement in Hilton-Stammbach's book. Actually they ask to describe the representative of zero in $Ext^n$ in Exercise 9.4 on page 155. Can you write me the definition? – guest31 May 12 '15 at 16:36
Equivalence is the smallest equivalence relation between extensions which extends the equivalence relation one uses for $Ext^1$ and such that it is compatible with Yoneda product. Maybe this is explained in these terms in MacLane's Homology. – Mariano Suárez-Álvarez May 12 '15 at 16:56

0 Answers0