2

Suppose that $f (x)$ and $g(x)$ are irreducible over $F$ and that $\deg f (x)$ and $\deg g(x)$ are relatively prime. If $a$ is a zero of $f (x)$ in some extension of $F$, show that $g(x)$ is irreducible over $F(a)$.

I understand that $f(a) = 0$ implies that $[F(a):F] =$ deg $f(x)$ and that all elements of $F(a)$ are of the form $c_{n-1}a^{n-1} + \dots + c_0$, but from here I can't to find a way to relate the coprimeness to show that $g(x)$ can only factored as a unit and a polynomial from $F(a)[x]$.

Anyone have any ideas?

user26857
  • 52,094
Oliver G
  • 4,792
  • 1
    Let $b$ be a root of $g$ in some extension of $F(a)$ (or in the algebraic closure of $F$, if you like it). Then $[F(a,b):F]=[F(a):F][F(b):F]=\deg f\deg g$ (since they are coprime), so $[F(a,b):F(a)]=\deg g$. – user26857 Sep 24 '16 at 18:30
  • @user26857 Why is $[F(a,b):F]=[F(a):F][F(b):F]=\deg f\deg g$? I see that $\deg g \deg f$ divides $[F(a,b):F]$ because of coprimality, but how did you get them to be equal? – Oliver G Sep 24 '16 at 18:41
  • 1
    @OliverG You can have a look as http://math.stackexchange.com/questions/1391210/let-u-v-be-algebraic-over-the-field-k-such-that-kuk-n-and-kvk/1391233#1391233 – mathcounterexamples.net Sep 24 '16 at 19:37

0 Answers0