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Let $D$ be an integral domain which is not a field and $Q=\text{Frac}(D)$ the field of fractions of $D$. Then $Q$ as a $D$-module has not a projective cover.

By Corollary 5.35 of Rotman's Homological Algebra we got that : If $D$ is an integral domain and $Q=\text{Frac}(D)$, then $Q$ is a flat $D$-module.

So we got it, $Q$ is a flat $D$-module, now my idea is to use Bass Theorem to kill this one, this theorem states (among other things) that $M_{R}$ has projective cover ($M_{R}$ is perfect) iff every $M_{R}$ flat module is projective.

So I reduced (or complicated :S) the problem to prove that the flat module $Q_{D}$ is not projective, which is the part I cannot prove. I tried to study the proof of $\mathbb{Q}_{\mathbb{Z}}$ is a flat module that is not projective but they use the fact $\mathbb{Q}_{\mathbb{Z}}$ is finitely generated and that $\mathbb{Z}$ is DIP but in my case I dont know if $Q_{D}$ is finitely generated, also $D$ is not be DIP. Any help in order to prove this problem in the direction I propose or any other will be apreciated. Thanks!

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2 Answers 2

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As a hint, note that $Q=\text{Frac}(D)$ is also injective as an $R$-module. Therefore to show that it is not projective you can show that there are no non-zero projective-injective modules over $D$.

There is also a solution to this problem on this site, which can be found here. This is actually exercise 3.18 in Rotman's Intro to Homological Algebra, 2nd edition, so you might want to try and prove it using his hints.

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  • $\begingroup$ Thanks a lot @zeek! Already tried the last problem I asked about semi-perfect cover with the definition of indecomposable in Kasch book that Rickard mentioned in the comments. Let me write that on the original post. $\endgroup$
    – Cos
    Apr 30, 2020 at 18:59
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As a non zero projective module has maximal submodules, if K had a projective cover, then K would have a maximal submodule (The maximal submodules of a projective cover of K are in bijective correspondence with the maximal submodules of K). Then it would be a divisible simple module, which can not exist.

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  • $\begingroup$ Thanks! But I got confused following you. Lets say $K=Frac(D)$ and suppose that $K$ has a proyective cover and also supposing $K$ is projective :/ ? Lets say $P$ then $P$ have a maximal submodule which is in correpondence with a maximal submodule of $K$. I know already that $K$ is divisible but why it would also be simple? @Hugo $\endgroup$
    – Cos
    May 15, 2020 at 18:44

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