Finite rings and subrings isomorphic to $\mathbb{Z}_n$

Question

My book has proven this:

Every ring with unity has a subring isomorphic to either $\mathbb{Z}$ or $\mathbb{Z}_n$.

The $\mathbb{Z}_n$ case arises if the parent ring has characteristic $r>0$.

I have this question:

Is it possible for a ring with characteristic $r>0$ to have two subrings isomorphic to $\mathbb{Z}_n$ and $\mathbb{Z}_m$ with $n \neq m$?

I believe so, yes. Consider $\mathbb{Z_8}$. clearly $\mathbb{Z_6} \subset \mathbb{Z_8}$ and $\mathbb{Z_6} \cong \mathbb{Z_6}$, and $\mathbb{Z_4} \subset \mathbb{Z_8}$ and $\mathbb{Z_4} \cong \mathbb{Z_4}$.

My question is how to prove this.

Answer 1

If your subrings are required to contained the unit element of the former ring, then the answer is NO.

Let $X$ be any ring, and let $A, B$ are its subrings, assume that $A \cong \mathbb{Z}_n$, and $B \cong \mathbb{Z}_m$, where $n \neq m$. From $A \cong \mathbb{Z}_n$, we can deduce that the unit element $e$ of $X$ has additive order $n$ (note that $e \in A$ will be mapped to $\overline{1} \in \mathbb{Z}_n$); whereas the fact that $B \cong \mathbb{Z}_m$ yields the unit element $e$ of $X$ has additive order $m$, which is a contradiction.

Answer 2

user49685 got there first and should get credit for his work. (I've upvoted his answer...)

Final answer:

That a subring must contain the additive identity and the multiplicative identity requires that all subrings contain the characteristic subring. (I.e., they must contain the subring generated by $0$ and $1$, and since $0$ isn't doing any heavy lifting there, is the subring generated by $1$ -- the characteristic subring.) If such a subring contains an element not in the characteristic subring, then it is not singly generated, so is not cyclic.

Original answer:

Yes. $\mathbb{Z}/6\mathbb{Z}$ contains an isomorphic copy of $\mathbb{Z}/2\mathbb{Z}$ and an isomorphic copy of $\mathbb{Z}/3\mathbb{Z}$. This idea works for any ring with non-prime order.

Updated answer:

Since subrings contain $1$, the method above does not work. However, any of $ \mathbb{Z}_p \times \mathbb{Z}_q$, $\mathbb{Z}_p \oplus \mathbb{Z}_q$, $ \prod_{i=2}^n \mathbb{Z}_i$, or $ \bigoplus_{i=2}^n\mathbb{Z}_i$, would work rather transparently. (Need to be careful in the product or direct summation that only a finite number of distinct finite characteristics appear in the list of multiplicands or summands, otherwise the resulting object has characteristic zero. If, however, the "finite spectrum of characteristics" condition is satisfied, the product or direct sum can be extended to an infinite index set.)

Update to the update: This is still wrong. Too many habits from the category of rings that may or may not contain $1$.

Answer 3

If $p$ and $q$ are coprime and greater than 1, $\mathbb Z_{pq}$ does contain a copy of $\mathbb Z_q$.

Proof: Bezout's identity says that $$ap-bq=1$$ for some positive integers $a,b$ and $a<q$. Say that $\mathbb Z_{pq}=\{0,1,\ldots,pq-1\}$. We shall see that the subset $A=\{0,ap,2ap,(q-1)ap\}$ is a subring. The products are computed in $\mathbb Z_{pq}$, of course.

$A$ is clearly a subgroup, and as a group is isomorphic to $\mathbb Z_q=\{0,1,\ldots,q-1\}$, the isomorphism being the function $f(k\cdot ap)=k$. So we must show that $f$ is also an isomorphism of rings. Since $$f(jap\cdot kap)=f(jkap\cdot ap)=f(jk(bq+1)\cdot ap)=f(jkabpq+jkap)=f(jk\cdot ap),$$ $f$ is an isomorphism of rings.

The statement doesn't hold if we remove the condition that $p$ and $q$ are coprime. Take $\mathbb Z_8$. If it contained a (proper) subring, there would be an element $x\neq 0,1$ such that $x^2=x$, that is, $x(x-1)=0$. Since $x$ or $x-1$ are odd, the other must be multiple of $8$, and that implies that $x=0$ or $x=1$, a contradiction.