Show that the sum of two conjugate complete sets is conjugate complete In the proof of the transcendence of $e^a$ for an algebraic nonzero $a$ Burger's Transcendence book gives a very unclear argument that the sum of two conjugate complete sets is a conjugate complete set but I can't understand it at all! For an example each of the two sets ${\{\sqrt{2},-\sqrt{2}}\}$ and ${\{\sqrt{3},-\sqrt{3}}\}$ are conjugate complete since each consists of all zeros of their irreducible polynomial. Consequently sum of the mentioned two sets, which is ${\{\sqrt{2}-\sqrt{3},-\sqrt{2}+\sqrt{3},-\sqrt{2}-\sqrt{3},\sqrt{2}+\sqrt{3} }\}$, is all of zeros of a single irreducible polynomial with integer quotient . So how to prove this in general, that is two sets of zeros of two arbitrary irreducible polynomials? I know a little abstract algebra.
 A: Let $\overline {\mathbb Q} / \mathbb Q$ be the field of algebraic numbers. Then the conjugates of an algebraic number $\alpha$ are just the images of $\alpha$ under automorphisms of $\overline {\mathbb Q}$, and a set $S$ of algebraic numbers is conjugate complete if and only if it is stable under the action of this automorphism group.
If $S$ and $T$ are conjugate complete sets, and $\sigma$ is an automorphism of $\overline{\mathbb Q}/ \mathbb Q$, then for any $s \in S$ and $t \in T$, we have
$$
\sigma(s + t) = \sigma(s) + \sigma(t) \in \sigma(S) + \sigma(T) = S + T ,
$$
and so $S+T$ is also conjugate complete.
Adding in response to questions:
In the context of your original question, we can think of all of this happening within a finite extension of $\mathbb Q$. Let $K$ and $L$ be the fields obtained by adjoining to $\mathbb Q$ all the roots of $f, g \in \mathbb Q[x]$. These are Galois extensions. The roots of $f$ are permuted by $G_K = \operatorname{Gal}(K/\mathbb Q)$, and the conjugates of a root $\alpha$ of $f$ are of the form $\sigma(\alpha)$ for $\sigma \in G_K$.
Similarly, the roots of $g$ are permuted by $G_{L} = \operatorname{Gal}(L/\mathbb Q)$.
Let $\alpha$ be a root of $f$ and $\beta$ be a root of $g$.
Then $\alpha + \beta$ lives in the composite field $KL$, and its conjugates are all of the form
$$
  \sigma(\alpha + \beta)
= \sigma(\alpha) + \sigma(\beta) 
$$
for some $\sigma$ in $G_{KL} = \operatorname{Gal}(KL/\mathbb Q)$.
Since $\sigma(\alpha) = \sigma|_K(\alpha)$ and $\sigma|_K$ is in $G_K$, $\sigma(\alpha)$ is a conjugate of $\alpha$.
Similarly, $\sigma(\beta)$ is a conjugate of $\beta$.
We conclude that the set of sums $\alpha + \beta$ is conjugate-closed.
