4
$\begingroup$

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.

  • 2
    Yes. http://mathoverflow.net/questions/62464/rings-with-all-modules-projective2011-12-22
  • 0
    A key search term is Global Dimension. http://en.wikipedia.org/wiki/Global_dimension2011-12-22
  • 0
    But the product of fields is not necessarily a field, is it? Does specifying that $R$ be integral make it the trivial product?2011-12-22
  • 0
    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.2011-12-22
  • 1
    Right, the product of fields is never an integral domain, so the only integral domains with your property are fields.2011-12-22

1 Answers 1

8

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.

  • 0
    :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 ?2017-07-06
  • 0
    @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 $ef=0$, $e\ne0\ne f$.2017-07-06
  • 0
    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 $R$ itself2017-07-06
  • 0
    @users - Also note that any correct argument must use the assumption that $R$ is a domain.2017-07-06
  • 1
    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 field2017-07-06