Let $M|F$ be a normal extension. Let $a,a'\in M$ be roots of minimal polynomial $min(F,a)$ and $b,b'\in M$ be roots of minimal polynomial $min(F,b)$. Does there exist $\sigma\in Gal(M|F)$ such that $\sigma(a)=a'$ and $\sigma(b)=b'$?
My attempt: My intuition says that it is not possible. I tried with $\mathbb{Q}(\sqrt{2},\sqrt{3})$. But since here we have maps , those interchange the roots of $min(\mathbb{Q},\sqrt{2}),min(\mathbb{Q},\sqrt{3})$ and every element of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is of the form $a+b\sqrt{2}+c\sqrt{3}$ there is no chance of getting a counter-example. I also tried with some other finite extension. Since any finite normal extension is separable for $char~F=0$ and hence simple, i think there is no chance of getting a counter-example here also.
I'm stuck with this problem for 3 days! can anyone please give some hint for proving it or some example for disproving it?
Here is a very explicit example: take $F = \mathbb Q$ and let $M$ be the splitting field over $\mathbb Q$ of any irreducible cubic whose discriminant is a square in $\mathbb Q$. ${\rm Gal}(M:F)$ is then the alternating group $A_3$, acting by permutation on the roots $\alpha, \beta, \gamma$ of the cubic. Now take $a = \alpha, a' = \beta$ and take $ b = \beta, b' = \alpha$, so $a$ and $b$ both have minimum polynomial equal to our chosen cubic. Since $(12) \notin A_3$, there is no $\sigma \in {\rm Gal}(M:F)$ such that $\sigma(a) = a'$ and $\sigma(b) = b'$.