Find all positive integer $n$ such that $\mathbb{Z}_{n}$ contains a subring isomorphic to $\mathbb{Z}_{2}$

A subring a subset of the ring such that it is closed under addition, contains the identity of the ring, closed under multiplication, and closed under additive inverses.

So we want a subring isomorphic to $(\mathbb{Z}_2,+)$, then $n = 2k$. The identity of $\mathbb{Z}_{2k}$ is $0$ since $a+0 = 0+a = a$. How do we describe the inverses for the subring?

  • $\begingroup$ in the last paragraph, shouldn't it be $\mathbb{Z_{2k}}$? in that case, the subring should be $ \{0,k]$ where $0$ is the identity and $k$ is its own inverse (under addition). $\endgroup$ – Shreya Dec 14 '15 at 12:16
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    $\begingroup$ If a subgroup of $\mathbb{Z}_n$ contains $1$, then it is equal to $\mathbb{Z}_n$. Can you conlcude now? $\endgroup$ – Crostul Dec 14 '15 at 12:16
  • $\begingroup$ @shrey I'm not familiar with what ${0,k]$ means. What is meant by the bracket? Crostul, you mean, if a subgroup of $\mathbb{Z}_{n}$ contains the identity element of $\mathbb{Z}_{n}$, then it is isomorphic to $\mathbb{Z}_{n}$? $\endgroup$ – abstractGone Dec 14 '15 at 12:24
  • $\begingroup$ There are two possible questions here, and it might be worthwhile to confirm which flavor is appropriate. Suppose that $R$ is a ring with identity and $S$ is a subring. Sometimes subrings are required to have the same identity element as the original ring, so that the identity of $S$ is the identity of $R$. On the other hand, sometimes subrings are just subsets which are also rings (with the same operation). In this case, the identity in $S$ (if it exists) does not need to be the same as the identity in $R$. $\endgroup$ – Michael Burr Dec 14 '15 at 12:24
  • $\begingroup$ @shrey Both ends should be curly brackets. It should read $\{0,k\}$. $\endgroup$ – Michael Burr Dec 14 '15 at 12:25

There is some confusion in the original posting. The question asked for a subring $S$ of $R=\mathbb{Z}_n$ that included the identity. Through the comments, it appears that the OP wanted the additive identity to be included in the subring, not the multiplicative identity. I will provide discussion of both ideas here.

If $S$ must contain the multiplicative identity, $1$, then $S$ contains all sums of $1$ with itself, in other words, all the elements of $R$. Hence, if $1\in S$, then $S=R$. The only way for $S\simeq\mathbb{Z}_2$ is if $R=\mathbb{Z}_2$.

If $S$ does not need to contain the multiplicative identity, then for $S$ to be isomorphic to $\mathbb{Z}_2$, then $S$ must have two elements. It has the $0$ of $R$ because $S$ is a ring and it has one other element, which we'll call $g\not=0$. Since $S\simeq\mathbb{Z}_2$ and $S$ is a subring of $\mathbb{Z}_n$, it must he that $g+g\equiv 0\pmod n$ and $g\cdot g\equiv g\pmod n$.

Since $g+g\equiv 0\pmod n$ and $g\not=0$, we know that $n\mid 2g$ and that $2\mid n$. Therefore, $n=2k$ for some $k$. Assuming, without loss of generality that $0<g<n$, then $2g<2n$, so the only way for $n$ to divide $2g$ is if $n=2g$. In this case, we know that $g\equiv k=\frac{n}{2}\pmod n$.

Since $g\cdot g\equiv g\pmod n$, $g\not=0$, $n=2k$, and $g\equiv k\pmod n$, we consider $k^2$. If $k$ is even, then $n=2k\mid k^2$, which cannot happen because $g\cdot g\equiv g\not\equiv 0\pmod n$. Therefore, $k$ is odd. Since $k$ is odd, $k=2l+1$ for some integer $l$, then $k^2=k(2l+1)=2kl+k=nl+k\equiv k\pmod n$. Therefore, when $k$ is odd, $k^2\equiv k\pmod n$.

Combining all of this, we have that when $n=2k$ where $k$ is odd. $R=\mathbb{Z}_n$ has the subring $S=\{0,k\}$, which is isomorphic to $\mathbb{Z}_2$.

  • $\begingroup$ Thanks for clearing up my confusion! I understand now. $\endgroup$ – abstractGone Dec 14 '15 at 14:01

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