Prove that I_6 is simultaneously an injective and a projective module over itself.

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- Dec 4th 2011, 08:23 PMjcir2826Projective Modules
Prove that I_6 is simultaneously an injective and a projective module over itself.

- Dec 4th 2011, 08:46 PMDrexel28Re: Projective Modules
As a $\displaystyle \mathbb{Z}_6$-module, $\displaystyle \mathbb{Z}_6$ is free. Surely you know that all free modules are simultaneously projective. Thus, all we have to show is that $\displaystyle \mathbb{Z}_6$ is injective. To do this we can use Baer's criterion, which says that $\displaystyle \mathbb{Z}_6$ will be injective over $\displaystyle \mathbb{Z}_6$ if and only if we can extend every $\displaystyle \mathbb{Z}_6$-homomorphism $\displaystyle \mathfrak{a}\to\mathbb{Z}_6$ can be extended to $\displaystyle \mathbb{Z}_6\to\mathbb{Z}_6$ for every $\displaystyle \mathfrak{a}$ a left ideal of $\displaystyle \mathbb{Z}_6$. So...