TITLE: A Galois theory sanity check about conjugates.
QUESTION [5 upvotes]: Here is my question...
If $L/K$ is an algebraic extension and $\alpha,\beta \in L$ are $K$-conjugates (that is, they have the same minimal polynomial), is it always true that there  exists some $\sigma \in $ Aut$(L/K)$ such that $\sigma(\alpha)=\beta$?
I have thought about this for a while and can neither come up with a proof nor a counterexample :( Of course this fails if the algebraicity condition is removed: consider $\mathbb{R}/\mathbb{Q}.$
Any hints will be much appreciated!

REPLY [7 votes]: Consider the equation $(x^2+x-i)(x^2+x+i)=0$ over $K=\mathbb{Q}$.
Now take $L=\mathbb{Q}(i, \sqrt{1+4i})$. That is I have added the roots of one of the factors but not the other. (Of course one must check that $\sqrt{1-4i} \not \in L$). In any case $i\mapsto -i$ sends an element to its conjugate. It also switches the two factors in the above equation. It cannot be extended since only one of these factors has a root, in this field.
This is a counterexample with $\alpha =i$ and $\beta =-i$