Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

2 Answers

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.

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

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.