Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.