2
$\begingroup$

I have a follow up question on this question of mine:

I can't reconstruct how I got $\operatorname{Im}{d_1^\ast} = 0$ from the following chain:

$0 \to \mathbb Z \otimes_{\mathbb Z} (\mathbb Z / 2 \mathbb Z) \xrightarrow{d_1^\ast = \cdot 284 \otimes id} \mathbb Z \otimes_{\mathbb Z} (\mathbb Z / 2 \mathbb Z) \xrightarrow{d_0^\ast=0} 0$

Now I think $\operatorname{Im}{d_1^\ast} = 284 \mathbb Z \otimes N$ and $\operatorname{Ker}{d_0^\ast} = \mathbb Z \otimes (\mathbb Z / 2 \mathbb Z)$.

And then $Tor^1 (\mathbb Z / 284 \mathbb Z, \mathbb Z / 2 \mathbb Z) = (\mathbb Z \otimes \mathbb Z / 2 \mathbb Z) / (284 \mathbb Z \otimes \mathbb Z / 2 \mathbb Z) $.


Is $\operatorname{Im}{d_1^\ast} = 284 \mathbb Z \otimes (\mathbb Z / 2 \mathbb Z) $ and $\operatorname{Ker}{d_0^\ast} = \mathbb Z \otimes (\mathbb Z / 2 \mathbb Z) \cong \mathbb Z / 2 \mathbb Z$ correct ?

And what does $(A \otimes B) / (C \otimes D)$ look like? Is it isomorphic to $(A/C) \otimes (B/D)$? Thanks for your help.

  • 1
    ... their kernels and images explicitly; not in terms of objects expressed in convoluted terms by various unsimplified tensor products. Regards,2012-07-28

1 Answers 1

2

Your description of the image is not correct.

You would do well to heed Jack Schmidt's warning in the comments: although $248\mathbb Z$ is a submodule of $\mathbb Z$, this is no longer true once you tensor with $\mathbb Z/2$. So your description of the image is not only incorrect, but the candidate image you have written down is not a subobject of the target.

I think you would also do well to follow my advice in the comments above, and to simplify the various tensor products in your complex , and then describe the maps in terms of the simplified objects, before you try to compute its cohomology.

  • 0
    Dear @MattE, thank you very much for checking my computation and teaching me how to compute $\mathrm{Tor}$.2012-07-29