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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?

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    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: http://math.stackexchange.com/questions/150114/showing-that-if-r-is-local-and-m-an-r-module-then-m-otimes-r-r-mathfr .2012-06-07
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    Well, the result is definitely wrong: $Z/aZ \otimes_Z Z/bZ =Z/(a,b)Z$2012-06-07
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    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.2012-06-07
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    And thanks @awllower.2012-06-07
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    Nothing to thank. Also, I encountered much difficulty as well when studying this part before. My way to hold control of it is to construct, by virtue of its linearity and universal paoperties, step by step, tensor products from the most simple ones: $M \otimes_A A \cong M$, and, for I focused on simple algebras, then use the structure theorems...^^.Hope this helps you.2012-06-07
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    @ClarkKent I understand, but do not get too frustrated if you find yourself confused by the construction! Most algebraists are also confused by it, so once they know it exists the actual construction no longer becomes interesting. The idea is to characterise the module $M \otimes_R N$ by a universal property (http://en.wikipedia.org/wiki/Tensor_product_of_modules#Definition) which is far more useful in practice.2012-06-07
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    @ClarkKent Tensor products are 1 million percent confusing when you first start learning!2012-06-07
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    The result I mentioned, plus (for when you learn about localisation) the following: $$S^{-1} M \cong M \otimes_R S^{-1} R$$ are extremely powerful for solving nice examples like this. I only started to get the hang of tensor products recently so if you look through my question history you can see more examples :)2012-06-07
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    @PaulSlevin Perhaps you should say that your isomorphism comes from tensoring the exact sequence of $R$ - modules $0 \rightarrow I \rightarrow R \rightarrow R/I \rightarrow 0$ with $M$....2012-06-07
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    @BenjaminLim I linked to the proof:) I'll put it in my answer though.2012-06-07
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    @PaulSlevin whoops :D2012-06-07

2 Answers 2

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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 $0$.

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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.