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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!

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  • $\begingroup$ looks good to me $\endgroup$ – zoidberg Dec 9 '18 at 19:04
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    $\begingroup$ 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. $\endgroup$ – Dietrich Burde Dec 9 '18 at 19:17
  • $\begingroup$ Try subrings of a polynomial ring over a field, $k[x]$ $\endgroup$ – R.C.Cowsik Dec 10 '18 at 6:48

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