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Note: $(d)$ is a nonzero proper ideal of $R$.

My attempt:

Consider $\pi:R\to R/(d)$, then $\ker(\pi)=(d)$. I want to prove that there is no $\sigma:R/(d)\to R$ such that $\pi\circ \sigma=id_M$.

Suppose that such a $\sigma$ exists, then $R\cong (d) \oplus R/(d)$. I'm not entirely sure if this is a contradiction. I know that the ideals of $R/(d)$ are the (principal) ideals of $R$ that contain $(d)$, but other than that I don't really see where I could use the fact that $R$ is a PID.

Thanks.

MyWorld
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1 Answers1

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Starting from where you ended, $R\cong (d)\oplus R/(d) $ you could refine it to say that actually $R=(d)\oplus N$ (an internal direct sum). Since a domain has no proper summands, $(d)=\{0\}$ or $(d)=R$.

Both cases are just fine, since both $\{0\}$ and $R$ are projective, but presumably you wanted a nonzero, proper ideal $(d)$.

In fact, the same reasoning shows you $R/I$ is not projective for any nontrivial ideal $I$ in a domain, or even more generally, any uniform ring.

Another way:

A projective module over a PID is free, and a nonzero free module is faithful (i.e. has trivial annihilator) but $M$ is clearly annihilated by $d$. (Again, I exclude $(d)$ equal to $\{0\}$ or $R$, which I guess you omitted unintentionally.)

rschwieb
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  • Thank you so much! The question does not state what $(d)$ can't be, but I also assumed that it had to be a nonzero, proper ideal; I'll edit the question for future readers. I am a little confused with the first paragraph though: 1) how did you go from a quotient to a module N (?), are you using the fact that $R/I$ ($I\trianglelefteq R$) is an $R$-module? 2) why can't a domain have proper summands? – MyWorld Jan 08 '20 at 17:31
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    @Zachary When the first map is the inclusion map, the splitting lemma says that when $0\to A\to B\to C\to 0$ splits, $A$ is actually an internal direct summand of $B$. What you said is true, of course, that $R/(d)$ is an (external) direct sum of $(d)$ and $R/(d)$, but I wanted to sharpen that to talk about an actual internal summand of $R$. A domain obviously can't be expressed as $A\oplus B=R$ for nonzero ideals $A,B$. How could $A\cap B={0}$ when $AB\neq{0}$? – rschwieb Jan 08 '20 at 17:34
  • Ok, I get it now! Thank you. – MyWorld Jan 08 '20 at 17:46
  • @Zachary No problem! – rschwieb Jan 08 '20 at 17:49