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Related earlier post:

Description of the kernel of the tensor product of two linear maps

Let $A$ be a commutative, unital ring.

Let $M$ and $N$ be finitely presented $A$-modules: we have exact sequences

$$A^r\xrightarrow{\phi} A^m \xrightarrow{\alpha} M \to 0,$$

$$A^s\xrightarrow{\psi} A^n \xrightarrow{\beta} N \to 0.$$

These give rise to an exact sequence

$$A^m\otimes_A A^n \xrightarrow{\alpha\otimes\beta} M\otimes_A N \to 0.$$

Consider the $A$-linear map

$$(A^r\otimes_A A^n)\oplus(A^m\otimes_A A^s)\xrightarrow{\Phi}A^m\otimes_A A^n$$

given by

$$(u\otimes v, x\otimes y) \mapsto \phi(u)\otimes v+ x \otimes \psi(y).$$

I want to show that

$$\mathrm{im}(\Phi)=\ker(\alpha\otimes\beta).$$

The inclusion from left to right is trivial but I'm stuck with the reverse inclusion.

I am looking for the most hands-on/explicit approach possible.

Many thanks!

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  • 1
    $\begingroup$ Bourbaki, Algebra II, § 3.6 Proposition 6. $\endgroup$ – A.G Aug 28 '16 at 11:28
  • $\begingroup$ Many thanks! :) $\endgroup$ – user350031 Aug 28 '16 at 14:12
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We prove a more general result.

Suppose we have two exact sequences of $A$-modules:

$$(1) \hphantom{1em} P'\xrightarrow{u} P \xrightarrow{v} P'' \to 0,$$

$$(2) \hphantom{1em} Q'\xrightarrow{s} Q \xrightarrow{t} Q'' \to 0.$$

We prove that

$$\ker(v\otimes t) = \mathrm{im}(u\otimes 1_Q)+\mathrm{im}(1_P\otimes s).$$


The inclusion from right to left is clear.

Pick any $z \in \ker(v\otimes t).$ We want to find an $x \in P'\otimes Q$ such that

$$(*) \hphantom{1em} z-(u\otimes 1_Q)(x)\in\mathrm{im}(1_P\otimes s)$$

As $(2)$ is exact, so is the sequence $P\otimes(2)$ and hence

$$\mathrm{im}(1_P\otimes s)=\ker(1_P\otimes t).$$

The condition $(*)$ is therefore equivalent to

$$(1_P\otimes t)(z)=(1_P\otimes t)(u\otimes 1_Q)(x) = (u\otimes t)(x) = (u\otimes1_{Q''})(1_{P'}\otimes t)(x).$$

Since $v\otimes t=(v\otimes 1_{Q''}) (1_P\otimes t),$ we have that

$$z \in \ker(v\otimes t) \Rightarrow (1_P\otimes t)(z) \in \ker(v\otimes 1_{Q''}).$$

As $(1)$ is exact, so is the sequence $(1)\otimes Q''$ and hence

$$\ker(v\otimes 1_{Q''})=\mathrm{im}(u\otimes 1_{Q''}).$$ Therefore, we have

$$z \in \ker(v\otimes t) \Rightarrow (1_P \otimes t)(z)\in\mathrm{im}(u\otimes1_{Q''}).$$

As $(2)$ is exact, so is the sequence $P'\otimes (2)$ and hence $1_{P'}\otimes t$ is surjective and so

$$\exists x \in P'\otimes Q \, : \, (u\otimes1_{Q''})(1_{P'}\otimes t)(x) = (1_P\otimes t)(z).$$

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