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Let $R$ be commutative, and $V,W\in R$-$\textbf{Mod}$. As the title states, I am seeking an example of an injective $R$-module $V$ where $V/W$ is not injective. There is a question/answer here about the topic, but to be honest it don't understand it enough to create a counter-example.


What I know:

A submodule of an injective module need not be injective. The first example that comes to mind is $\Bbb Z$ as a submodule of $\Bbb Q$ (as $\Bbb Z$ modules).

A quotient of a divisible module is divisible (you can simply push through the quotient map). Thus, any counterexample cannot be over a ring where divisible $\Rightarrow$ injective. In particular, the ring must not be a PID.


Ideas:

Let $R=\Bbb F_2[\Bbb Z/2]$. Then the regular module is of course projective, and thus projective since $R$ is a group algebra and the regular module is finitely generated. It seems like $\Bbb F_2$ is a quotient of $R$ which is not a summand, but I keep getting confused proving this.


I would love a straightforward example. The only route I have uses the fact that, for modules over a group algebra, injective$\Leftrightarrow$injective, and I would rather not have to use this fact. Thanks.

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  • $\begingroup$ Try a ring of global dimension $\ge2$. $\endgroup$ – Lord Shark the Unknown Jul 27 '18 at 18:52
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A ring is called hereditary if every quotient of an injective module is injective or, equivalently, if every submodule of a projective module is projective.

We need a ring with a projective module with a nonprojective submodule. Here it is: $\mathbb{Z}/4\mathbb{Z}$. The submodule $2\mathbb{Z}/4\mathbb{Z}$ is not projective.

Now, how to find a quotient of an injective module over $\mathbb{Z}/4\mathbb{Z}$ which is not injective? Hint: consider the same modules as before.

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  • $\begingroup$ I was able to just use Baer's criterion here, but do you have any hint on how to prove the equivalence you mentioned for hereditary rings? $\endgroup$ – Elliot G Jul 28 '18 at 17:56
  • $\begingroup$ @ElliotG That’s a pretty exercise. No time now, sorry. $\endgroup$ – egreg Jul 28 '18 at 18:23

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