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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.

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  • $\begingroup$ Ewan, I changed mistaken tags "algebraic" and "numbers" by "algebraic # theory". The composite tags (like "algebraic numbers") use dash. $\endgroup$
    – Wadim Zudilin
    Commented Jul 1, 2010 at 8:04
  • $\begingroup$ Ewan, I retagged galois-theory based on answers; and I don't really see what it has to do with commutative algebra. If I am mistaken on the second point, feel free to put it back. $\endgroup$
    – Willie Wong
    Commented Jul 1, 2010 at 10:31
  • $\begingroup$ @ Willie : I agree with you, although a non-galois-theoretic solution is not excluded a priori. $\endgroup$
    – Ewan Delanoy
    Commented Jul 1, 2010 at 11:14

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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.

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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.

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    $\begingroup$ Another way to see it is that $X^6-2$, the annihilator of $a$, factors as $(X-b^2)(X^2+b^2X-b^4)$. The minimal polynomial of $a$ over ${\mathbb Q}(b)$ is $X^2+b^2X-b^4$. So the degree of $a$ over ${\mathbb Q}(b)$ is 2. $\endgroup$
    – Ewan Delanoy
    Commented Jul 2, 2010 at 8:40
  • $\begingroup$ @Ewan: Your comment isn't another way, by itself, to see the degree is 12, since the degree of $a$ over ${\mathbf Q}(b)$ does not on its own tell you the degree of $a+b$ over ${\mathbf Q}$, which is what your question was about. The point is that ${\mathbf Q}(a+b)$ need not be ${\mathbf Q}(a,b)$. For example, let $r$ and $s$ be two roots of $x^4+8x+12$. Both $r$ and $s$ have degree 4 over ${\mathbf Q}$ and $r$ has degree 3 over ${\mathbf Q}(s)$, but $r+s$ has degree 6, not 12, over ${\mathbf Q}$. In your case, ${\mathbf Q}(a+b) = {\mathbf Q}(a,b)$, but that needs to be checked separately. $\endgroup$
    – KConrad
    Commented May 14, 2012 at 18:27

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