Is $\mathbb{Z}_2$ an injective module over $\mathbb{Z}_4$ and $\mathbb{Z_6}$?

Question

I don't know which definition should I use to answer this question. Maybe I should use that $\mathbb{Z}_6 \equiv \mathbb{Z}_2 \oplus \mathbb{Z}_3$ or should I approach it with Baer's criterion?

Thanks in advance for any help.

@DietrichBurde But $\mathbb Z_6$ is a semisimple ring, and all modules are injective and projective. — rschwieb, Sep 28 '17 at 14:48
Yes, you are right. But for an integral domain $R$, not a field, the only projective and injective $R$-module is the zero module, right? — Dietrich Burde, Sep 28 '17 at 14:50
@DietrichBurde Yes, I think so. That question was also asked not long ago. — rschwieb, Sep 28 '17 at 14:50

rschwieb · Accepted Answer · 2017-09-28T14:58:40.173

4

$\mathbb Z_6$ is a semisimple ring, so all of its modules are injective. This is easy to see since it is isomorphic to the product of two fields. Using the Baer criterion immediately shows you why all modules are injective.

$\mathbb Z_2$ isn't an injective module of $\mathbb Z_4$ for the following reason: $\mathbb Z_4$ does not have any nontrivial summands, but it contains a copy of $\mathbb Z_2$ in the form of $(2+4\mathbb Z)$. But an injective module must split out of every module that contains it, and it clearly doesn't split out of $\mathbb Z_4$, so it's not injective.

edited Sep 28 '17 at 14:58

answered Sep 28 '17 at 14:50

rschwieb

160,592

Can you tell me a bit more about this answer? Why is $\mathbb{Z}_6$ semisimple? What exactly do you denote with $(2)$? – Atvin Sep 28 '17 at 14:57
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@Atvin I revised it to be $(2+4\mathbb Z)$ by which i mean the principal ideal generated by $2+4\mathbb Z$ in the quotient ring. – rschwieb Sep 28 '17 at 14:59
Thank you for the answer! – Atvin Sep 28 '17 at 15:00

Is $\mathbb{Z}_2$ an injective module over $\mathbb{Z}_4$ and $\mathbb{Z_6}$?

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