It is known that simple modules are either projective or singular. Is it true that simple projective modules over (commutative) rings are injective ?
$\begingroup$
$\endgroup$
5
-
1$\begingroup$ No, cf. T.Y.Lam, Lectures on modules and rings, Springer GTM 189, Exercise I.3.3. $\endgroup$– Fred RohrerCommented Sep 28, 2013 at 20:00
-
$\begingroup$ what is a singular module? $\endgroup$– Mariano Suárez-ÁlvarezCommented Sep 28, 2013 at 20:18
-
$\begingroup$ @Mariano: An $R$-module $M$ is called singular if $R$ is an essential extension of $(0:_Rx)$ for every $x\in M$. $\endgroup$– Fred RohrerCommented Sep 28, 2013 at 20:22
-
$\begingroup$ @FredRohrer, there are two questions in the question and your comment anwers «No»... to which of the two? :-) $\endgroup$– Mariano Suárez-ÁlvarezCommented Sep 28, 2013 at 21:02
-
$\begingroup$ @Mariano: I see only one question. (Well, if we also look at the title then I actually see two questions that are the same.) $\endgroup$– Fred RohrerCommented Sep 28, 2013 at 21:06
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
3
If $P$ a projective $R$-module is simple, then it is a direct summand of $R$ itself (indeed, any nonzero map $R\to P$ is surjective and therefore splits) and is then isomorphic to a minimal ideal $I$ which is projective. $I$ is generated by an idempotent, which is central if the ring is commutative. It follows that $I$ is a direct factor of $R$ as a ring, and therefore $I$ is also injective.
answered Sep 28, 2013 at 21:06
-
5$\begingroup$ If the ring is not commutative this breaks down. For example, the ring of $2\times2$ upper triangular matrices over a field has a simple projective which is not injective. $\endgroup$ Commented Sep 28, 2013 at 21:20
-
$\begingroup$ +1 but I cannot see why $I$ is injective. $\endgroup$– user37834Commented Sep 30, 2013 at 6:11
-
1$\begingroup$ Your ring $R$ is of the form $I\times J$. Since $I$ is a minimal ideal, it is a simple commutative ring, so a field. $\endgroup$ Commented Sep 30, 2013 at 7:25
$\begingroup$
$\endgroup$
4
If we assume that $\text{Spec}(R)$ is connected, then $R$ is always a field (note the spectrum of a local ring is always connected). This is equivalent to the nonexistence of nontrivial idempotents. Indeed, over a commutative ring any simple $R$-module is also of the form $R/\mathfrak m$ for a maximal ideal $\mathfrak m$ of $R$. Write $R = R/\mathfrak m \oplus I$, for some finitely generated ideal $I$ of $R$. By the determinant trick, there is an $x\in R$ with $x-1\in I$ such that $xI = I$, hence $I = I^2$. This implies $I$ is generated by an idempotent, hence $I = 0$, since $R/\mathfrak m$ is nonzero.
-
2$\begingroup$ Not quite true, $\mathfrak{m}= 0$ does not follow. The ring just has to contain a field as a direct factor, as per Mariano's answer. $\endgroup$ Commented Oct 20, 2013 at 14:18
-
1$\begingroup$ Once corrected, this is exactly the same as what I did, no? :-) $\endgroup$ Commented Oct 21, 2013 at 21:25
-
$\begingroup$ Yes, it is! Hah, alas, I guess that's what you get when rushing through things. $\endgroup$– FloreszaCommented Oct 22, 2013 at 0:59
-
$\begingroup$ Though a few added details are never a bad thing :) $\endgroup$– FloreszaCommented Oct 22, 2013 at 1:00
$\begingroup$
$\endgroup$
1
No, the cyclic group of order $p$ is not a direct summand in the cyclic group of order $p^2$ ($p$ a prime).
-
12$\begingroup$ How does this answer the question? $\endgroup$ Commented Sep 28, 2013 at 20:45