# Help understanding tensoring of exact sequences

I'm reading the proof of the following fact:

Let A be a ring, a an ideal of A, M an A-module. Then $(A/a) \otimes_{A} M$ is isomorphic to M/aM.

So the solution is to tensor the exact sequence $0 \rightarrow a \rightarrow A \rightarrow A/a \rightarrow 0$ with M.

This induces an exact sequence:

$a \otimes_{A} M \rightarrow A \otimes_{A} M \rightarrow (A/a) \otimes_{A} M \rightarrow 0$

One question:

Why is $a \otimes_{A} M \cong aM$ ? Why is this "clear and obvious" i.e how do we see it quickly?

I just started reading about the tensor product of modules and have trouble seeing why such isomorphisms are natural, I'm sure there's a "quick" way to see them, so why this is clear? is there any other way to think about the tensor product besides been the quotient of a free module by a set of generators?

Thanks

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Your induced sequence is not exact on the left. – curious Mar 1 '11 at 2:59
So I am not sure you want $a\otimes M \cong aM$ – curious Mar 1 '11 at 3:13
Rather, the image of the map from $a\otimes M$ to $M$ is $aM$. – curious Mar 1 '11 at 3:17
Just to reiteratre the comment of @curious: curious is correct. The sequence you get after tensoring up is typically not exact on the left, and it is not true in general that $a \otimes M \cong a M.$ What is true, as Soarer explains in an answer below, is that $a\otimes M$ maps onto $a M$ under the natural map. (Think about the case $A =\mathbb Z$, $a = 2 \mathbb Z$, and $M = \mathbb Z/2 \mathbb Z$ to see how exactness on the left can fail, and (hence) why you don't always get an isomorphism.) – Matt E Mar 1 '11 at 3:47
BTW, as to your last question: yes, the other way to think about the tensor product is via its defining universal mapping property. This may seem very pie-in-the-sky at first, but with practice it gives you a good handle on things. In particular, it forces you to think about the functorial properties of your homomorphisms, which is a healthy practice. – Pete L. Clark Mar 1 '11 at 4:04

Since $a \otimes_{A} M \rightarrow A \otimes_{A} M \rightarrow (A/a) \otimes_{A} M \rightarrow 0$ is exact,
the second function in the sequence $g:A \otimes_{A} M \rightarrow (A/a) \otimes_{A} M$ is surjective. Moreover, the image of $f:a \otimes_{A} M \rightarrow A \otimes_{A} M$, that is $f(a \otimes_{A} M)$, is the kernel of $g$.
By the first theorem of isomorphism for modules,

$A \otimes_{A} M/f(a \otimes_{A} M) \cong (A/a) \otimes_{A} M$ (1)

Now, if $h: S \rightarrow T$ is an isomorphism of modules, and $S'$ is a submodule of $S$, we have $S/S'\cong T/h(S')$ (2)
which can be proved by applying again the theorem of isomorphism to the projection function $S\rightarrow T/h(S')$.

Since $h:A \otimes_{A} M\rightarrow M$ , defined by $h(a\otimes_{A} m)=am$ is an isomorphism and $h(f(a \otimes_{A} M))=aM$, it follows by (2) that

$A \otimes_{A} M/f(a \otimes_{A} M) \cong M/aM$ (3)

Finally, (1) and (3) prove your requirement.

Another proof can be given using the universal property of the tensor product mentioned above.

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$a \otimes_A M \to A \otimes_A M$ may NOT be injective. – Soarer Mar 1 '11 at 5:38
Thanks. I modified it. – Theta33 Mar 1 '11 at 6:27
thanks! – user6495 Mar 3 '11 at 19:18

The isomorphism $A \otimes_A M \to M$ is by sending $a \otimes m \to am$. Under this isomorphism, the image of $\mathfrak{a} \otimes M$ would be $\mathfrak{a}M$.

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That the natural map $a \otimes_A M \rightarrow aM$ be an isomorphism is very far from being "clear and obvious". As others have noted, it is not generally true. But even more, it may be viewed as the entire obstruction to the functor "tensoring with $M$" being exact. In other words, if for all ideals $a$ of $A$ (in fact it suffices even to restrict to finitely generated ideals) the map $a \otimes_A M \rightarrow aM$ is an isomorphism, then the module $M$ is flat and tensoring any injection with $M$ gives an injection. See for instance $\S 3.11$ of my commutative algebra notes, where I state and prove this (well-known) result, which I call the Tensorial Criterion For Flatness.

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thanks, I still don't see why the conclusion of the problem follows, namely why $M/aM \cong (A/a) \otimes_{A} M$? sorry I'm new with this stuff. The solution is based on the hint in Atiyah's book. – user6495 Mar 1 '11 at 4:35

As mentioned by curious, the induced sequence is in general not exact on the left, though it is exact at the other two places. But as both curious and Soarer point out, the image of the map on the left is $aM$ under the isomorphism $A\otimes_A M \cong M$, and then exactness at the other two places gives you the isomorphism that you want.

is there any other way to think about the tensor product besides been the quotient of a free module by a set of generators?

Yes! In fact, there is a much more useful, if more abstract, way to think of tensor products. Let $A$ be a commutative ring, and let $M,N,Z$ be $A$-modules. (Analogous statements hold over noncommutative rings, but the formulation is not quite as clean.) An $A$-bilinear map $b:M\times N\to Z$ is a map of sets such that for each $m\in M$ and each $n\in N$, the maps $b(m,-):N\to Z$ and $b(-,n):M\to Z$ are $A$-linear.

Bilinear maps appear all over the place; you can think of them as "generalized multiplications" -- for any $A$-algebra $R$, its multiplication $R\times R\to R$ is $A$-bilinear. Note that a bilinear map $M\times N\to Z$ is not $A$-linear with respect to the usual $A$-module structure on $M\times N$. That's too bad, because it's always nice to have things be linear. To "correct" this, the tensor product $M\otimes_A N$ is exactly the $A$-module such that bilinear maps $M\times N\to Z$ are the same thing as linear maps $M\otimes_A N\to Z$.

More precisely, $M\otimes_A N$ is characterized by the following universal property: if $Q$ is an $A$-module and $b:M\times N\to Q$ is an $A$-bilinear map, then there is a unique $A$-linear map $\tilde b:M\otimes_A N\to Q$ such that $\tilde b(m\otimes n) = b(m,n)$ for all $(m,n)\in M\times N$.

So, for example, if $R$ is an $A$-algebra, then the multiplication of $R$ is an $A$-linear map $R\otimes_A R\to R$. Indeed, this lets you phrase the axioms of an associative algebra completely in terms of linear maps, which is a useful thing to do if, for example, you want to talk about "algebra objects" in categories other than the category of $A$-modules.

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