If $P$ is projective and $A,B$ are direct summands of $P$ then $A\cap B$ is a direct summand of $P$

Question

I need to prove that if $P$ is a projective module over $\mathbb Z$ and $A,B$ are direct summands of $P$ then $A\cap B$ is a direct summand of $P$.

This in turn would imply if $A$ and $B$ are projective then $A\cap B$ is projective. (Since $A$ and $B$ would be direct summands of $A+B$ which is clearly also projective.

score 2 · Accepted Answer · answered Nov 01 '15 at 22:40

2

We have an exact sequence $0\to P/A\cap B\stackrel{f}\to P/A\oplus P/B$. (Define $f(x)=(x,-x)$.) Then $P/A\cap B$ is a submodule of a projective module which over $\mathbb Z$ is free, so $P/A\cap B$ is also free.

answered Nov 01 '15 at 22:40

user26857

53,190

Thanks, we had started to see this in with the TA but I forgot. Just out of curiosity, did you take this out of some book or notes or something? – Asinomás Nov 01 '15 at 22:41
The short exact sequence $0\to P/A\cap B\to P/A\oplus P/B\to P/(A+B)\to 0$ is well known. – user26857 Nov 01 '15 at 22:43
One question, how does the fact that $P/A\cap B$ is free imply $A\cap B$ is a direct summand of $P$? – Asinomás Nov 02 '15 at 02:17
Free $\implies$ projective and use the short exact sequence $0\to A\cap B\to P\to P/A\cap B\to 0$. – user26857 Nov 02 '15 at 02:30
Ok, I know every summand of a projective module is projective, but does the converse hold? – Asinomás Nov 02 '15 at 13:05
I've thought that this property is well known. – user26857 Nov 02 '15 at 15:02
Why is a submodule of a projective module over $\mathbb Z$ free? – Asinomás Nov 02 '15 at 20:22
@dREaM See http://math.stackexchange.com/questions/162945/submodule-of-free-module-over-a-p-i-d-is-free-even-when-the-module-is-not-finit. So, over $\mathbb Z$ (or over a PID) any submodule of a free (=projective) module is free. – user26857 Nov 02 '15 at 20:53

If $P$ is projective and $A,B$ are direct summands of $P$ then $A\cap B$ is a direct summand of $P$

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