Existence of certain autormorphism in Normal extensions

Question

Let $F \subset L$ be and normal extension. If $\alpha_1, \alpha_2$ are in $L$ and $p(x) \in F[x]$ are they mininal polynomial, and $\beta_1, \beta_2$ with $g(x) \in F[x]$ his minimal prove that exists $\sigma \in Gal(L,K)$ such that $\sigma(\alpha_1)=\alpha_2$ and $\sigma(\beta_1)=\beta_2$.

First, ($p$ and $g$ roots are distinct) If this roots are equal, the identity map $ id \in Gal(L,K)$ will do. So supose the roots are distinct, then $[L,K]\geq 2$.

I know that if exists, $\sigma$ must take roots of minimal polynomials to another root and since the extension is normal $p,g$ splits over $L$, so the result is easy to see, but my dificult is to see why there must be this autormorphism? why can't $Gal(L,K)= \{id\}$?

Answer 1

It is not clear that such a $\sigma$ will exist in general.

Some obvious problematic cases are:

1. $\alpha_1 = \beta_1$ and $\alpha_2 \neq \beta_2$. This is immediately problematic because if $\sigma(\alpha_1) = \alpha_2$, then $\sigma(\beta_1) = \sigma(\alpha_1) = \alpha_2 \neq \beta_2$.

2. $\alpha_1 = \alpha_2$ and $\beta_1 \neq \beta_2$. Consider @DerekHolt's example in the comments. Take $F = \mathbb{Q}$, $L = F(\sqrt{2})$, $\alpha_1 = \alpha_2 = \sqrt{2}$, $\beta_1 = 1 + \sqrt{2}$ and $\beta_2 = 1 - \sqrt{2}$. Obviously if $\sigma(\alpha_1) = \alpha_2$ then $\sigma(\beta_1) = \beta_1 \neq \beta_2$.

3. $\alpha_1 = \beta_2$ and $\alpha_2 = \beta_1$. If $Gal(L/F)$ is cyclic of degree $3$, for instance, then there cannot exist any $\sigma$ satisfying the conditions we seek.

Even neglecting such scenarios, suppose $\alpha_1,\alpha_2,\beta_1,\beta_2$ are all distinct. It is still not necessary that such a $\sigma$ will exist. For example, consider $p(X) = g(X) = Irr(\zeta,\mathbb{Q},X)$, where $\zeta$ is a primitive $7$th root of unity. We know that $Irr(\zeta,\mathbb{Q},X) = 1 + X + X^2 + \dots + X^6$, and the roots of this polynomial are $\zeta^j$ for $j = 1,\dots,6$.

Let $\alpha_1 = \zeta$, $\alpha_2 = \zeta^2$ and $\beta_1 = \zeta^3$, $\beta_2 = \zeta^4$. If $L = \mathbb{Q}(\zeta)$ is the splitting field of $Irr(\zeta,\mathbb{Q},X)$, then there is no $\sigma \in Gal(L/\mathbb{Q})$ such that $\sigma(\alpha_1) = \alpha_2$ and $\sigma(\beta_1) = \beta_2$. Because, if $\sigma(\zeta) = \zeta^2$, then $\sigma(\zeta^3) = \zeta^6 \neq \zeta^4$.

Perhaps we want that the 4 elements $\alpha_i,\beta_j$ are all distinct, and additionally that $p$ and $g$ are distinct? But even that doesn't work. Let $F = \mathbb{Q}$. Let $p(X) = X^4 - 2X^2 + 9$ and $g(X) = X^2 - 2$. Let $\alpha_1 = \sqrt{2}+i$, $\alpha_2 = \sqrt{2}-i$ and let $\beta_1 = \sqrt{2}$ and $\beta_2 = -\sqrt{2}$. Let $L$ be a splitting field of $\{p,g\}$. There is no $\sigma \in Gal(L/F)$ such that $\sigma(\alpha_1) = \alpha_2$ and $\sigma(\beta_1) = \beta_2$.

So, the only reasonable answer to the question of whether such a $\sigma \in Gal(L/F)$ exists is that it does not exist except under very special circumstances.