As stated in the title, given a principal ideal domain $R$, is the polynomial ring $R[x]$ necessarily a principal ideal domain? In particular, is the polynomial ring $(\mathbb{Z}[i])[x]$ over the Gaussian integers a principal ideal domain?

I have tried the usual method of showing this (choose an ideal $I$ of $R[x]$ and some $p(x) \in R[x]$ with the lowest degree and show that $p(x)q(x) \in I$ implies that $q(x)$ is a multiple of $p(x)$). However, I can't seem to get anywhere with this. Any help would be appreciated.

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    $\begingroup$ $R[X]$ is a PID iff $R$ is a field. $\endgroup$ – user26857 Dec 6 '15 at 8:48

In $\mathbb{Z}[x]$, the ideal generated by $2$ and $x$ is not principal. You can produce a similar example over the Gaussian integers.

  • $\begingroup$ So, for instance, could we say that $I=\langle 1+i,x\rangle$ is not principal, since $I = (1+i)f(x) + xg(x)$ for $f,g \in (\mathbb{Z}[i])[x]$ and if $I = \langle f(x) \rangle$, we have if $f(x)$ is constant, every polynomial has coefficients a multiple of $1+i$ and so we do not get $x$, while if $\deg(f) \geq 1$, every polynomial has degree at least 1 and so we do not get $1+i$? Is this a valid argument? $\endgroup$ – mathsman Dec 6 '15 at 7:19
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    $\begingroup$ Yes, exactly. Of course $1+i$ can be replaced by many constants, anything but a unit. $\endgroup$ – André Nicolas Dec 6 '15 at 7:26
  • $\begingroup$ Thank you very much for your comment! I was so stuck on trying to show that it worked that I didn't stop to think that it didn't. $\endgroup$ – mathsman Dec 6 '15 at 7:27

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