Proper subring of $\mathbb{Z}_{12}$? Can someone explain why there are no proper subrings of $\mathbb{Z}_{12}$? My explanation is that any proper subset of $\mathbb{Z}_{12}$ would have a different 0 element. Thus, it would not be a subring. I'm not sure how accurate this is, though.
 A: Depends on what you mean by "subring".
A subring of $\mathbb{Z}_{12}$ must be a subgroup. The subgroups of $\mathbb{Z}_{12}$ are $\mathbb{Z}_{12}$, $\{0,2,4,6,8,10\}$, $\{0,3,6,9\}$, $\{0,4,8\}$, $\{0,6\}$, and $\{0\}$.

*

*If by "subring" you mean "subring with the same identity", then the only one that includes $1$ is $\mathbb{Z}_{12}$, so none of the other ones are subrings.


*If by "subring" you mean "subgroup that is closed under multiplication, has a multiplicative identity, but the identity need not be the same as that of the original group", then it is easy to verify that $\{0,2,4,6,8,10\}$ does not have a multiplicative identity (since $2\times x\neq x$ for all $x\neq 0$); that $9$ is a multiplicative identity for $\{0,3,6,9\}$ ($9\times 3 = 27\equiv 3\pmod{12}$, $9\times 6 =54 \equiv 6\pmod{12}$, $9\times 9=81\equiv 9\pmod{12}$); that $4$ is a multiplicative identity for $\{0,4,8\}$ ($4\times 4=16\equiv 4\pmod{12}$, $4\times 8 = 32\equiv 8\pmod{12}$); that $\{0,6\}$ does not have a multiplicative identity; and that $\{0\}$ does have a multiplicative identity (namely, $0$ itself).
So under this definition, $\{0\}$, $\{0,3,6,9\}$, and $\{0,4,8\}$ are all subrings.


*Finally, if by "subring" you mean "subgroup that is closed under multiplication", then all of the above are subrings.
A: The issue here is with the element $1$, not $0$. By definition, a subring must have the same multiplicative identity element as the whole ring (as well as the same 0 element and addition and multiplication operations). But any subset of $\mathbb{Z}_{12}$ which contains $1$ and is closed under addition must be the whole ring.
A: I don't know for you, but my definition of a subring is just a subgroup (with respect to addition) of the ring which is closed under multiplication (c.f. Dummit & Foote's Abstract Algebra if someone doesn't agree, page 228). There is no need for the subring to contain the identity. Thus the subgroups
$$
\{0, 2, 4,6,8,10\}, \{0,3,6,9\}, \{0,4,8\}, \{0,6\},
$$
do indeed give you subrings, when equipped with multiplication $\mathrm{mod} \, 12$, but they are not isomorphic to, say, $\mathbb Z_6$, because $\mathbb Z_6$ has the identity for multiplication and the guy with $6$ elements here does not. 
The reason why I think something's wrong in your way of defining things is this : ideals in general do not contain the identity element (when they are non-trivial), and they're defined as subrings which are closed under multiplication by every element of the original ring. It would make non-sense for them to contain the identity.
Hope that helps,
A: There is not universal agreement on the definition of ring, subring, or ring homomorphism. Some require a ring to have a $1$ and others not, some require a subring to contain the $1$ of the parent ring, others do not. Some require that a homomorphism of rings map the $1$ of one ring to the $1$ of the other, others do not. It's mostly a matter of convention. So, Alex and Patrick are both correct according to their respective definitions of subring. Most important for you, Greg, is under what definitions of ring and subring you are working. This whole business reminds me of the old saying that the Americans and the English are two people divided by a common language.
