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Let $E$ be a field extension of the field $F$. Let $a,b \in E$ and suppose that the minimal polynomial of $a$ over $F$ has degree m, the minimal polynomial of $b$ over $F$ has degree n and that gcd($m,n$)$=1$. Show that $[F(a,b):F]=m\cdot n$.

Here's my attempt at a solution, I just need some clarification:

We have that $[F(a,b):F]=[F(a,b):F(a)][F(a):F]$. Furthermore we are given that $[F(a):F]=m$. Now suppose that $b \in F(a)$, then deg(b,F)|deg(a,F). However we are given that gcd(m,n) = 1 and thus $b$ cannot be in F(a). My question is why does this mean (or does this even mean?) that $[F(a,b):F(b)] = n$ (i.e. why cant $[F(a,b):F(b)] < n$)?

Zachary Luety
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1 Answers1

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We have $[F(a,b): F] = [F(a,b): F(a)] [F(a):F]$, hence $m$ divides $[F(a,b) : F]$. Similarly $n$ also divides $[F(a,b):F]$. But then $m$ and $n$ divide $[F(a,b):F]$, hence $mn$ also divides it since $\mathrm{gcd}(m,n) = 1$. Therefore $mn \le [F(a,b):F] \le mn$ (the last inequality being because the set of all $a^i b^j$ span $F(a,b)$ as an $F$-vector space).

Hope that helps,

Patrick Da Silva
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