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

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.

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Any subring must be an additive subgroup! – Sigur Nov 9 '12 at 0:53
With my assumption about what you define to be a subring, you should also include the case $n=0$; i.e. $S= \{0\}$ is a subring. Or should I say subrng... – Jason Polak Nov 9 '12 at 1:03
You can prove this using the division algorithm and the well ordering principle. The general proof is just Jason Polak's sketch made rigorously. The if part is easy. The only if part is obtained by arguing by contradiction as he does. – Pedro Tamaroff Nov 9 '12 at 1:31

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?

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Hint: consider a minimal positive element in the subring.

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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 $\rm\:m_{\:1}.$

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