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Is it possible to describe all the index 2 subgroups of the group $G = \prod_{i\in \mathbb{N}}\; \mathbb{Z}/2\mathbb{Z}$?

For example, one can take the kernel of the $i$-th projection map $\pi_i\colon G\to \mathbb{Z}/2\mathbb{Z}$. More generally, by forming linear combinations of these projection maps we get other surjective homomorphisms $G\to \mathbb{Z}/2\mathbb{Z}$.

But still, there are many more index 2 subgroups which do not appear as kernels of the maps described above. One easy argument to see this is to equip $G$ with the product ring structure so that the group structure is the underlying additive group of this ring. Then the direct sum $I = \bigoplus_{i\in \mathbb{N}} \mathbb{Z}/2\mathbb{Z}$ is a proper ideal of $G$, so by Krull's theorem we can find a prime ideal $\mathfrak p$ of $G$ containing $I$. Since $G$ is Boolean every prime ideal is maximal and of index 2. But $\mathfrak p$ cannot be obtained as the kernel of one of the projection maps given before.

I am aware of the fact that $\text{Spec}(G)$ can be described using ultrafilters. So now we also have these subgroups of index 2. But are there still more of them?

Any information related to the above is appreciated. Thanks in advance, AYK.

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  • $\begingroup$ Every subgroup of index $2$ is a prime ideal —maximal, in fact. What you are asking, therefore, is for a descrition of the elements sof the spectrum, and you know the standard description. $\endgroup$ – Mariano Suárez-Álvarez Jul 29 '15 at 7:33
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    $\begingroup$ $G$ is a vector space over $\Bbb{Z}/2\Bbb{Z}$. If you believe in Choice, then it has vector space bases. Furthermore, any linearly independent subset can be extended to a basis. Therefore we can construct subspaces of codimension one missing a non-zero element of our choice. $\endgroup$ – Jyrki Lahtonen Jul 29 '15 at 7:49
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    $\begingroup$ @MarianoSuárez-Alvarez I do not agree with your statement. The kernel of the map $\pi_1 - \pi_2$ (notation as above) is the set of elements $(x_1,x_2,\ldots)\in G$ for which $x_1 = x_2$. This is not an ideal, let alone a prime ideal. $\endgroup$ – AYK Jul 29 '15 at 9:06
  • $\begingroup$ @JyrkiLahtonen Yes thanks, I realised this a bit later, so in any case this gives us the cardinality of the set of such subgroups. $\endgroup$ – AYK Jul 29 '15 at 9:24

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