Let $k$ be a commutative ring and let $C$ be a flat module over $k$. Let $M$ be a module and let $A,B \subseteq M$ be two submodules. We get a pullback diagram:

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where $s, i, j, t$ are inclusions. If we tensor by $C$ we get the diagram: enter image description here However is this a pullback diagram? I cannot work out how to define the unique morphism.

Sorry about the size of the pictures.

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    $\begingroup$ Remember: $C$ flat means that $C \otimes -$ preserves all finite limits. And pullbacks are finite limits. $\endgroup$ – Martin Brandenburg Oct 24 '12 at 10:29

Yes. Consider the following conversion of pullback into kernel:

$0\to A\cap B\to A\oplus B\stackrel{(i,-j)}{\to} Im(i,-j)\to 0$

is exact iff $A\cap B$ is the pullback of $i$ and $j$ (it satisfies the same universal property). Since $C$ is flat the following sequence is also exact:

$0\to (A\cap B)\otimes C\to (A\otimes C)\oplus (B\otimes C)\stackrel{(i\otimes C,-j\otimes C)}{\to} Im(i,-j)\otimes C\to 0$

Hence by the same argument as above $(A\cap B)\otimes C$ is the pullback of the two given maps, hence $(A\cap B)\otimes C\cong (A\otimes C)\cap (B\otimes C)$.

  • $\begingroup$ Doesn't the square need to be a pushout for $(i,-j)$ to be the cokernel of $A \cap B \to A \oplus B$ ? By the way I am assuming that map is the one that is given by the product structure of $A \oplus B$ and the maps $s,t$ $\endgroup$ – Paul Slevin Oct 22 '12 at 16:02
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    $\begingroup$ You are right, $(i,-j)$ need not be surjective. Edit. $\endgroup$ – Julian Kuelshammer Oct 22 '12 at 16:11
  • $\begingroup$ OK I see what's going on. It's because the functor $- \otimes C$ is left exact $\endgroup$ – Paul Slevin Oct 22 '12 at 16:14
  • $\begingroup$ That's the crucial point, yes. The requirement that $C$ is flat has to be used at some point. $\endgroup$ – Julian Kuelshammer Oct 22 '12 at 16:14

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