# Proof that $\Bbb Z$ has no other subring than itself

I'm asked to prove that the ring of integers $\Bbb Z$ admits no other subring than itself.

I'm no too sure about how to prove it. I started using the minimality argument, but I found a counter-example - which must be wrong but I would like to know why. Actually, I thought to the set of all even number of $\Bbb Z$. This set is not empty, close under addition and multiplication. Where is my mistake in this example, and what could be a rigorous proof of this statement ?

Thanks.

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Are your rings unital, i.e. do they have a multiplicative identity? – Isaac Solomon Oct 26 '13 at 21:44
... in particular, the same multiplicative identity of the overring? – Hurkyl Oct 26 '13 at 21:49
If you don't require an identity or that the identity in the subring is the same as in the ring, then $\{0\}$ is a counterexample. If for a subring the same identity as the ring is required, then the statement is true. – egreg Oct 26 '13 at 22:21

You are correct: $2\mathbb{Z}$ is a non-unital subring of $\mathbb{Z}$.
In fact, even if you know what a ring is, "subring" can have multiple meanings. It is possible to have $A\subset B$, where $A$ and $B$ are both unital rings with the same operations, but without the same unity! So when working with unital rings, it is normal to define a subring in such a way as to ban those examples. So, again, it is very important to make sure that you understand what definitions the author/professor is working with, as subtle variations can completely change what is or isn't true.
For the curious: $2\mathbb{Z}/6\mathbb{Z}$ is a ring, and many would hastily conclude that it has no unity. But it does; its unity is $\overline{4}$! – Slade Oct 26 '13 at 22:24