# When does a finitely generated projective module surjects onto the ring

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.

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 $$I\to R$$.
• Why is there no $R$-linear surjection $I\to R$ ? – uno May 1 '20 at 16:17
• 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 $1$... – uno May 1 '20 at 16:28
• A homomorphism $I\to R$ must be given by $x\mapsto ax$ for some $a$ in the fraction field of $R$. @uno – Angina Seng May 1 '20 at 16:33