Thread: Borel field Algebra

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

" alt="\mathbb{F}/(\;r(x)\" /> is a field so, it has only two ideals:


\quad (ii)\quad\mathbb{F}/(\;r(x)\=(\;1+(r(x))\" alt="(i)\quad\{0+(r(x))\}=(\;0+(r(x))\\quad (ii)\quad\mathbb{F}/(\;r(x)\=(\;1+(r(x))\" /> .

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.