To check a subset of a given ring is a subring,
is it enough to check that the subset is closed under induced operations(multiplication and addition) or
do I also need to show that it contains 0 and additive inverses of each element?
[Math] How to check a set of ring is a subring
abstract-algebra
Related Solutions
If your book defines (like most books do I think) rings as unitary rings, in other words requires them to contain a neutral element for multiplication, then ideals are almost never subrings. Indeed, for unitary rings one requires subrings of$~R$ to contain the element $1\in R$ (rather than some element that is neutral for mulitplication restricted to the subring), and an ideal that contains $1$ must contain its multiples by elements of $R$, which gives all of $R$; the only ideal of $R$ that is a subring is $R$ itself.
If on the other hand your book says nothing of neutral elements for multiplication, then it deals with a larger category of structures that are often called "rngs", to indicate the (possible) lack of $1$. Then indeed all ideals are subrngs.
Finally I note that sometimes a non-trivial ideal can be considered a ring in itself by choosing a different element to be $1$ (this was the reason for my parenthesised remark above). For instance in $\def\Z{\mathbf Z}\Z/10\Z$ you can make the ideal of "even" classes into a ring (using the same addition and multiplication) by electing the class of $6$ to be the neutral element for multiplication. This situation arrives typically in rings that can be decomposed as a product of rings; in the current example $\Z/10\Z\equiv (\Z/5\Z)\times(\Z/2\Z)$ by the Chinese remainder theorem. The factors in such a decomposition are naturally quotients of the original ring, but also correspond to an ideal of that ring (namely the kernel of projecting to the other factor(s)); such an ideal is not a subring, but can be made into a ring by choosing the projection of $1$ onto the ideal as new neutral element.
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 $\mathbb{Z}_2$.
Best Answer
Let's denote this set by $S$. It's a subring if: