Q:

algebraic conjugates of a sum

Given algebraic numbers $a$ and $b$, is it the case that all algebraic conjugates of $a+b$ take the form $a'+b'$ where $a'$ and $b'$ are algebraic conjugates of $a$ and $b$ respectively?

A:

There is a characterization of conjugates by means of morphism of fields.
Let $K$ be a finite normal extension of $\Bbb Q$, and $a \in K$. Then, all conjugates of $a$ are
$$\{ \sigma (a) : \sigma \in \mathrm{Gal} (K / \Bbb Q) \}$$
Now, if you have two algebraic numbers, say $a,b$, consider the normal closure of $\Bbb Q (a,b) / \Bbb Q$, and call it $K$. Then the conjugates of $a+b$ have the form
$$\sigma (a+b)= \sigma (a) + \sigma (b)$$
where $\sigma \in \mathrm{Gal} (K / \Bbb Q)$. This means that the conjugates of $a+b$ are exactly sums of conjugates of $a$ and $b$.
Let's make an example. The conjugates of $\sqrt 2 + \sqrt 3$ are exactly
$$\sqrt 2 + \sqrt 3 , \ \sqrt 2 - \sqrt 3 , \ -\sqrt 2 + \sqrt 3 ,  \ -\sqrt 2 - \sqrt 3$$

