For extension fields, does $[F(a,b):F(a)]=[F(b):F]$?

Question

sorry if this question seems obvious,

For a field F and $a,b\notin F$, does $[F(a,b):F(a)]=[F(b):F]$?

If so, how do you prove it or is there a counter example?

Answer 1

No, let $F=\Bbb Q$, $a=\sqrt[3]{2}, b=\zeta_3\sqrt[3]{2}$

Then $[F(a,b):F(a)]=2$, but $[F(b):F]=3$.

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This is because both are distinct roots of the irreducible polynomial $x^3-2$ over $\Bbb Q$, so they both generate degree $3$ extensions, but then over $F(a)$ the polynomial has a single root, so the degree goes down to at least $2$, so that $[F(a,b):F(a)]\le 2$ but $[F(b):F]=3$ is clear. (In this way you don't even need to explicitly compute the degree either!) And it also shows that even if you strengthen your assumption to saying $a\ne b$ it's still hopeless.

Answer 2

The easiest example you can think of gives a counter example : take $a=b$. More generally, you can pick $a,b$ such that $F(a) = F(b)$, so that $F(a,b) = F(a) = F(b)$ and you get more counter examples.

Explicitly, take $a \notin F$ and for $b$, take a polynomial in $a$ with coefficients in $F$ of degree less than the minimal polynomial of $a$ over $F$. For instance, if $a = i$ and $F = \mathbb Q$, you can take $b = 1+i$.

However, under certain conditions you can recover the statement. Your statement is actually equivalent to $$ [F(a,b) : F] = [F(a):F][F(b):F] $$ (think about this). To obtain this equation for particular values of $a$ and $b$, you can choose them for instance such that $([F(a):F],[F(b):F]) = 1$. An example would be with $\sqrt 2$ and $\sqrt[3]2$ over $\mathbb Q$.

Hope that helps,

Answer 3

For a counterexample, take $a=b$.