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Let $A$ be a commutative ring with unity and $\mathfrak a$ be an ideal of $A$. Further let $M$ be an $A$-module. Show that $A/ \mathfrak a \otimes_A M$ $\cong$ $M/ \mathfrak a M$.

One possible way of doing this is as follows: Consider the bilinear map $ B: A/ \mathfrak a \times M \to M/\mathbb a M$ defined as $(a \bmod \mathfrak a ,m) \to am \bmod \mathfrak a M$. By universal property of tensor product this map induces a linear map $B$: $A/\mathfrak a \otimes_A M \to M/\mathfrak a M$ whose inverse map $ M/\mathfrak a M \to A/\mathfrak a \otimes_A M $ is $m \bmod \mathfrak a M\to (1 \bmod \mathfrak a) \otimes m $. Hence the induced map is an isomorphism. Here are two questions:

$1$. Is there any problem in my solution?

$2$. What are other ways of solving this problem?

Edit: One other way of proving this might be as follows: Consider the exact sequence $ 0 \to \mathfrak a \to A \to A/\mathfrak a \to 0 $. Now applying the functor $ (- \otimes M)$ gives a exact sequence $ \mathfrak a \otimes M \to A \otimes M \to A/ \mathfrak a \otimes M \to 0$. How to conclude from here?

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    $\begingroup$ Offhand, it looks like a correct solution and uses the approach I would use, namely the universality of the tensor product. $\endgroup$
    – Moya
    Sep 11, 2015 at 4:46
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    $\begingroup$ Careful: The tensor functor you are applying to that exact sequence will only yield a right exact sequence in general; $a \otimes M \to A \otimes M$ need not be injective. $\endgroup$
    – Max
    Sep 11, 2015 at 5:31

2 Answers 2

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The first attempt is fine. The inverse you describe is well defined and the maps are in fact mutually inverse.

Let's have a look at the second approach. (Edited for clarification.)

Tensoring the exact sequence $$0\to \mathfrak{a}\to A\to A/\mathfrak{a}\to 0$$ with $M$ yields the exact sequence $$\mathfrak{a}\otimes_A M\to A\otimes_A M\to A/\mathfrak{a}\otimes_A M\to 0.$$

After identifying $A\otimes_A M$ with $M$ the obvious way, the homomorphisms $\mathfrak{a}\otimes_A M\to M$ and $M\to A/\mathfrak{a}\otimes_A M$ are given as follows. The first, $\mathfrak{a}\otimes_A M\to M$, maps homogenous elements $f\otimes m$ to their product $fm$. It may not be injective, but its image is easily seen to be $\mathfrak{a}M\subset M$. The second map $M\to A/\mathfrak{a}\otimes_A M$ just maps an element $m$ to $1\otimes m$; that the sequence is exact just means that this map is surjective and its kernel is the image of the former map; hence it's kernel is $aM$.

Now, quite generally, if $\varphi\colon N\to N'$ is an $A$-module homomorphism, then the homomorphism $\overline{\varphi}\colon N/\ker{\varphi}\to N'$ mapping a residue class $[n]$ to $\varphi(n)$, is well defined and injective: we have $\varphi(n) = \varphi(m)$ if and only if $n - m\in \ker(\varphi)$. Note that here $\mathrm{im}(\varphi) = \mathrm{im}(\overline{\varphi})$.

Applying this to the map $M\to A/\mathfrak{a}\otimes_A M$, of which we know it is surjective and has kernel $\mathfrak{a}M$, we get the desired isomorphism $M/\mathfrak{a}M\to A/\mathfrak{a}\otimes_A M$. This map, in fact, is nothing but the inverse described in the question.

Arguing via right exactness of taking tensor products is just another way of getting that this map is well defined and bijective. Of course, there are more proofs, mostly because you can make things more sophisticated if you wish, but in their heart, eventually, they will all produce (at least) one of the maps and show that it's an isomorphism.

