# Some counterexamples in basic ring theory

Give an example if possible, and if not possible explain why not.

a) A subring of a PID that is not PID.

b) A PID that is a subring of a non-PID.

c) A subring of a PID that is not UFD.

My approach:

a) Since $$\mathbb{Q}$$ is field then $$\mathbb{Q}[x]$$ is PID. But it's subring $$\mathbb{Z}[x]$$ is not PID because the ideal $$(2,x)$$ in $$\mathbb{Z}[x]$$ is not principal.

b) Consider the subring $$\mathbb{Z}$$ of the ring $$\mathbb{Z}[x]$$. We know that $$\mathbb{Z}$$ is euclidean domain and hence is PID.

c) Consider the subring $$\mathbb{Z}[\sqrt{-5}]$$ of the ring $$\mathbb{Q}[\sqrt{-5}]$$. I know that $$\mathbb{Z}[\sqrt{-5}]$$ is not UFD, but $$\mathbb{Q}[\sqrt{-5}]\cong \mathbb{Q}[x]/(x^2+5)$$ and since $$x^2+5$$ is irreducible in $$\mathbb{Q}[x]$$ then it's field $$\Rightarrow$$ $$\mathbb{Q}[\sqrt{-5}]$$ is also field $$\Rightarrow$$ is PID.

Are my examples correct?

Would be very grateful!

• looks good to me – zoidberg Dec 9 '18 at 19:04
• Each question has been answered at this site already. You could also compare. For example, b) is here. For a), see here. Actually, you have the same example. – Dietrich Burde Dec 9 '18 at 19:17
• Try subrings of a polynomial ring over a field, $k[x]$ – R.C.Cowsik Dec 10 '18 at 6:48