prove that any F[x]/ <r(x)>, r(x) being irreducible is a PID

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- Apr 2nd 2011, 01:53 AMkarlito03Borel field Algebra
prove that any F[x]/ <r(x)>, r(x) being irreducible is a PID

- Apr 2nd 2011, 03:21 AMFernandoRevilla
$\displaystyle \mathbb{F}/(\;r(x)\;)$ is a field so, it has only two ideals:

$\displaystyle (i)\quad\{0+(r(x))\}=(\;0+(r(x))\;)\quad (ii)\quad\mathbb{F}/(\;r(x)\;)=(\;1+(r(x))\;)$ . - Apr 3rd 2011, 02:45 AMDeveno
and any field is a PID. perhaps what you meant to ask is: prove that if F is a field, and thus F[x] is a PID, then F[x]/<r(x)> is a field when r(x) is irreducible.

the proof might go something like r(x) irreducible --> <r(x)> is maximal, so F[x]/<r(x)> has no proper ideals.