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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.

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4 Answers 4

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 $\{0\}$.

  1. 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.

  2. 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.

  3. Finally, if by "subring" you mean "subgroup that is closed under multiplication", then all of the above are subrings.

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The definition of subring that I am using (Undergraduate Algebra by Serge Lang (not too fond of this book)) is the following: "By subbing R' of R one means a subset of R such that the unit element of R is in R', and if x,y are in R', then -x, x+y, and xy are in R'." I guess my confusion lies in what they mean by unit element. I figured that since R is an additive group, that it would be the additive identity. They don't say anything about the multiplicative identity in this definition. –  Greg Dec 2 '11 at 18:22
    
I happened to have a copy of that book close to hand - on p.84, Lang defines "unit element" to be the unique multiplicative identity. –  Alex Kruckman Dec 2 '11 at 18:26
    
Alright this clears everything up for me. Thanks for pointing that out! –  Greg Dec 2 '11 at 18:30
    
@Greg In ring theory, a unit is any invertible element, and the unit denotes the (unique) multiplicatively neutral element, i.e. the identity of the multiplicative monoid. The unit is unique since $1' = 11' = 1$. –  Bill Dubuque Dec 2 '11 at 18:48

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.

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By what definition? For all I know $2\mathbb Z$ is a subring of $\mathbb Z$ (plus, it's an ideal), even though $2 \mathbb Z$ has no identity. –  Patrick Da Silva Dec 2 '11 at 17:57

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,

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This is a fine definition of subring, if you don't require your rings to contain a multiplicative identity element. If you do require a distinguished identity element, the correct definition of substructure requires any subring to contain that element. All due respect to Dummit & Foote, but as far as I'm concerned, the standard definition of ring includes a multiplicative identity, and it appears to me that this assumption is implicit in the question. I remember a lengthy discussion of the proper definition of ring/subring on this site a while back... maybe someone can provide a reference? –  Alex Kruckman Dec 2 '11 at 18:20
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Well, if this is the case, then perhaps the context in which OP's question arose is probably more important than the discussion we're having here, because I also agree with what you're saying. (Saying that the standard definition of a ring includes a $1$ is maybe a little rough though, if you work in different areas of algebra rings may or may not have a $1$. It only depends on what you want to do...) –  Patrick Da Silva Dec 2 '11 at 18:22

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.

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Thanks everybody for the insightful discussion. I understand the question and definitions much more clearly now. –  Greg Dec 2 '11 at 18:31

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