Let $F$ be a field and $F[x]$ the univariate polynomial ring over $F$. Is it the case that every quotient ring ${F[x]}/{(p(x))}$, where $p(x) \in F[x]$, is a PID (principal ideal domain)?

This is just a minor issue that's been bothering me for the last couple of hours. I'm aware that when the ideal $(p(x))$ is prime (i.e. $p(x)$ irreducible) then ${F[x]}/{(p(x))}$ is a PID.

  • $\begingroup$ By an isomorphism theorem, it's easy to see that if $R$ is a commutative principal ideal ring then so is $R/I$ for every ideal $I\lhd R$. The quotient is additionally a domain if the ideal happens to be prime. $\endgroup$ – rschwieb Nov 15 '13 at 13:59

If $p(x)$ isn't irreducible (and isn't the zero polynomial), then $F[x]/(p(x))$ isn't an integral domain, much less a PID. (And when $p(x)$ is irreducible, then $F[x]/(p(x))$ is a field.)

  • $\begingroup$ Thanks, Andreas. But what about the quotient ring $S = \mathbb{Z}_3[x]/(x^4+x^3+x^2)$ for example? Are the ideals in S not all principal i.e. $(1), (x), (x^2), (x^2+x+1), (x^3+x^2+x), (x^4+x^3+x^2)$? I got miffed reading the first line of text on this webpage: math.uiuc.edu/Software/magma/text318.html $\endgroup$ – RedOrange Nov 15 '13 at 2:29
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    $\begingroup$ Dear @IsaacZebulunBurke, For any polynomial $p(x)\in F[x]$, every ideal of the ring $F[x]/(p(x))$ is principal, but this doesn't mean the quotient ring is a principal ideal domain. You might just call it a principal ideal ring. The quotient is a domain (and therefore a principal ideal domain) if and only if $p(x)$ is irreducible. $\endgroup$ – Keenan Kidwell Nov 15 '13 at 2:32
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    $\begingroup$ Ok, of course I see it now, thanks, S is crawling with zerodivisors! $\endgroup$ – RedOrange Nov 15 '13 at 2:36
  • $\begingroup$ Isaac, the quotient ring is not an integral domain. For instance, $x^2 \cdot (x^2 + x + 1) \equiv 0$, yet both factors are nonzero. $\endgroup$ – Sammy Black Nov 15 '13 at 2:36

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