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Let $A$ be a PID. Then, an $A$-module $Q$ is injective iff $Q=rQ$ for every $r\neq 0$ in $A$.

My question is, where is the property "A is a domain" used in the proof of the above? Can someone please provide a suitable proof to point this out?

jimm
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  • $Q$ also has to be torsion-free. – Bernard May 21 '20 at 17:52
  • Okay, where does the problem arise if its not? – jimm May 21 '20 at 18:00
  • I didn't check, but I guess you can't satisfy Baer's criterion. – Bernard May 21 '20 at 18:04
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    Torsion free is not required. Since the D in PID stands for domain, are you asking whether a similar result can be proved for non-domains? If so, what type of rings do you want to consider other than PIDs? – Mohan May 21 '20 at 18:25
  • To elaborate on Mohan's comment re: Bernard's comment, $\mathbb{Q}/\mathbb{Z}$ is a torsion injective $\mathbb{Z}$-module. – diracdeltafunk May 21 '20 at 20:46
  • More generally, if $R$ is a ring such that its total ring of fractions, $T(R)$, is a finite product of fields, then $R$ is hereditary if and only if divisible $R$-modules are injective. Among domains, this characterizes Dedekind domains. – Badam Baplan May 22 '20 at 00:55
  • (cont.) The first result I mentioned does not, as far as I know, even extend to $T(R)$ being Von Neumann Regular. It seems to be key that every dense ideal of $R$ contains a nonzerodivisor, and that $(Ann(I), I)$ is a dense ideal for every ideal $I$ (aka that $R$ is reduced). My main point is that being a domain, and also having principal ideals, is way overpowered for "divisible $\implies$ injective." – Badam Baplan May 22 '20 at 00:55

2 Answers2

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Suppose $R$ is not a domain. Then $rQ=Q$ cannot hold for all $r\ne0$, unless $Q=\{0\}$.

Indeed, if $r,s\in R$ are nonzero with $rs=0$, we have $$ \{0\}=0Q=rsQ=rQ=Q $$

On the other hand, it's a standard result of module theory that every module can be embedded in an injective module.

In general, a module $Q$ over a commutative ring $R$ is called divisible if, for every nonzero-divisor $r\in R$, we have $rQ=Q$ (note that $0$ is a zero-divisor). Every injective module is divisible: indeed, if $r$ is a nonzero-divisor and $x\in Q$, the homomorphism $rR\to Q$ defined by $rs\mapsto sx$ is well defined and it extends to a homomorphism $R\to Q$. If the image of $1$ is $y$, then it's clear that $ry=x$. Therefore $rQ=Q$.

On the other hand, being divisible does not generally imply that the module is injective even over domains, see Divisible module which is not injective

egreg
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  • This shows that there cannot be non-trivial divisible module over non-integral rings. To be 100% complete, it should also be mentioned that there are non-trivial injective modules over such rings. – Captain Lama May 21 '20 at 19:56
  • @CaptainLama I added the mention – egreg May 21 '20 at 20:44
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    @egreg I think this is really demonstrating why nobody uses this as the definition for divisible over general rings. A more common definition would be that $rM =M$ for every non-zerodivisor $r \in R$, which is still true for every injective module. I recognize that you are exactly answering the question as asked, but still thought this might be worth pointing out. – Badam Baplan May 21 '20 at 21:53
  • @BadamBaplan Is the addition fulfilling your request? – egreg May 21 '20 at 22:12
  • @egreg yes, to me that is a valuable addition! – Badam Baplan May 21 '20 at 22:18
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This is an extension of the comments I left on your post.

Let $R$ a PID and $M$ a divisible $R$-module. We invoke Baer's criterion to show that $M$ is injective. Given any homomorphism $\phi: I \rightarrow M$ where $I$ is an ideal of $R$, we need to show that $\phi$ extends to a homomorphism $R \rightarrow M$. But $I = aR$ for some $a \in R$. We know that $M = aM$, so we find $m \in M$ such that $\phi(a) = am$. Then the map $\phi': R \rightarrow M$ defined by $\phi'(r) = rm$ extends $\phi$.

Where did $R$ being a domain come into this? To be able to use Baer's criterion, we needed to be able to work with an arbitrary ideal $I$. If $R$ was just a Principal Ideal Ring, then we might have $I = aR$ for a zero-divisor $a$. As egreg pointed out, the condition that $M = aM$ for all $0 \not= a \in R$ is ridiculous for rings with zero divisors, because it forces $M = 0$. So we usually generalize the notion of "divisible" modules to mean something like $M = IM$ for every ideal $I$ that contains a non-zerodivisor — in any case if we focus on principal ideal rings most reasonable notions of module divisibility will converge.

So how to move past the domain case? Let's show that things still work out fine when $R$ is a reduced principal ideal ring. Then $\operatorname{Ann}(I) \cap I = 0$ for every ideal $I$, and given a homomorphism $\phi: I \rightarrow M$ we can extend it to a homomorphism $\phi^*: (I, \operatorname{Ann}(I)) \rightarrow M$ by defining $\phi(\operatorname{Ann}(I) = 0$. If $R$ is a Principal Ideal Ring then $(I, \operatorname{Ann}(I)) = aR$ for $a$ necessarily not a zerodivisor, and we can proceed as above.

More generally, there is the following result:

Let $R$ be a ring such that $T(R)$, the total ring of fractions of $R$, is a finite product of fields. $R$ is hereditary iff divisible $R$-modules are injective.

This is I think due to Levy in Torsion-free and divisible modules over non-integral domains (1963), and can be generalized to noncommutative rings just as well.

Badam Baplan
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