# Subrings of Intergers mod 8

I have to find all the subrings of $Z_8$. I've already deduced that since all subgrings are on subgroups of the additive group, that each must have zero in it, as it acts as the identity element in the additive group.

Obviously the original ring and the trivial ring are both subgroups. I easily found that $({[0]_8, [4]_8}, +, *)$ and $({[0]_8, [2]_8, [4]_8, [6]_8}, +, *)$ are also subgroups. And I think this is all of them, but I can't think of a way to be sure without ruling out all other options by trial error.

Am I missing any??

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You do not require unity (the multiplicative identity) to exist in the "subring". Therefore you can consider the cases which do not contain it (since the one which does is all of $Z_8$). Classify according to the least residue mod 8 greater than zero. – hardmath Nov 14 '12 at 19:59
Unless, of course, you are using the other common meaning of "ring" to refer to an algebraic theory that does require the multiplicative unit to exist in the "subring", or maybe to the less common algebraic theory that requires a "subring" to have a multiplicative unit, but it doesn't have to be the same one as the whole ring. Each of the three cases has different answers. – Hurkyl Nov 14 '12 at 20:07

You can use Bézout's identity to prove that if there is an element $[n]_8\in Z_8$ such that $n$ and $8$ are coprime, then the subring will contain $[1]_8$ and therefore will be the whole of $Z_8$.