Determine whether or not there is an automorphism $\sigma \in Gal(M/F)$ with $\sigma (a) = a'$ and $\sigma(b) = b'$.

Question

Let $M$ be a normal extension of $F$. Suppose that $a, a' \in M$ are roots of $min(F, a)$ and that $b, b'$ are roots of $min(F, b)$. Determine whether or not there is an automorphism $\sigma \in Gal(M/F)$ with $\sigma (a) = a'$ and $\sigma(b) = b'$.

Difficulty in figuring out a counter example.

Answer 1

There are lots of counterexamples for the case $b=a'$, $b'=a$. For example, if $M$ is a Galois extension of degree $3$, then it is impossible for any element to have order $2$ under the action of the Galois group.

For example, let $M$ be the splitting field of $X^3-3X-1$. Then we cannot swap two of the roots of this polynomial.

If you want an example where all four elements are distinct, consider a cyclic Galois group of order $4$.

Answer 2

There are many simple counterexamples, for example consider $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$. Then for $a = \sqrt{2}$ and $b = \sqrt{2}+1$ we have $$\min(F, \sqrt{2}) = X^2-2,$$ and $$\min(F, \sqrt{2}+1) = X^2-2X-1.$$ Now, note that any $\sigma \in Gal(\mathbb{Q}(\sqrt{2})/\mathbb{Q})$ is determined by $\sigma(\sqrt{2})$. Thus for each $a' = \sigma(a)$, there's a unique $b' = \sigma(b)$. In general case, a similar argument works for any $a$ and $b = a+1$.