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Let $R$ be an integral domain, $F$ be a flat $R$-module, and $A$ and $B$ are two $R$-submodules of $Q$, where $Q$ is the quotient field of $R$. How can we show that $F\otimes (A \cap B) = (F\otimes A) \cap (F\otimes B)$?

Thank you.

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If $A, B$ are submodules of $Q$ (it is not important that $Q$ is the fraction field of $R$) we have an inclusion $0\to Q/A\cap B\to Q/A\oplus Q/B$. Tensoring with $F$, using flatness, we have $0\to F\otimes Q/A\cap B\to F\otimes (Q/A\oplus Q/B)$ to be exact. Now, one easily checks that $F\otimes Q/A\cap B=F\otimes Q/F\otimes (A\cap B)$ and similarly, $F\otimes Q/A=F\otimes Q/F\otimes A$ and same for $B$. Rest should be clear.

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Consider the exact sequence $0\to A\cap B\to Q\to Q/A\oplus Q/B$. Tensor this sequence with $F$ to get an exact sequence due to flatness, your statement follows from this resulting exact sequence.

Here we are just assuming that $A$, $B$ are $R$ submodules of the $R$ module $Q$.

ADDED LATER: after tensoring with $F$, the rightmost term will be $Q/A\otimes F\oplus Q/B\otimes F$. Now consider the exact sequence $0\to A\to Q\to Q/A\to 0$, tensoring this sequence with $F$ yields $Q/A\otimes F\cong (Q\otimes F)/(A\otimes F)$. Similarly for $Q/B\otimes F$. Putting this together we have the following exact sequence: $$0\to A\cap B\otimes F\to Q\otimes F\to (Q\otimes F)/(A\otimes F)\oplus (Q\otimes F)/(B\otimes F),$$ which yields your result.

Note: To conclude it will be necessary that you keep track of the maps, so that you can find the kernel of the map on the right of the last exact sequence.

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    $\begingroup$ The above sequence is not exact on the right in general. $\endgroup$ – Mohan May 28 '16 at 20:29
  • $\begingroup$ oh yes, you are right, sorry that was in a hurry. I will make the change and add a few more details. $\endgroup$ – user114539 May 29 '16 at 4:52

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