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We know that every module $M$ is embedded in an injective module $D$. Is it true that the module $D/M$ is torsion?

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No, let $A = \frac{k[x, y]}{x^2, y^2}$ where $k$ is a field of characteristic $2$. This is a Frobenius algebra, hence it's self injective which means that $A$ is injective as a module over itself. As a module, the socle of $A$ is the principal ideal generated by $xy \in A$. This is $1$-dimensional so $A$ is indecomposable and hence the injective envelope of it's socle.

The quotient, $\frac{k[x, y]}{(x, y)^2}$, of $A$ by its socle is not a torsion module.

Remember: For an element $m \in M$ in a module to be torsion we must have $rm = 0$ for some nonzero $r \in A$ which is, additionally, not a zero-divisor.

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Good example, assuming this is what was intended by torsion. I wish the OP would clarify :S – rschwieb Nov 8 '13 at 14:02

Since $M$ is an essential submodule of $E(M)$, the quotient $E(M)/M$ is in fact a singular module.

So if by "torsion" you mean "for every nonzero $x\in E(M)/M$ there exists a nonzero $r\in R$ such that $xm=0\in E(M)/M$," then yes, you're in luck, because singular modules have that property.

$N/M$ is singular for any essential extension $M\subseteq_e N$.

It's true in the example that Jim gave that there aren't any regular elements in the annihilators (because the nonunits are all zero divisors.) However, in $(k[x,y]/(x^2,y^2))/((xy)/(x^2,y^2))$, all elements do have nonzero annihilators.

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