# 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 $$(\mathbb{Z},+)$$ ?

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 $$\Bbb Z$$.

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.

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