While checking many sites about subrings and how to prove a subset of a ring $R$ is also a subring of it, I saw many sites asking if a set is a subring of a ring but even the operation in these two sets are different, and this is contradictory to the definition of a subring!

For example, I saw a question that asks if $\mathbb{Z_6}$ is a subring of $\mathbb{Z_{12}}$, and justifies that's not by showing that it's not closed under addition $mod$ $12$: $5+5= 10$ $mod$ $12$, but $10$ is not in $\mathbb{Z_6}$. I can't understand this way of disproving anymore, since $10$ in $\mathbb{Z_{12}}$ is different from $10$ in $\mathbb{Z_6}$ which's $4$ $mod$ $6$, we are talking about different operations!! ( addition $mod$ $12$ and addition $mod$ $6$ ), so this is clear that $\mathbb{Z_6}$ is not a subring of $\mathbb{Z_{12}}$ without the need of a proof! So why are many sites dealing the same way with such questions?!

I have another one:

Can we consider $\mathbb{Z_p}$ a $subset$ of $\mathbb{Z_q}$ if $p\leq q$??

I thought this is right because while deciding whether a set is a subset of another set, we don't look at their operations, but I was shocked that one site writes: $\mathbb{Z_{2}}$ is $not$ $a$ $subset$ of $\mathbb{Z_{6}}$!

I feel like I have many new problems in abstract algebra, please help me to reorder my info !

  • $\begingroup$ so $\mathbb{Z_{6}}$ is a subring of $\mathbb{Z_{12}}$ and $\mathbb{Z_{2}}$ is a subring of $\mathbb{Z_{6}}$ regardless the different operations?? $\endgroup$
    – Nour
    Sep 30, 2017 at 10:57
  • $\begingroup$ Oh I didn't take ring homomorphisms till now, can you illustarte in other ways? $\endgroup$
    – Nour
    Sep 30, 2017 at 10:58
  • $\begingroup$ @Nour You don't have to think of it as a monomorphism of $\Bbb Z/n \Bbb Z$ into $\Bbb Z/m \Bbb Z$ if you don't want to. Just look at the elements of $\Bbb Z/n\Bbb Z$ and $\Bbb Z/m \Bbb Z$, in the same way you would if you were trying to list subgroups of $\Bbb Z/m \Bbb Z$ $\endgroup$ Sep 30, 2017 at 11:00
  • $\begingroup$ @ÍgjøgnumMeg, aha, $\mathbb{Z_{12}}$ has 6 subgroups, none of them is $\mathbb{Z_{6}}$, so $\mathbb{Z_{6}}$ is not a subring of $\mathbb{Z_{12}}$. Right ? $\endgroup$
    – Nour
    Sep 30, 2017 at 11:38
  • $\begingroup$ @Nour I didn't really mean it like that, I was merely setting up an analogy; looking at elements is usually a lot less convenient than looking at the structure as a whole, but it's good to build up intuition when you're starting out. Regarding your comment, a ring IS an abelian group under addition, and also has some further multiplicative structure which is compatible with the addition operation. If you have a candidate for a subring, you certainly want it to also be a subgroup of your ring when considered as an additive abelian group. $\endgroup$ Sep 30, 2017 at 11:45

1 Answer 1


Being a subring requires sharing the operations.

Any (nonempty) subset $S$ of a given ring $R$ can be considered as a ring on its own with different operations. Just use the fact that there is a ring $A$ with the same cardinality of $S$ and transfer the operations on $S$.

What would be the usefulness of this? None. Changing the bijection between $A$ and $S$ would change the ring structure, but there will be no connection whatsoever with the operations in $R$, in general.

About $\mathbb{Z}_6$ being a subset of $\mathbb{Z}_{12}$, the situation is similar. You can define a map $\mathbb{Z}_6\to\mathbb{Z}_{12}$ by decreeing that $[x]_6$ is mapped to the residue class modulo 12 that contains the least nonnegative element in $[x]_6$; so \begin{align} [0]_6&\mapsto[0]_{12} & [1]_6&\mapsto[1]_{12} & [2]_6&\mapsto[2]_{12} \\ [3]_6&\mapsto[3]_{12} & [4]_6&\mapsto[4]_{12} & [5]_6&\mapsto[5]_{12} \end{align} but you could also decide to map $[x]_6$ to the residue class modulo $12$ of the greatest negative element in $[x]_6$, so \begin{align} [0]_6&\mapsto[-6]_{12}=[6]_{12} & [1]_6&\mapsto[-5]_{12}=[7]_{12} & [2]_6&\mapsto[-4]_{12}=[8]_{12} \\ [3]_6&\mapsto[-3]_{12}=[9]_{12} & [4]_6&\mapsto[-2]_{12}=[10]_{12} & [5]_6&\mapsto[-1]_{12}=[11]_{12} \end{align} Both maps are injective, but they're quite different from each other. Which one is “right”? Neither: both depend on a somewhat arbitrary choice of an element in $[x]_6$.

A good question might be: is there an injective map $f\colon\mathbb{Z}_6\to\mathbb{Z}_{12}$ such that the image is a subring? Well, this depends on what you mean by subring. If you require that subrings also share the identity with the overring, then the answer is no: $\mathbb{Z}_{12}$ has no proper subring in this sense.

So let's not bother with the identity. Then, yes: $\mathbb{Z}_{12}$ has a unique additive subgroup of order $6$, namely $2\mathbb{Z}_{12}$, which happens to be an ideal of $\mathbb{Z}_{12}$ (hence a subring, under the stated convention). More precisely, the map $$ [x]_{6}\mapsto [2x]_{12} $$ is actually a ring homomorphism (when not requiring the identity is mapped on the identity). The map is well defined: indeed, if $x\equiv y\pmod{6}$, then $2x\equiv 2y\pmod{12}$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .