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'm reading the proof of the existence of the tensor product. If $M,N$ are two $R$-modules then we can construct the tensor product $T$ as the quotient $C/D$ where $C$ is the free module over $M \times N$ and $D$ is the submodule generated by the set of all elements in $C$ of the form $$(m+m^\prime, n) - (m,n) - (m^\prime, n)$$ $$ (m, n+n^\prime) -(m,n) - (m,n^\prime) $$ $$ (am, n) - a(m,n)$$ $$ (m,an) - a(m,n)$$

I use $(m,n)$ to denote the element $e_{(m,n)} \in F(M\times N)$. Since $F(S) \cong \bigoplus_{s \in S} R$ I picture these elements as $e_{(m,0)} = (0, \dots, 0,1, 0, \dots)$ where the $1$ here is at position $m$ and $e_{(m,n)}$ the sequence with $1$ at position $m \cdot |M| + n$ and so forth.

Is this correct so far?

Now I wanted to see what this looks like. So I computed the tensor product of $M = N = \mathbb Z / 2 \mathbb Z$ over $R=\mathbb Z$. For $C$ I get that $C \cong \mathbb Z^4$. Then I computed all the elements above and noticed that $D \cong \langle \{(1,0,0,0), (0,1,0,0), (0,0,1,0)\} \rangle$. Hence $$M \otimes N = \mathbb Z / 2 \mathbb Z \otimes_{\mathbb Z} \mathbb Z / 2 \mathbb Z = \langle (0,0,0,1) \rangle \cong \mathbb Z$$

Is this correct?

share|cite|improve this question
It is better to study the tensor product abstractly instead of worrying about the specific construction. If you do that then you get the identity $$R / I \otimes_R M \cong M / IM$$ which should help you here (where $I \lhd R$ is an ideal). It's proved here:… . – Paul Slevin Jun 7 '12 at 14:06
Well, the result is definitely wrong: $Z/aZ \otimes_Z Z/bZ =Z/(a,b)Z$ – awllower Jun 7 '12 at 14:07
Thanks @PaulSlevin. Though I think to complement my studies it's good to look at concrete examples, too. Perhaps I should post my sums then it should become apparent where I messed up. – Rudy the Reindeer Jun 7 '12 at 14:10
@ClarkKent Tensor products are 1 million percent confusing when you first start learning! – user38268 Jun 7 '12 at 14:25
@BenjaminLim I linked to the proof:) I'll put it in my answer though. – Paul Slevin Jun 7 '12 at 14:34
up vote 2 down vote accepted

I will add to the comment I gave beneath your question. I know you said that you were interested in the specific construction, but perhaps later you will find this answer useful too.

We know that for any ideal $I \lhd R$, and any $R$-module $M$, we have the isomorphism:

$$ R/I \otimes_R M \cong M/IM$$

To see this, take the exact sequence $$0 \to I \to R \to R/I \to 0$$ and tensor by $M$ (recalling that tensoring by $M$ preserves right-exact sequences), then use the standard isomorphism theorem for modules.

So in your case it follows that

$$ \mathbb{Z}/2 \mathbb Z \otimes_\mathbb{Z} \mathbb{Z}/2 \mathbb Z \cong \frac{\mathbb{Z}/2\mathbb Z }{ \langle 2 \rangle \mathbb{Z} / 2\mathbb{Z}} \cong \mathbb{Z} / 2 \mathbb{Z} $$

Since the bottom part of the quotient is just $0$.

share|cite|improve this answer

Your last isomorphism above cannot hold; the guy on the left is a finite group but the one on the right is 100% not a finite group!

What it should be isomorphic to is $\Bbb{Z}/d\Bbb{Z}$ where $d$ is the greatest common divisor of $2$ and $2$, in this case $2$ itself so that $\Bbb{Z}/2\Bbb{Z} \otimes_\Bbb{Z} \Bbb{Z}/2\Bbb{Z} \cong \Bbb{Z}/2\Bbb{Z}$ . To see this, given any elementary tensor $a \otimes b$ in the tensor product, there are only 4 possibilities: $a$ odd $b$ even, $a$ odd $b$ odd, $a$ even $b$ odd, $ a$ odd $b$ even. But the cases where you have an even appearing are just zero because

$$\begin{eqnarray*} 0 \otimes 1 &=& (0 + 0) \otimes 1 \\ &=& 0 \otimes 1 + 0 \otimes 1 \\ \implies 0 \otimes 1 &=& 0 \end{eqnarray*} $$

and similarly $1 \otimes 0 = 0$. Hence there are only two distinct elements that appear, namely $1 \otimes 1$ and $0$ so that your tensor product is isomorphic to the cyclic group of order 2.

share|cite|improve this answer

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.