$a,b$ algebraic over $F$, $([F(a):F], [F(b):F]) = 1 \implies m_b(X)$ irreducible in $F(a)[X]$.

Question

Let $a,b$ be two algebraic numbers over $F$ and suppose that $\deg f$ is relatively prime to $\deg g$, where $f = \text{Irr}(a,F,X), \ g = \text{Irr}(b,F,X)$ are the minimal monic irreducible polynomials of the two numbers respectively. Then $g$ as a polynomial in $F(a)[X]$ is still irreducible.

I would think the proof would have something to do with $[F(a,b) : F(a)][F(a):F] = [F(a,b) : F] = [F(b,a):F(b)][F(b):F]$ but I don't see how to apply it.

Hints please.

Answer 1

$([F(a):F],[F(b),F])=1$ implies that $lcm([F(a):F],[F(b),F])=[F(a):F]\cdot [F(b),F]$.

Now suppose, $m_b(X)$ is not irreducible in $F(a)[X]$, then $[F(a,b),F(a)]<deg(m_b(X))$. But then, $[F(a,b),F]$ is a common multiple of $[F(a):F]$ and $[F(b),F]$, and $[F(a,b),F] <[F(a):F]\cdot [F(b),F]$. Contradiction with $lcm([F(a):F],[F(b),F])=[F(a):F]\cdot [F(b),F]$.

Answer 2

You're on the right track.

$[F(a,b) : F]$ is a multiple of both $[F(a):F]$ and $[F(b):F]$ and so $[F(a,b) : F]\ge\operatorname{lcm}([F(a):F],[F(b):F])=[F(a):F][F(b):F]$.

On the other hand, $b$ satisfies an equation of degree $[F(b):F]$ with coefficients in $F$ and so a fortiori with coefficients in $F(a)$.

This means that $[F(a,b) : F(a)]\le [F(b):F]$ and so $[F(a,b) : F] = [F(a,b) : F(a)][F(a):F] \le [F(b):F] [F(a):F]$.

Thus $[F(a,b) : F(a)] = [F(a):F][F(b):F]$ and so $[F(a,b) : F(a)]= [F(b):F]$, which implies the result.

Answer 3

Thanks @Nitrogen & @lhf.

• $[F(a,b):F]$ is a multiple of both $[F(a):F]$ and $[F(b):F]$ and so is $\geq \text{lcm}(\text{the two})$.
• If $g$ were not irreducible in $F(a)[X]$ then $[F(a,b):F(a)] < [F(b):F]$.
• So $[F(a,b):F] = [F(a,b):F(a)][F(a):F] < [F(b):F][F(a):F]$, which contradicts the first bullet.

Yay!