2
$\begingroup$

We know that an integral domain is a commutative ring with unity and no zero-divisors. It is obvious that if $R$ is an integral domain and $S$ is a subring of $R$ that $S$ must also be commutative, and if $a,b\in S$ and $ab=0$, then $a=0$ or $b=0$. But I'm having trouble proving that $S$ has unity, does the unity of $R$ have to be the unity of $S$?

Also what's a good counterexample for a subring of a field that is not a field.

$\endgroup$
  • $\begingroup$ A subring WITH unity of an integral domain is a integral domain $\endgroup$ – L.F. Cavenaghi Nov 14 '15 at 16:28
2
$\begingroup$

Usually one requires a subring of a unital ring to contain the unit. If you remove this requirement, the result does not hold. For example, $\mathbb{Z}$ is an integral domain, but if we do not require subrings to contain the unit, $2\mathbb{Z}$ is a subring which is not an integral domain.

As for your second question, $\mathbb{Z}$ is a subring of the field $\mathbb{Q}$, but is not itself a field.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.