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  • $\begingroup$ Sorry..How do you get the map from $M \to A/ \mathbb a \otimes M$ ? $\endgroup$ Sep 11, 2015 at 14:43
  • $\begingroup$ It goes under the name homomorphism throrem. If you mod out the kernel of a homomorphisms, then on the quotient there is an induced injective homomorphism. Here $ aM $ is the image of $ a\otimes M $ in $ M $, and so it is the kernel of the map $ M\to A/a\otimes M $ by right exactness of $-\otimes M $. $\endgroup$
    – Ben
    Sep 11, 2015 at 16:15
  • $\begingroup$ I'm bit confused.Initally we'had a surjective homomorphism from $ A \otimes M \to A/a \otimes M$,then how do we get map from $M \to A/a \otimes M$ ? $\endgroup$ Sep 11, 2015 at 16:22
  • $\begingroup$ I'm far from a keyboard right now. I'll edit the answer and expand as soon as possible. $\endgroup$
    – Ben
    Sep 11, 2015 at 16:33
  • $\begingroup$ Dear Ben,now everything is very clear.Thanks for your time.Best Regards, $\endgroup$ Sep 15, 2015 at 8:27
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I offer a solution that is very similar to the one given by Ben, but I cite specific results used from Chapter 2 of Atiyah-Macdonald and give a slightly different presentation. I leave this here only because it may be useful to some readers who are struggling with this chapter, like I was when I first worked through it.

Problem:

Let $A$ be a commutative ring with unity, $\mathfrak{a}$ an ideal, $M$ an $A$-module. Show that $(A/\mathfrak{a}) \otimes_A M \cong M/\mathfrak{a}M.$

Solution:

Proceeding by way of the hint, consider the exact sequence $$0 \to \mathfrak{a} \xrightarrow{\pi} A \xrightarrow{\phi} A/\mathfrak{a} \to 0.$$

The map $\pi$ is the standard inclusion, $\pi(a) = a,$ for all $a \in \mathfrak{a}$. It is easy to see that $\pi$ is injective and $\mathrm{Im}(\pi) = \mathfrak{a}$. The map $\phi$ is the standard surjection, $\phi(a) = a + \mathfrak{a},$ for all $a \in A$. It is easy to see that $\mathrm{Ker}(\phi) = \mathfrak{a}$. Since $\mathrm{Im}(\pi) = \mathrm{Ker}(\phi)$, $\pi$ is injective, and $\phi$ is surjective, we are satisfied that this is in fact a short exact sequence.

Since each object in the short exact sequence can be seen as an $A$-module, we can tensor with the $A$-module $M$, which by $\textbf{Proposition 2.18}$ (page 28) induces a new exact sequence $$ \mathfrak{a} \otimes_A M \xrightarrow{\pi \, \otimes \, 1} A \otimes_A M \xrightarrow{\phi \, \otimes \, 1} A/\mathfrak{a} \otimes_A M \to 0.$$

Referring to $\textbf{Proposition 2.14, iv)}$ (page 26) we know we have an isomorphism \begin{align}\varphi &\colon A \otimes_A M \to M\\ \varphi &\colon a \otimes x \mapsto ax. \end{align}

Since our goal is to make a claim $M/\mathfrak{a}M$ is isomorphic to $A/\mathfrak{a} \otimes_A M$ it is a natural thought, with the first isomorphism theorem in mind, to try and use what we have so far to cook up a surjective map from $M$ to $A/\mathfrak{a} \otimes_A M$, and hope the kernel is $\mathfrak{a}M$. Thus the natural candidate is to consider $$(\phi \otimes 1) \circ \varphi^{-1} \colon M \to A/\mathfrak{a} \otimes_A M.$$

First see that by exactness of the tensored sequence $\phi \otimes 1$ must be a surjective morphism. Since $\varphi$ is an isomorphism, $\varphi^{-1}$ must also be a surjective morphism, so $(\phi \otimes 1) \circ \varphi^{-1}$ is a composition of surjective morphisms and thus is also a surjective morphism. But we also have

\begin{align*} \mathrm{Ker}\left((\phi \otimes 1) \circ \varphi^{-1}\right) &= \varphi \left( \mathrm{Ker}(\phi \otimes 1) \right) \tag{prove this if it is not obvious why}\\ &= \varphi(\mathfrak{a} \otimes_A M) \tag{$\mathrm{Im}(\pi \otimes 1) = \mathrm{Ker}(\phi \otimes 1)$}\\ &=\mathfrak{a}M \tag{definition of $\varphi$}. \end{align*}

Thus by the first isomorphism theorem for modules, we conclude that $(A/\mathfrak{a}) \otimes_A M \cong M/\mathfrak{a}M.$

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