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The following result is basic ( P.J.Hilton, U.Stammabach, a course in homological algebra ).

Let $A$ be a principal ideal domain. Then a $A$ module is injective iff it is divisible.

Now if the condition is "Let $A$ be a domain", does the result hold ? I think that it is probably wrong. Can anyone give me a counterexample?

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  • $\begingroup$ The classical example is $\mathrm{Frac}(A)/A$. It's divisible, but, as far I remember, usually not injective. $\endgroup$
    – YCor
    Commented Oct 12, 2016 at 5:06
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    $\begingroup$ Corollary 3.24 In Lam's GTM book: books.google.fr/…: all divisible $A$-modules are injective iff $A$ is Dedekind. $\endgroup$
    – YCor
    Commented Oct 12, 2016 at 5:23
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    $\begingroup$ What is your definition of "divisible" for a module over an arbitrary ring? (There seem to be different variants in use.) $\endgroup$
    – Fred Rohrer
    Commented Oct 12, 2016 at 5:51
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    $\begingroup$ This seems to be a rather strange notion of divisibility. For example, if your ring is not reduced, then the zero module is the only divisible module. I suggest you consider only non-zerodivisors, or (and probably better) you take the definition given in Lam's book mentioned by Yves. (The @-notification works only if there is no space right after the @.) $\endgroup$
    – Fred Rohrer
    Commented Oct 12, 2016 at 7:32
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    $\begingroup$ Oh, I just realised that I misread the question. You do not want to omit "domain", but only "principal". Sorry! $\endgroup$
    – Fred Rohrer
    Commented Oct 12, 2016 at 7:34

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Let $A$ be a domain. Then an $A$-torsion-free $A$-module is injective if and only if it is divisible. This is well-known. As mentioned in one of the comments, an arbitrary divisible $A$-module is injective if and only if $A$ is a Dedekind domain.

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  • $\begingroup$ Does this also hold in the graded case? If $A$ is a graded domain and $M$ a graded torsion free module, is $M$ injective in the graded modules iff $M$ is graded divisible? Does this imply that $k[x,x^{-1}]$ is graded injective? $\endgroup$
    – Bubaya
    Commented Oct 4, 2023 at 16:17

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