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Let $R$ be an integral domain with the property that all modules over $R$ are projective. Does it follow that $R$ is a field? Obviously the converse is true.

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    $\begingroup$ Yes. mathoverflow.net/questions/62464/… $\endgroup$ – Qiaochu Yuan Dec 22 '11 at 19:18
  • $\begingroup$ A key search term is Global Dimension. en.wikipedia.org/wiki/Global_dimension $\endgroup$ – jspecter Dec 22 '11 at 19:22
  • $\begingroup$ But the product of fields is not necessarily a field, is it? Does specifying that $R$ be integral make it the trivial product? $\endgroup$ – Lepidopterist Dec 22 '11 at 19:22
  • $\begingroup$ If this is true I would like to know a little bit about how technical the proof is. I have tried proving this using basic facts about projective modules but can't seem to do it. $\endgroup$ – Lepidopterist Dec 22 '11 at 19:24
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    $\begingroup$ Right, the product of fields is never an integral domain, so the only integral domains with your property are fields. $\endgroup$ – Thomas Andrews Dec 22 '11 at 19:32
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If $R$ is not a field, it has a nonzero proper ideal $I$, and $R/I$ is not projective, because it is a nonzero torsion module.

Variation: The canonical projection $R\to R/I$ doesn't split.

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  • $\begingroup$ :how to see easily that the canonical projection doesn't split ? If it splits then there is a module homomorphism $h:R/I \to R$ such that $h(1+I)-1 \in I$ , but then what ? $\endgroup$ – user Jul 6 '17 at 15:54
  • $\begingroup$ @users - If $R\to R/I$ split, there would be an ideal $J$ such that $R=I\oplus J$ (direct sum of $R$-modules). Let $e,f\in\text{End}_R(R)$ be the associated projections. View $e$ and $f$ as elements of $R$ (note $\text{End}_R(R)\simeq R$). Then we would have $ef=0$, $e\ne0\ne f$. $\endgroup$ – Pierre-Yves Gaillard Jul 6 '17 at 16:14
  • $\begingroup$ ah thanks . I have also been able to get it like this : As I said , we get a module morphism $h:R/I \to R$ such that $\pi \circ h = 1_{R/I}$ , so $h(1+I)+I=1+I$ , so $h(1+I)=1+i$ for some $i \in I$ , so $h(r+I)=r(1+i) , \forall r \in R$ . Now as $I$ is non-zero , so $\pi : R \to R/I $ is not injective , so $h \circ \pi$ is not injective , so for some non-zero $s \in R$ , $0=h \circ \pi (s)=h(s+I)=s(1+i)$ , and then since $R$ is a domain , so $i=-1 \in I$ , hence $I=R$ . Thus the only non-zero ideal of $R$ is $R$ itself $\endgroup$ – user Jul 6 '17 at 16:31
  • $\begingroup$ @users - Also note that any correct argument must use the assumption that $R$ is a domain. $\endgroup$ – Pierre-Yves Gaillard Jul 6 '17 at 17:02
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    $\begingroup$ yes I can see that ... cause in general the ring would be semisimple artinian and then we need the domain property to conclude it is a field $\endgroup$ – user Jul 6 '17 at 17:10

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