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I was reading this Lang's book where he says

F3: For every injection $0\rightarrow E'\rightarrow E$, we have $0\rightarrow E'\otimes F\rightarrow E\otimes F$

F1: For every exact sequence $E'\rightarrow E\rightarrow E''$, we have an exact sequence $E'\otimes F\rightarrow E\otimes F\rightarrow E''\otimes F$.

How do I show from F3 to F1? Lang says consider the kernel and image of the mapping $E'\rightarrow E$, but I couldn't figure out why. Thanks!

enoughsaid05
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    Tensor product is right exact, so F3 says that tensoring with flat modules preserves short exact sequences. Can you split up a 3-term exact sequence into some short exact sequences so that kernel and image can be analyzed? –  Nov 06 '12 at 05:46

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After spending some time, this is the best I can come up with: suppose we have $\phi:E'\rightarrow E$ and $\psi:E\rightarrow E''$, with $E'\rightarrow E\rightarrow E''$ exact. Then,

\begin{eqnarray} 0\rightarrow \ker\phi \rightarrow E'\rightarrow E\rightarrow \mbox{coker}\phi\rightarrow 0 \end{eqnarray}

So using your right-exactness of tensor and F3, I have

\begin{eqnarray} 0\rightarrow \ker\phi\otimes F \rightarrow E'\otimes F\rightarrow E\otimes F\rightarrow \mbox{coker}\phi\otimes F\rightarrow 0 \end{eqnarray}

This means that

\begin{eqnarray} E'\otimes F\rightarrow E\otimes F\rightarrow \mbox{coker}\phi\otimes F \end{eqnarray}

us exact, but $\mbox{coker}\phi\cong E''$, so I guess this should solve the problem? Or am I missing something?

enoughsaid05
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  • You have to say that $coker\phi\otimes F\to E''\otimes F$ is injective because you what the kernel of $E\otimes F\to E''\otimes F$ to be isomorphic to the image of $E'\otimes F$. –  Nov 06 '12 at 14:03
  • Why is $coker \phi \cong E’’$? – dahemar Jan 05 '23 at 18:27