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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.

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Your last paragraph might not be a good idea to consider. There are several problems with it. If you keep track of the maps, then things are better. –  Jack Schmidt Jul 28 '12 at 19:53
Basically, do not use $C \otimes D$ to mean a subgroup of $A \otimes B$ even if $C \leq A$ and $D \leq B$. Even if enough modules are flat that you could get a monomorphism, it can be dangerous to think like this. For instance, what is $2\mathbb{Z} \otimes \mathbb{Z}/2\mathbb{Z}$? As a tensor product, it is an abelian group of order 2. As a subgroup of $\mathbb{Z} \otimes \mathbb{Z}/2\mathbb{Z}$ it has order 1. –  Jack Schmidt Jul 28 '12 at 20:04
Dear Matt, The only tensor products you seem to have in sight are of the form $\mathbb Z\otimes N$. Since I imagine that the tensor product is also taking place over $N$, these are canonically isomorphic to $N$, and so can certainly be simplified. Regards, –  Matt E Jul 28 '12 at 20:10
Dear Matt, I don't understand your remark about needing to do other things before you simplify. When computing Tor, you take a free resolution of one of your modules, which involves terms like $\mathbb Z^n$. You then tensor with your second module, say $N$. The very first step is to replace all the expressions $\mathbb Z^n \otimes_{\mathbb Z} N$ by $N^n$; so in your case you certainly will want to replace all that $\mathbb Z\otimes_{\mathbb Z} \mathbb Z/2$'s by $\mathbb Z/2$'s. Once you have the terms of the complex computed, you will then want to describe the maps, and then compute ... –  Matt E Jul 28 '12 at 22:50
... their kernels and images explicitly; not in terms of objects expressed in convoluted terms by various unsimplified tensor products. Regards, –  Matt E Jul 28 '12 at 22:51

1 Answer 1

up vote 2 down vote accepted

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.

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Dear MattE, thank you so much for your help and patience. I have tried again, here is what I did: (The ring over which we tensor is $\mathbb Z$.) I start with $$ 0 \to \mathbb Z \otimes \mathbb Z / 2 \mathbb Z \to \mathbb Z \otimes \mathbb Z / 2 \mathbb Z \to 0$$ Then I used $R \otimes_R M \cong M$ to get $$ 0 \to \mathbb Z / 2 \mathbb Z \to \mathbb Z / 2 \mathbb Z \to 0$$ Then I used the isomorphism $R \otimes_R M \cong M$, $m \mapsto 1 \otimes m$, to compute the new maps: $$ m \in \mathbb Z / 2 \mathbb Z \mapsto 1 \otimes z \mapsto 284 \otimes z \mapsto 284z \equiv_2 0$$ –  Rudy the Reindeer Jul 29 '12 at 7:11
So that the new chain complex is $$ 0 \to \mathbb Z / 2 \mathbb Z \xrightarrow{0} \mathbb Z / 2 \mathbb Z \to 0$$ –  Rudy the Reindeer Jul 29 '12 at 7:12
From which I compute the homology (I thought it was homology groups, but maybe it's cohomology?) as $$ Tor^1 (M,N) = \mathrm{Ker}(0) / \mathrm{Im}(0) = \mathrm{Ker}(0) = \mathbb Z / 2 \mathbb Z$$ –  Rudy the Reindeer Jul 29 '12 at 7:15
Is this correct? Again, I apologise for the typos in my OP yesterday, it was a long day. –  Rudy the Reindeer Jul 29 '12 at 7:16
@MattN : Dear Matt, Your chain complex in the comments is correct, and so is your computation of $Tor_1$. This is right way to go --- simplify everything in sight, at which point you can hope that the relevant images and kernels are easy to compute, as they were in this example. [Sorry, I wrote cohomology rather than homology just out of habit; I'm used to thinking of things as cochain complexes.] Best wishes, –  Matt E Jul 29 '12 at 13:05

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