degree of a sum of two algebraic numbers

Question

Let $a=\sqrt{2}$ and $b=\sqrt{3}$, then $a$ and $b$ are algebraic numbers of degree 2, while the degree of $a+b$ is not 2, actually it is 4, by the standard argument: if $x=a+b$, then we can rationalize it completely and get $x^4-10x^2+1=0$, so $a+b$ is of degree 4, since the polynomial is irreducible.

Now take $a=\sqrt[2019]{2}$ and $b=\sqrt[2019]{3}$. How to prove that the degree of $a+b$ is:

• greater than 2019? (if it can be done with less advanced methods then the second part)

• equal $2019^2$? (if this value is correct)

Answer 1

Let $\alpha=2^{1/2019}$, $\beta=3^{1/2019}$, $\zeta=\exp(2\pi i/2019)$, $K=\mathbb{Q}(\alpha)$, and $L=\mathbb{Q}(\zeta,\beta)$.

Here, we prove the following statement: $\alpha\notin L$ and the degree of $\alpha$ over $L$ is $2019$.

Then it implies the validity of the second claim of mathprincess since $$\frac{(\zeta^m - 1)\alpha}{(1-\zeta^n)\beta} = 1,$$ cannot hold if $\alpha\notin L$. This also leads to the validity of the first claim and a full solution to this question.

It is known that $[L:\mathbb{Q}]=2019\phi(2019)$ (see Jacobson, Velez cited in this answer: Computing the Galois group of polynomials $x^n-a \in \mathbb{Q}[x]$). Thus, the basis of $\mathbb{Q}(\zeta)$ over $\mathbb{Q}$ are also a basis of $L$ over $\mathbb{Q}(\beta)$. Then $$ \textrm{disc}_{\mathbb{Q}(\beta)}L=\textrm{disc}_{\mathbb{Q}}{\mathbb{Q}(\zeta)} \bigg\vert (2019)^{\phi(2019)}. $$

It is well-known that a rational prime $p$ does not divide the discriminant $\textrm{disc}_{\mathbb{Q}}L$ if and only if $p$ is unramified in $L$. Then we have by (Exrcise 23b of Marcus 'Number Fields'), $$\begin{align} \textrm{disc}_{\mathbb{Q}}L &\bigg\vert (\textrm{disc}_{\mathbb{Q}}{\mathbb{Q}(\beta)})^{\phi(2019)} N_{\mathbb{Q}}^{\mathbb{Q}(\beta)}\textrm{disc}_{\mathbb{Q(\beta)}}L\\ &\bigg\vert (3^{2018}\cdot 2019^{2019})^{\phi(2019)}((2019)^{\phi(2019)})^{2019}.\end{align} $$ This gives $2\nmid \textrm{disc}_{\mathbb{Q}}L$. Thus, the rational prime $2$ is unramified in $L$. Now, take any prime $\mathcal{P}\subseteq L$ lying above $2$. Then $\mathcal{P}$ divides any coefficient of $x^{2019}-2$ except for the leading coefficient, and $\mathcal{P}^2$ does not divide the constant coefficient $2$ since $2$ is unramified in $L$. Hence, by Eisenstein, $x^{2019}-2$ over $L$ is irreducible. This shows that the degree of $\alpha$ over $L$ must be $2019$. Moreover, the same argument also yields that the degree of $\alpha$ over $\mathbb{Q}(\beta)$ is $2019$.

As in mathprincess's answer, we may proceed with $\mathbb{Q}(\alpha+\beta)=\mathbb{Q}(\alpha,\beta)$. Then $$\begin{align} [\mathbb{Q}(\alpha+\beta):\mathbb{Q}]&=[\mathbb{Q}(\alpha+\beta):\mathbb{Q}(\beta)][\mathbb{Q}(\beta):\mathbb{Q}]\\ &=[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\beta)][\mathbb{Q}(\beta):\mathbb{Q}] =2019^2.\end{align} $$

Answer 2

I claim the order is $2019^2$, but I'm not going to prove some of my sub-claims (and I'm not 100% certain they're all true). So it's not quite a full answer, but maybe it's a new way for you to look at the problem.

Throughout, let $\zeta$ denote a primitive $2019^{\tiny\mbox{th}}$ root of unity, let $\alpha = \sqrt[2019]{2}$, and let $\beta = \sqrt[2019]{3}$.

The first sub-claim is that $\mathbb{Q}(\alpha, \beta)$ has degree $2019^2$ over $\mathbb{Q}$. Let $f(x) = x^{2019}-2$ and $g(x)=x^{2019}-3$. If $g(x)$ splits over $\mathbb{Q}(\alpha)$, then, since all roots of $g$ in $\mathbb{C}$ have the form $\zeta^k \beta$, then $\beta \in \mathbb{Q}(\alpha)$ (which I sub-claim to be false without proof), as the constant term of any factor of $g$ will be of the form $\zeta^\ell \beta$, which is not real unless $\ell = 0 \pmod{2019}$, in which case $\beta \in \mathbb{Q}(\alpha)$. So $g(x)$ is the minimal polynomial of $\beta$ over $\mathbb{Q}(\alpha)$, and thus $\mathbb{Q}(\alpha,\beta)$ has degree $2019^2$ over $\mathbb{Q}$.

The second sub-claim is that $\alpha+\beta$ generates $\mathbb{Q}(\alpha,\beta)$ over $\mathbb{Q}$, i.e., is a primitive element. This can be done by showing that $$\frac{(\zeta^m - 1)\alpha}{(1-\zeta^n)\beta} \ne 1,$$ for any choice of $m,n$ (which I sub-claim to be true without proof). To see why, see the constructive proof of the primitive element theorem. This implies that $\mathbb{Q}(\alpha+\beta)=\mathbb{Q}(\alpha,\beta)$ has degree $2019^2$ over $\mathbb{Q}$.

Note, as @ÍgjøgnumMeg indicates in the comments, there is an algorithm for finding the minimal polynomial of $\alpha + \beta$. Since it lies in $\mathbb{Q}(\alpha,\beta)$, which has basis $\{\alpha^i\beta^j : 0\le i,j\le 2018\}$ (this hinges upon my first sub-claim), and so you can raise $\alpha+\beta$ to higher and higher powers, writing these powers in terms of the basis, and look for linear dependency among the thus-far-computed powers until there is one.