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A standard fact in Commutative Algebra is that a Projective $A$-module is flat.

The converse is false.

Can someone show me an example of a Flat Non Projective $A$-Module?

Thank you!

user26857
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Sabino Di Trani
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  • @user26857 thanks, done. – quid Jul 12 '15 at 22:00
  • Proof that $\mathbb{Q}$ is flat $\mathbb{Z}$-module: See Problem 4 (part b) in this link. – Prism Jul 13 '15 at 00:42
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    Joseph, here is another example of flat module that is not projective. Take $A=\mathbb{Z}$ and $M=\mathbb{Z}_{(2)}$ (localization at the prime ideal $(2)$). Then $M$ is a flat $A$-module, but it is not projective. See the last page (problem 14) of this document for the details. – Prism Jul 13 '15 at 19:52

2 Answers2

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An elementary proof that $\mathbf Q$ is not projective over $\mathbf Z$: if $\mathbf Q$ were projective, it would be a direct summand of a free $\mathbf Z$-module $L$, hence there would be an injective homomorphism from $\mathbf Q$ into $L$.

However the only homorphism from $\mathbf Q$ into a free module is the null homomorphism: indeed, for any $n$, and any homomorphism $f\colon \mathbf Q\to L$, we have $$f(1)=2^nf\Bigl(\dfrac1{2^n}\Bigr)\in 2^n L,\quad\text{hence}\quad f(1)\in\bigcap_{n\ge 0}2^n L=0.$$ Since $f(1)=0$, it is easy to deduce $f\Bigl(\dfrac ab\Bigr)=0\;$ for any $\;\dfrac ab\in\mathbf Q$.

user26857
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Bernard
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  • This looks like a nice proof. Could you justify why $\bigcap_{n\geq 0} 2^{n}L=0$? If $x\in 2^{n}L$ for each $n\geq 1$, this means $x=2^{n} a_{n}$ for some $a_n\in L$. How do we get a contradiction if $x\neq 0$? – Prism Jul 12 '15 at 23:45
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    This is just because on each component of $L\simeq \mathbf Z^{(I)}$ the $i$-th coordinate of $x$ lies in $\bigcap\limits_n 2^n\mathbf Z$, and this one is $0$. – Bernard Jul 13 '15 at 00:04
  • Thanks! Makes sense. +1. Also, we could have written $n$ instead of $2^{n}$ throughout the solution, right? – Prism Jul 13 '15 at 00:09
  • Yes, we could, but I think it would have been more complicated, as the ideals $n\mathbf Z$ are not linearly ordered. – Bernard Jul 13 '15 at 00:28
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The rationals as a module over the integers is flat but not projective.

quid
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