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I know that if the direct sum of countably many copies of $\mathbb Z/2\mathbb Z$, $I$ were a direct summand of the direct product of countably many copies, $R$ (direct summand as $R$-modules), then the complement would be a direct summand isomorphic to $R/I$. But I can't think of any properties of the quotient that would preclude it from being a subgroup. Could I have a hint?

Nishant
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    Pretty sure it is a direct summand... ? Both groups are just vector spaces over $\mathbb F_2$. – Dustan Levenstein Jul 07 '14 at 00:19
  • Is the infinite sum you refer to the obvious embedding into the infinite product, or are you after something more general, where there is some subgroup of the infinite product that is isomorphic to the infinite sum? – 2'5 9'2 Jul 07 '14 at 00:20
  • It's the obvious embedding. – Nishant Jul 07 '14 at 00:29
  • And how are we supposed to give a proof for something which is wrong? :-) Please address Dustan's comment. – Martin Brandenburg Jul 07 '14 at 00:40
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    @Martin OP hides a critical fact within the text: "direct summand as $R$-modules," where the sentence structure indicates what $R$ is. Surely this could be stated more clearly and up-front, though... – anon Jul 07 '14 at 00:45
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    If the question is "not a direct summand as an $\prod \mathbb{F}_2$-module," you can completely classify all such direct summands by classifying the idempotents in $\prod \mathbb{F}_2$, then looking at their images. (Hint: every element is idempotent.) – Qiaochu Yuan Jul 07 '14 at 00:51
  • ^Why does that work? I know that all finitely generated ideals of $R$ are principal and thus generated by an idempotent, but this $I$ is not finitely generated. – Nishant Jul 07 '14 at 00:56
  • That is the laziest way of addressing a comment I have ever seen. – Dustan Levenstein Jul 07 '14 at 01:13
  • What do you mean? – Nishant Jul 07 '14 at 01:16
  • I mean that you also didn't bother to post a comment explaining that your edit was meant to address my comment, leaving others to do that hard work for you. – Dustan Levenstein Jul 07 '14 at 01:22
  • Oh, I see. blue said that we should clean up the comments after I edit the OP, but I guess I shouldn't have deleted my comment anyway. – Nishant Jul 07 '14 at 01:23
  • Ah, I see. My bad then, I just missed a conversation apparently. :) – Dustan Levenstein Jul 07 '14 at 01:24

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Let $R=\prod k$ (infinitely many copies) be considered a module over itself, $k$ a field. Then $\bigoplus k$ is a submodule of $R$. Suppose that $N$ is any other nontrivial submodule (equivalently, ideal) of $R$.

Show that $N$ and $\bigoplus k$ intersect nontrivially: take an arbitrary nonzero element of the former, and multiply it by an appropriately coordinate-annihilating element of $R$, and you remain in $N$...

anon
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