Describe all subrings of the ring of integers I know that every subset $S \subseteq \mathbb{Z}$ is is a subring if it is a ring under the operations defined on the ring $(\mathbb{Z},+,*)$ 
that is, $( S ,+)$ is an abelian subgroup of $( \mathbb{Z,+} )$ and $\forall \  x,y,z \in S $ we have that
$$x*y \in S \  \text{closure} $$ $$ x*(y*z) = (x*y)*z \text{ associativity } $$ and $$ x*(y+z) = x*y+x*z \ , (y+z)*x = y*x+z*x  \ \text{distributivity}$$
I have read on this  website https://www.quora.com/How-to-describe-all-the-subrings-of-the-ring-of-integers that is enough to describe all subgroups of integers under addition, I think that is because if $x,y,z \in S \subseteq Z$
and $(\mathbb{Z},+,*)$ is a ring then we have that associativity and distributivity holds( Am I right? ). But what about closure? 
or, why is enough to describe all additive subgroups of $(\mathbb{Z},+)$ ? 
Thanks for your help!
 A: I am going to admit non-unital subrings of $\Bbb Z$ to my discussion; if every subring contains $1$, then every subring is equal to $\Bbb Z$ and there is not much more to be written.
I claim that if
$S \subseteq \Bbb Z \tag 1$
is an additive subgroup, then it is of the form
$S = s \Bbb Z, \; \tag 2$
for some $s \in \Bbb Z$; for if
$S \ne \{ 0 \}, \tag 3$
there is some $0 \ne s \in S$; since $S$ is a subgroup,
$s \in S \Longleftrightarrow -s \in S, \tag 4$
so we may without loss of generality assume that
$s > 0; \tag 5$
since $S$ has positive elements, it has a smallest such; we may assume that $s$ is the same.  Then clearly
$ns \in S, \; \forall n \in \Bbb Z; \tag 6$
this may be seen by simply adding $s$ or $-s$ to itself $n$ times.  Thus
$s\Bbb Z \subset S; \tag 7$
now if there is some 
$t \in S \setminus s \Bbb Z, \tag 8$
we may let $m \in \Bbb Z$ be the largest integer with
$ms < t; \tag 9$
then
$0 < t - ms < s; \tag{10}$
we cannot have $t - ms = 0$ or $t - ms = s$ since then $t \in s \Bbb Z$; but since $t \in S$, 
$t - ms \in S, \tag{11}$
which contradicts the hypothesis that $s$ is the smallest positive element of $S$.  Therefore (2) binds.  
Since every subring of $\Bbb Z$ is also an additive subgroup, we have shown that every subring of $\Bbb Z$ is of the form $s \Bbb Z$, which sets are clearly closed under addition and multiplication.
Also, every additive subgroup $S$, being of the form $S = s \Bbb Z$, is a subring as well since it is multiplicatively closed:  if $as, bs \in S$, then
$(as)(bs) = abs^2 = (abs)s \in s \Bbb Z$; the other ring axioms, associativity, commutativity, etc.,  are simply inherited from the ring $\Bbb Z$.
For the same reasons, that every subring is an additive subgroup etc., it suffices to find the additive subgroups of $\Bbb Z$.
I guess it is worth pointing out the the sets $S = s \Bbb Z$ are also the ideals of the principal ideal domain $\Bbb Z$.
A: All the subgroups of $\mathbb{Z}$ have the form $m\mathbb{Z}$ when $0\leq m\in\mathbb{Z}$.  It is pretty easy to see that every such subgroup is a subring. If $x,y\in m\mathbb{Z}$ then you can write $x=mp,y=mq$ when $p,q\in\mathbb{Z}$. And then:
$xy=mpmq=m^2pq=m(mpq)\in m\mathbb{Z}$
So $m\mathbb{Z}$ is closed under multiplication. 
