12
$\begingroup$

For rings with unit there are at least three ways to define a projective module:

  1. The universal property, i.e. a module $P$ is projective if for any epimorphism $M\to N$ and any morphism $P\to N$ there exists a morphism $P\to M$ such that the diagram commutes.
  2. A projective module is a direct summand of a free module.
  3. Any epimorphism onto $P$ splits.

My question is what relation there is between these three conditions when the rings does not have a unit. Of course 1. implies 3., but what about the other directions? I'm most interested in "3. implies 1.?".

Edit: @t.b. has answered the 3. implies 1. In searching the web I found the claim that not even the rng (as the most basic free module) itself is projective. (see e.g. Anh, Marki: Morita equivalence over rings without identity and Arando Pino, Rangaswamy, Siles Molina: Weakly regular and self-injective Leavitt path algebras over arbitrary graphs.) So 2. seems not to imply 1. However I didn't find a counterexample in those papers. So, what is a counterexample for that?

$\endgroup$
3
  • 1
    $\begingroup$ For 3. implies 1. take the pull-back $Q = P \mathop{\times_N} M$ of $P \to N \gets M$, then choose a splitting of $Q\to P$ and observe that $P \to Q \to M$ is a lift of $P \to N$. To see that 2. is equivalent to 3. note that free modules have property 3 and that property 3 is stable under retractions and for the converse note that every module is a quotient of a free module. $\endgroup$
    – t.b.
    Commented Mar 15, 2012 at 11:27
  • $\begingroup$ Thanks for 3. implies 1. But how is every module a quotient of a free module if you don't have a unit? $\endgroup$ Commented Mar 15, 2012 at 11:45
  • 1
    $\begingroup$ Oh, you're right. Sorry in my work I almost always assume that modules are essential in the sense that $R \times M \to M$ is epi (since the rngs I consider have surjective multiplication $R \times R \to R$, this is a reasonable assumption for me to make). I was assuming this carelessly. I don't know about the more general situation. Sorry for the confusion. $\endgroup$
    – t.b.
    Commented Mar 15, 2012 at 12:00

1 Answer 1

4
$\begingroup$

For a simple example where (2) does not imply (1), let $R$ be any nonzero ring and let $S$ be the rng with the same underlying abelian group as $R$ but with the multiplication $ab=0$ for all $a,b\in R$. Consider the $S$-module $M=R\times R$, with $S$ acting by $a\cdot(b,c)=(0,ab)$. There is an epimorphism of $S$-modules $p:M\to S$ given by the projection onto the second coordinate, and $\ker p$ is isomorphic to $S$. The epimorphism $p$ cannot split, since otherwise we would have $M\cong S\oplus S$ and the action of $S$ on $M$ would be trivial. Thus $S$ does not satisfy (3) (or equivalently, (1)) as a module over itself.

For an example where (1) does not imply (2), as well as a broader understanding of projective modules over rngs, let us back up a bit and understand what a module over a rng really is. If $R$ is a rng, then there is a ring $U(R)$ obtained by freely adjoining a unit to $R$: the underlying abelian group of $U(R)$ is $\mathbb{Z}\oplus R$, and multiplication is given by $(n,a)(m,b)=(nm,nb+ma+ab)$. You should think of $(n,a)$ as being $n\cdot 1+a$ (since $(1,0)$ is the unit of $U(R)$).

It is easy to see that an $R$-module is in fact the same thing as a (unital) $U(R)$-module: given an $R$-module $M$, let $U(R)$ act on it by $(m,a)\cdot x=mx+ax$, and conversely, given a $U(R)$-module, just consider the action of elements of the form $(0,a)$. It is also easy to see that under this correspondence, a homomorphism of $R$-modules is the same thing as a homomorphism of $U(R)$-modules.

It follows that definitions (1) and (3) are the same thing for $R$-modules and $U(R)$-modules. Since $U(R)$ is unital, they also coincide with (2) for $U(R)$-modules. Thus we can say: an $R$-module $M$ is projective (i.e., satisfies (1)) iff as a $U(R)$-module, $M$ is a direct summand of a free $U(R)$-module. In particular, $U(R)$ (considered as an $R$-module) is always projective. But, for instance, if $R$ satisfies $ab=0$ for all $a,b\in R$, then $U(R)$ cannot be a direct summand of a free $R$-module (since $R$ acts nontrivially on $U(R)$). So for any rng $R$ with trivial multiplication, $U(R)$ is an $R$-module which satisfies (1) but not (2).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .