Let $a$ and $b$ belong to some extension of $F$ and let $b$ be algebraic over $F$. Prove that $[F(a, b):F(a)] \leq [F(a, b):F]$.

Question

I need help with this excercises

Let $a$ and $b$ belong to some extension of $F$ and let $b$ be algebraic over $F$. Prove that $[F(a, b):F(a)] \leq [F(a, b):F]$.

I try, How $b$ is algebraic exist polynomial $f(x)$ such that $f(b)=0$

Then $[F(b):F]=deg(f(x))$

o could use that: $[F(a,b):F]=[F(a,b):F(b)][F(b):F]$ then

$[F(b):F] \leq [F(a,b):F] $

Answer 1

You don't need anything about algebraic extensions for this problem. If $K \subseteq E \subseteq L$ are any fields whatsoever, then $[L : K] = [E : K][L : E]$ (even for infinite cardinalities), so of course $[L : E] \leq [L : K]$.

Here $[L : K]$ means the dimension of $L$ as a vector space over $K$.

Just take $K = F, E = F(a), L = F(a,b)$.

Answer 2

Yes, just use $[F(a,b):F]=[F(a,b):F(b)][F(b):F]$. Since $[F(b):F]\geqslant1$, $[F(a,b):F(b)]\leqslant [F(a,b):F]$