algebraic numbers of degree 3 and 6, whose sum has degree 12 This question is related to Degree of sum of algebraic numbers. Forgive me if
this is a dumb question, but are there two algebraic numbers $a$ and $b$ of degree $3$ and $6$ respectively, such that the sum $a+b$ has degree $12$ ?
Intuitively it would seem that the degree of $a+b$ should divide $3 \times 6=18$, but I was unable to prove this. Hence my question. 
 A: The splitting field $K$ of $x^6-2$ has degree 12 over the rationals. Its Galois group is the dihedral group 
$D_{12}$. This group has subgroups $A$ and $B$, where $A$ has order 2, $B$ has order 4, and $B$ does not contain 
$A$; the subgroup generated by $A$ and $B$ together is all of $D_{12}$. Now let $E$ and $F$ be the fixed fields of $A$ and $B$ respectively; then $E$ has degree 6, $F$ has degree 3, and $F$ is not contained in $E$, so the smallest field containing both $E$ and $F$ is $K$. Now if you pick pretty much any $a$ in $F$ and pretty much any $b$ in $E$, you should have what you want. 
A: Here's an example making Gerry's suggestion explicit (I was curious what an example would look like): $a = {\omega}2^{1/3}$, $b = 2^{1/6}$, where $\omega$ is a nontrivial cube root of unity. By PARI, $a + b$ has minimal polynomial 
$$
x^{12} - 8x^9 + 18x^8 + 12x^7 + 20x^6 - 72x^5 + 276x^4 - 232x^3 + 180x^2 + 24x + 4.
$$  
Without giving it a lot of thought, you might think the minimal polynomial might be something like 
$$
((x-2^{1/3})^6-2)((x - {\omega}2^{1/3})^6-2)((x - {\omega}^22^{1/3})^6-2)
$$
and this polynomial of degree 18 does have rational coefficients, but it factors as
$$
(x^6 - 4x^3 - 18x^2 - 12x + 2)
(x^{12} - 8x^9 + 18x^8 + 12x^7 + 20x^6 - 72x^5 + 276x^4 - 232x^3 + 180x^2 + 24x + 4).
$$
Yes, the second factor is the polynomial above.  The multiplicative relations among 
$a$ and $b$ account for such breaking up.
