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It is well known that $\mathbb{Q}$ is an injective $\mathbb{Z}$-module, and more generally if $R$ is a domain, then its field of fractions $\mathrm{Frac}(R)$ is an injective $R$-module. Now my question: Let $R$ be a commutative ring with unit and let $Q(R)$ be its total ring of quotients.

Is $Q(R)$ an injective $R$-module?

I think it is not, but I don't have a counterexample.

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Note that $Q(R)$ is an essential extension of $R$, the question is whether $Q(R)$ is the maximal essential extension. – user58764 Jan 21 '13 at 8:41

Not necessarily. Let $R={\Bbb Q}[x,y]/(x^2,xy,y^2)$. Then every element of $R$ is either a zero-divisor or invertible, so $R=Q(R)$. However $R$ is not injective as the $R$-module homomorphism from the ideal ${\Bbb Q}x+{\Bbb Q}y$ to $R$ which sends $x$ to $y$ and $y$ to $x$ cannot be extended to $R$.

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@Moews, thanks, this is a very nice example. Do you know any conditions one must impose in order to have $Q(R)$ injective as an $R$ module, ofcourse when $R$ is a domain then that is true. – user58764 Jan 21 '13 at 11:19

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