# Subring of a commutative ring with unity implies ring is an integral domain

Let $$R$$ be a commutative ring with unity. If a subring $$S$$ of $$R$$ is an integral domain containing the unity of $$R$$ (that isn’t $$\{0,1\}$$), does this imply that $$R$$ is too an integral domain? I tried to find a counterexample but I did not find one:

I figured maybe the subring of $$Z_{10}$$, $$(2)$$, would work since I figured it would be isomorphic to $$Z_5$$ (and thus an integral domain), but then I realized it has no multiplicative identity so that’s a no.

Any help would be appreciated.

• @AtticusStonestrom this doesn't contain the unit of $\mathbb Z\times\mathbb Z$. You could instead look at the diagonal inside $\mathbb Z\times\mathbb Z$.
– Dave
Commented Sep 10, 2021 at 1:32
• @Dave oops, you are of course right! am running on v little sleep lolol Commented Sep 10, 2021 at 2:01
• @AtticusStonestrom it's no problem, you're example is essentially right: you want to look at a copy of $\mathbb Z$ inside $\mathbb Z\times\mathbb Z$; it's just a small detail.
– Dave
Commented Sep 10, 2021 at 2:14
• In itself the ideal of $\Bbb Z/10\Bbb Z$ generated by the class of $2$ (which is I think what you meant be "$(2)$"), equipped with the restricted arithmetic operations, does have a multiplicative identity, namely the class of $6$. Nonetheless it is not a subring according to most texts, since (for them) by definition a (unitary) subring must share its mutliplicative identity with the containing ring. Commented Sep 10, 2021 at 9:42
• Every single algebra over a field contains a copy of that field... but they're not all domains... Commented Sep 10, 2021 at 18:36

Another good class of rings for examples is polynomial rings and their quotients. For this problem, consider the subring $$\mathbb C$$ inside $$\mathbb C[x]/\langle x^2\rangle$$.
If the implication of your title were true, it would be impossible to extend an integral domain with zero divisors. But that clearly is possible. For instance in $$\Bbb Z[X]$$ if $$P,Q$$ are any non-constant polynomials, their images in $$\Bbb Z[X]/(PQ)$$ will be zero divisors, but the quotient ring still contains (the isomorphic image of) the integral domain$$~\Bbb Z$$. So no, the implication does not hold.