-1

Let $K$ be a field and $K[x]$ be the ring of polynomials over $K$ in a single variable $x$ for a polynomial $f$ belong to $K[x]$. Let $(f)$ denote the ideal in $K[x]$ generated by $f$. show that $(f)$ is a maximal ideal in $K[x]$ if and only iff $f$ is an irreducible polynomial over $K$.

DrinkingDonuts
  • 1,257
  • 6
  • 18
dev
  • 1
  • And what tools are you allowing yourself to use? Do you have Euclidean division, or do you want to prove that for yourself along the way? – Lubin Jul 23 '17 at 13:31
  • In this link you will find useful stuff. – DrinkingDonuts Jul 23 '17 at 13:33
  • For $"\rightarrow"$, let $f$ generate a maximal ideal of $K[X]$. Try a proof by contradiction by considering $f=pq$.

    For $"\leftarrow"$, use the definition of maximal ideal.

     – Joe Jul 23 '17 at 13:35
  • @AlgebraicallyClosed next time you edit, consider to include more than one tag. – Xam Jul 23 '17 at 17:29

1 Answers1

0

Suppose $(f)$ maximal and suppose $f=gh$ for certain polynomial $g,h$. In particular, $(f)\subset (g)$ and by maximality of $(f)$, you get $(f)=(g)$, i.e. there is $k$ s.t. $kf=g$ and thus $$kgh=g\implies (kh-1)g=0\implies kh=1,$$ because $g\neq 0$, and thus $h$ is a unit. Therefore $f$ is irreducible.

Conversely, if $(f)$ is irreducible, then $K[X]/(f)$ is a field and thus $(f)$ is maximal.

Surb
  • 57,262
  • 11
  • 68
  • 119