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Let $0\to M_1\xrightarrow{f} M_2\to M_3\to 0$ be an exact sequence of $R$-modules. Show that if $N$ is a free $R$-module, then $-\otimes_R N$ is exact.

This amounts to proving that $f\otimes id:M_1\otimes_R N\to M_2\otimes_R N$ is injective.

Let $\{e_i\}_{i\in I}$ be a $R$-basis for $N$. Let $x\in {\rm Ker}(f\otimes id)$, we write $x=\sum x_i \otimes e_i$. Then $(f\otimes id)(x)=\sum f(x_i)\otimes e_i$. I want to conclude that $f(x_i)=0$ for all $i$.

I am following a course on modules and categories, but I still don't feel too comfortable with modules.

We can write $N\cong R^{(I)}$. I tried using the following isomorphisms

$M\otimes_R R^{(I)}\cong (M\otimes_R R)^{(I)}\cong M^{(I)}$ where $x\otimes (r_i)_{i\in I}\mapsto (x\otimes r_i)_{i\in I}\mapsto (xr_i)_{i\in I}$.

So $\Sigma f(x_i)\otimes e_i\mapsto \left(\Sigma f(x_i)\right) \otimes e_i\mapsto \Sigma f(x_i)e_i=0$, so $f(x_i)=0$.

I am not sure about this last line in particular. Could someone point me in the right direction...?

Dr. Heinz Doofenshmirtz
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1 Answers1

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Basically everything you say is correct. I will try to phrase in different termes to make it more clear.

You are using two things in your attempted solution: First, you use that a general element of $M_1 \otimes_R N$ can be written in the form $\sum x_i \otimes e_i$ and second that if an element $\sum y_i \otimes e_i$ vanishes in $M_2 \otimes_R N$, then all the coefficients $y_i$ vanish (you want to apply this with $y_i = f(x_i)$).

Both of these follow from your observation that for every $R$-module $M$ the map $$M^{(I)} \to M \otimes_R N, \qquad (x_i)_{i \in I} \mapsto \sum_{i \in I} x_i \otimes e_i$$ is a bijection (this is the inverse map to the chain of isomorphisms that you gave).