# (Z/n)^(I) is a direct summand of (Z/n)^I

Dear group theorists,

Let $n \geq 1$ and $I$ be an infinite set (you may assume $I$ to be countable). Is the abelian group $(\mathbb{Z}/n)^{(I)}$ (direct sum of copies $\mathbb{Z}/n$) a direct summand of $(\mathbb{Z}/n)^I$ (direct product of copies $\mathbb{Z}/n$)? This question is motivated by that one.

If $n$ is prime, this follows from Linear Algebra. Of course, this is not constructive at all. Thus it's also true when $n$ is squarefree (use the Chinese Remainder Theorem). What happens otherwise? The smallest example is $n=4$. I don't know how to start ...

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## 1 Answer

Yes. The ring ${\mathbb Z}/n {\mathbb Z}$ is injective over itself, and over a Noetherian ring, direct limits of injectives are again injective; thus ${\mathbb Z}/n{\mathbb Z}^{(I)}$ is injective over ${\mathbb Z}/n{\mathbb Z}$. Finally, any embedding of an injective splits, as follows directly from the property of being injective. We can now apply this to the embedding $({\mathbb Z}/n{\mathbb Z})^{(I)} \hookrightarrow ({\mathbb Z}/n{\mathbb Z})^I.$

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Thank you, that's great :). I used Baer's Criterion twice to understand your statements. – Martin Brandenburg Jan 21 '10 at 6:29