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I want to find an injective but not divisible $R$-module.

If $R$ is integral domain, every injective is divisible so it should be $R$ is not an integral domain. Is there any example?

Gobi
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  • An $R$-module $M$ is divisible if for every $0 \neq r \in R$ and $x \in M$, there exists $y \in M$ such that $ry=x$. – Gobi Dec 05 '13 at 10:44
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    Yeah, this seems to be the "natural" generalization of divisibility, but... What if $r$ is a zerodivisor in $R$, that is, there exists $a\in R$, $a\ne 0$, with $ar=0$ and however $ax\neq 0$? This is why Lam, Lectures on Modules and Rings, Definition 3.16, adds an extra-condition: ann$(r)\subset $ ann$(x)$, and with this condition one can prove that injective $\Rightarrow$ divisible. –  Dec 05 '13 at 11:23

2 Answers2

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Consider the ring $R=F_2[Z]/(Z^2)=M$ where $F_2$ is the field of two elements. This is a self-injective ring, so $M$ is an injective $R$-module.

But now consider $x=1$ and $r=Z$, where I abuse notation for the images of $1$ and $Z$ in this ring. Saying that there exists $y\in R$ such that $yZ=1$ implies that $Z$ is a unit, but it is clearly not since it is a nilpotent element. So this module is not divisible in your sense.

What user says is good information: while this definition of "divisible" seems like the simplest extension, it turns out to be undesirable. T.Y. Lam's definition cited by user is the nicest one I know.

Notice that in the example above, $ann(Z)=(Z)\not\subseteq ann(1)=\{0\}$, so this is no longer a counterexample for Lam's definition. Indeed, all injective modules are divisible by that definition.

For domains this is not an issue since $ann(r)=\{0\}$ for every nonzero $r$, and therefore the condition that $ann(r)\subseteq ann(x)$ is met automatically. That's how this definition reduces to the domain definition.

rschwieb
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For each integer $n>1$, $\mathbb{Z}/n\mathbb{Z}$ is a self-injective ring. However, $\mathbb{Z}/n\mathbb{Z}$ as its own module is not divisible if $n$ is not prime.

Kevin
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