# Corollary 2.13 of Atiyah - Macdonald

I just started learning about tensor products and I have some trouble understanding this corollary in Atiyah - Macdonald. All modules are assumed to be $A$ - modules for $A$ a commutative ring.

Corollary 2.13. Let $x_i \in M, y_i \in N$ such that $\sum x_i \otimes y_i = 0$ in $M \otimes N$. Then there exist finitely generated modules $M_0$ and $N_0$ of $M,N$ respectively such that $\sum x_i \otimes y_i = 0$ in $M_0 \otimes N_0$.

Now I know that if $\sum x_i \otimes y_i = 0$ in the tensor product of $M$ and $N$, then this means that $\sum (x_i,y_i)$ is in the submodule $D$ generated by elements of the form

$$\begin{eqnarray*} &(x_1 + x_2,y) - (x_1,y) - (x_2,y)&\\ &(x,y_1 + y_2) - (x,y_1) - (x,y_2)& \\ &(ax,y) - a(x,y)&\\ &(x,ay) - a(x,y)\\ \end{eqnarray*}$$

where $x_i \in N$, $y_i \in N$ and $a \in A$. It follows that $\sum (x_i,y_i)$ is a finite linear combination of elements of $D$. Now in the proof in AM, it says

"Let $M_0$ be the submodule of $M$ generated by the $x_i$ and all elements of $M$ which occur as first coordinates in these generators of $D$, and define $N_0$ similarly."

I don't understand what they mean in the definition of $M_0$ above. Don't all the $x_i$'s already appear as first coordinates in the generators of $D$? Besides why can't we just define $M_0$ to be generated by the $x_i$, $N_0$ generated by the $y_i$ ? Also I am not sure about how one defines $M_0 \otimes N_0$. Am I right to say that $$M_0 \otimes N_0 = F(M_0 \times N_0)/D$$

where $D$ is the submodule generated by the same relations above and $F(M_0 \times N_0)$ the free module on the set $M_0 \times N_0$?

Thanks.

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## 1 Answer

$\sum(x_i,y_i)$ can be written as a finite linear combination of certain elements of $D$. Let $M_0$ be the submodule of $M$ generated by the first coordinates of the elements appearing in that finite linear combination and similarly for $N_0$.

Remark. One has to do this, for otherwise the result is not true. For example, let $R=\mathbb Z$, $M=\mathbb Z$ and $N=\mathbb Z/2\mathbb Z$, and let $x=2\in M$ and $y=1\in N$. Then $x\otimes y=0$ in $M\otimes N$, yet the element $x\otimes y$ is not zero in $M_0\otimes N_0$ with $M_0=(x)\subseteq M$ the submodule of $M$ generated by $x$ and $N_0=(y)\subseteq N$ the submodule of $N$ generated by $y$ (which happens to be equal to $N$, of course).

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Thanks for clarifying. However how does $\sum x_i \otimes y_i = 0$ in $M_0 \otimes N_0$? Perhaps I don't know well what is $M_0 \otimes N_0$. –  user38268 Mar 20 '12 at 5:19
Indeed, you should read up a bit again on how to construct the tensor product, probably :) –  Mariano Suárez-Alvarez Mar 20 '12 at 5:23
The idea in the construction of $M_0 \otimes N_0$ is that you include enough that that same sum is an element of our tensor product, and also enough that it still dies (that's what the generators from D do). –  user23214 Mar 20 '12 at 5:26
@MarianoSuárez-Alvarez Am I right in saying if we write $(2,1) = (7,1) - (2,1) - (3,1)$ and let $M_0 = (2,3,7)$, $N_0 = N$ then I can write $2 \otimes 1 = 2(1 \otimes 1)$ because $1 = \operatorname{gcd}(2,3,7)$. How can I show this works in general? –  user38268 Mar 20 '12 at 5:40