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Let $M$ be a non-zero, finitely generated projective module over a commutative integral domain $R$.

Is it necessarily true that there is an $R$-linear surjection $M\to R$ ? If this is not true in general, what if we also assume $R$ is Noetherian ?

I know this is true if $R$ is local as projective modules over local rings are free.

Of-course this is not true if we drop the integral domain assumption on $R$, for example $M=\mathbb Z/2\times 0$ is a finitely generated projective module over the Noetherian ring $R=\mathbb Z/2\times\mathbb Z/2$, but definitely $M$ cannot surject onto $R$ for cardinality reason. But with $R$ an integral domain, I have no counterexample.

Please help.

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If $R$ is a Dedekind domain, and $I$ a non-principal ideal of $R$, then $I$ is a projective module, but there is no $R$-module epimorphism $I\to R$.

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  • $\begingroup$ Why is there no $R$-linear surjection $I\to R$ ? $\endgroup$
    – uno
    May 1, 2020 at 16:17
  • $\begingroup$ Is the reason as follows ? If $I$ surjects onto $R$, then since $R$ is free, so $I\cong R\oplus N$ for some projective $R$-module $N$ and $N\ne 0$ since $I$ is not principal , so then $N$ has some positive constant rank say $n\ge 1$, and then $I_P\cong R_P^{n+1}$ for every prime ideal $P$ of $R$, contradicting that an ideal can only have rank $1$... $\endgroup$
    – uno
    May 1, 2020 at 16:28
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    $\begingroup$ A homomorphism $I\to R$ must be given by $x\mapsto ax$ for some $a$ in the fraction field of $R$. @uno $\endgroup$
    – Angina Seng
    May 1, 2020 at 16:33

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