Timeline for Künneth formula for Ext groups
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 10, 2013 at 9:38 | comment | added | Martin Brandenburg | I don't understand. The image of $\alpha$ cannot be $1$-dimensional. | |
| Sep 6, 2013 at 4:51 | comment | added | Tyler Lawson | @Martin, I was too casual. The kernel of $\beta$ consists of all elements in the product which are annihilated by all $x_j$, and thus consists of the product over $j$ of the kernel $I$ of $B/(x_1) \to k$ (and in the other case, the product of $A \otimes I$). The map $\alpha$ annihilates all $x_j$, and so its image is a one-dimensional vector space $k$ (respectively image $A$). So what I should really say is that $A\otimes \prod I \to \prod A \otimes I$ is not an isomorphism. | |
| Sep 5, 2013 at 8:56 | comment | added | Martin Brandenburg | Thank you. The complex which computes $\mathrm{Ext}^2$ is $B/(x_1) \xrightarrow{\alpha} \prod_i B/(x_1) \xrightarrow{\beta} \prod_{ij} B/(x_1)$, where $\alpha([b])=([b x_i])_i$ and $\beta(([a_i])_i)=([x_j a_i])_{ij}$, right? Why do we get $\prod_i k$? | |
| Sep 5, 2013 at 2:48 | history | answered | Tyler Lawson | CC BY-SA 3.0 |