Proving that the tensor product is right exact

Solution 1:

The proof mentioned by Frederik and Loronegro is great because it provides a first example of how it can be useful to know that two functors are adjoint: left adjoints are right exact. However, you can also argue as follows. Let $D$ be the image of $\alpha \otimes \operatorname{id}$. You get an induced map $(B \otimes M)/D \to C \otimes M$. Let's try to define an inverse: if $(c, m) \in C \times M$ then choose a $b \in B$ such that $\beta(b) = c$, and send $(c, m)$ to $b \otimes m \bmod D$. You can check that this is well defined using the exactness of the original sequence.

Solution 2:

First of all, if you start with an exact sequence $A\to B\to C\to 0$ of left $R$-modules, then $M$ should be a right $R$-module, so that the tensor products $M\otimes A$, etc. are well defined.

Second, it happens that for the proof that I will explain, it is easier to consider the functor $M\otimes\underline{}$ which is applied to the exact sequence. Then we can use the isomorphism $M\otimes A\cong A\otimes M$ to prove the exactness of the sequence $A\otimes M\to B\otimes M\to C\otimes M\to 0$, in case that $A,B,C$ are right $R$-modules and $M$ is a left $R$-module.

$\DeclareMathOperator{\Hom}{Hom}$ I don't know a direct proof of the proposition and I think it may be difficult. The proof I know uses indeed the natural isomorphism mentioned by @Frederik (I think that in his comment there is a misorder of the modules involved). With the notation used by @Klaus, the natural isomorphism that is convenient is $\Hom(M\otimes A,Q)\cong \Hom(A,\Hom(M,Q))$, where $Q$ is an injective cogenerator right $R$-module (for example, the injective hull of the direct sum of a complete set of non-isomorphic simple modules). We can consider the functor $(\underline{})^*=\Hom(\underline{},Q)$, so the latter natural isomorphism can be stated as $(M\otimes A)^*\cong \Hom(A,M^*)$. This functor $(\underline{})^*$, which is contravariant, so that it reverses the direction of morphisms, has the following property:

For $R$-modules $K,N,L$, the sequence $K\to M\to N\to 0$ is exact if, and only if, the sequence $0\to N^*\to M^*\to K^*$ is exact.

Therefore, the sequence $M\otimes A\to M\otimes B\to M\otimes C\to 0$ is exact if, and only if, $0\to (M\otimes C)^*\to (M\otimes B)^*\to (M\otimes A)^*$ is exact, if and only if, $0\to \Hom(C,M^*)\to \Hom(B,M^*)\to \Hom(A,M^*)$ is exact.

But the contravariant functor $\Hom(\underline{},M^*)$ is left exact, that is, if the sequence $A\to B\to C\to 0$ is exact, then the sequence $0\to \Hom(C,M^*)\to \Hom(B,M^*)\to \Hom(A,M^*)$ is exact, and this is very much easier to prove directly, rather than the right exactness of the functor $M\otimes\underline{}$, which @Klaus was trying.

Solution 3:

Here is a direct proof of exactness at $B \otimes_R N$. Summary: it's very annoying to prove things in detail using the generators-and-relations definition of tensor products. Even in this long-winded version, significant detail is omitted. I would be curious to know if anyone has typed something like this into Coq.

This is usually treated using the universal property of the tensor product, as in the other answers, or in Dummit and Foote, or left as an exercise. In my opinion, leaving this as an exercise is mostly about not wanting to write this much detail. I have written this out because I can't find another source that does so.

First we state two clarifying lemmas:

Lemma 1 If $f \colon S \rightarrow T$ is a map of sets, then the kernel of the induced map of free $R$-modules $f \colon R[S] \rightarrow R[T]$ is generated by $\{s-s' \in R[S] \mid f(s) = f(s')\}$.

Lemma 2 If $$X \xrightarrow{f} Y \xrightarrow{g} Z \rightarrow 0$$ are maps of $R$-modules and $\mathrm{Ker}(f)$ is generated by $\{x_i\}$, $\mathrm{Ker} (g)$ is generated by elements $\{y_j\}$, and we have $x'_j \in X$ such that $f(x'_j) = y_j$, then $\mathrm{Ker} (gf)$ is generated by $\{x_i\} \cup \{x'_j\}$.

To prove that $\mathrm{Ker}(\beta \otimes \mathrm{Id}) \subset \mathrm{Im}(\alpha \otimes \mathrm{Id})$ we work with the definition of the tensor products as quotients of the free modules $R[B \times M]$ and $R[C \times M]$. If $(\beta \otimes \mathrm{Id} )(\sum m_i \otimes n_i) = 0 \in C \otimes_R M$, then the composition $$R[B \times M] \xrightarrow{ \beta \times \mathrm{Id}} R[C \times M] \xrightarrow{\pi} C \otimes_R M$$ sends $\sum (m_i ,n_i)$ to zero, where $\pi$ is the quotient map defining the tensor product.

Our goal is to show that any such element $\sum (m_i ,n_i)$ can be expressed ($\star$) as a sum $\sum_j (\alpha(a_j),m_j)$ plus a linear combination of tensor-product relation elements in $R[B \times M]$ (there are 4 types). This is equivalent to showing that $\sum m_i \otimes n_i$ is in the image of $\alpha \otimes \mathrm{Id}$.

That is to say, we want to show that the kernel of the composition $\pi \circ (\beta \otimes \mathrm{Id})$ is generated by 5 types of elements. Lemma 1 tells us that the kernel of $\beta \times \mathrm{Id}$ is generated ($\triangle$) by $\{(b,m)-(b',m) \in R[B \times M] \mid \beta(b) = \beta(b')\}$. The kernel of $\pi$ is generated by tensor-product relation elements by definition. Further, any tensor-product relation element in $R[C \times M]$ is the image of a tensor-product relation element in $R[B \times M]$, because $ \beta $ is onto.

We can write the annoying expression $$ (b,m)-(b',m) = (b-b', m) - [(b+(-b'),m)-(b,m)-(-b',m)]+[(-1)(b',m)-(-b',m)] $$ Since $\beta(b-b') = 0$, $b-b' = \alpha(a)$, so the first term on the RHS is an element in the image of $\alpha \times \mathrm{Id}$ and the other two terms are tensor product relations terms for $B \otimes_R M$.

By Lemma 2, we know a set of generators of $\mathrm{Ker}(\pi \circ (\beta \otimes \mathrm{Id}))$. We want to show that all of these can be expressed as described above ($\star$). The generators arising from tensor-product relation elements of $R[C \times M]$ are tensor-product relation elements in $R[B \times M]$, so there is nothing to prove. The other type of generator $\triangle$ is also expressible in the form $\star$, because of the annoying expression above.