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!


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

$ms < t; \tag 9$


$0 < t - ms < s; \tag{10}$

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 $\Bbb Z$.


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.

  • 2
    $\begingroup$ Unless, of course, your definition of subring includes having the same unit. $\endgroup$
    – Lubin
    Sep 23 '18 at 21:48
  • $\begingroup$ By definition a ring might not have a multiplication identity. It's not a field. Anyway, I followed OP's definition in my answer. $\endgroup$
    – Mark
    Sep 23 '18 at 21:50

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