Prove that $(n\mathbb Z, +, \times )$ are the only subrings of $(\mathbb Z, +, \times)$

Question

I had to find all the subrings of the integers and then prove that there aren't any more. It's clear to me the $(n\mathbb Z, +, \times )$ is a subring of the integers for all $n$ element of the natural numbers, $n>1$, but I'm not sure how to prove that there aren't any more.

Answer 1

I assume you assume that subrings don't have to contain the identity. If you can't figure out the proof right away, you should try an example like this:

A nonzero subring $S$ must have a nonzero element. We can choose the smallest positive one. In our example suppose it is $5$. If $S$ contains a number that is not an integer multiple of five like $12$, then we can do the following: $S$ contains $10$ because it contains $5$, and so it contains $12 - 10 = 2$. But we said $5$ was the smallest positive number in $S$!

Can you generalise this to a proof?

Answer 2

Hint: consider a minimal positive element in the subring.

Answer 3

By the Subring Test, a nonempty subset $\rm\:S\subset \Bbb Z\:$ is a subring iff it's closed under multiplication, and also closed under subtraction (i.e. $\rm\,S\,$ is an additive subgroup, by the Subgroup Test). But in $\Bbb Z$ multiplication reduces to (iterated) addition since $\rm\:n\cdot m = m +\,\cdots\,+m,\:$ with $\rm\,n\,$ summands. Hence subrings are the same as additive subgroups, which have form $\rm\,m\,\Bbb Z\,$ by this fundamental

Lemma $\ \ $ If a nonempty set of positive integers $\rm\: S\:$ satisfies $\rm\ n > m\ \in\ S \ \Rightarrow\ \: n-m\ \in\ S$
then every element of $\rm\:S\:$ is a multiple of the least element $\rm\:m_{\:1} \in S\:.$

Proof $\ \: $ If not there is a least nonmultiple $\rm\:n\in S\:,$ contra $\rm\:n-m_{\:1} \in S\:$ is a nonmultiple of $\rm\:m_{\:1}.$