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Let $E=\mathbb{Z}_4\oplus\mathbb{Z}_4$, regarded as a module over $\mathbb{Z}_4$.

Let $M= \{(0,0),(2,2)\}\subseteq E$ be a submodule.

Then how can I show that $M$ is the intersection of $2$ injective submodules of $E$?

I tried to find divisible submodules of $E$, as I know these are precisely the injective submodules over $\mathbb{Z}$, but I couldn't find any. Any suggestions?

tkf
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wanheda
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1 Answers1

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Working over $\mathbb{Z}$ you are right that there are no non-trivial divisible submodules of $E$, so there cannot be a solution.

However if we instead work over $\mathbb{Z}_4$, we can find a solution. In this case we can use this result to deduce that $\mathbb{Z}_4$ is injective over itself, and we can set $$M_1=\{(0,0),(1,3),(2,2), (3,1)\},\qquad M_2=\{(0,0),(1,1),(2,2), (3,3)\},$$ so that $M_1\cong M_2\cong \mathbb{Z}_4$ is injective, and: $$M=M_1\cap M_2.$$

tkf
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