Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there a way to show that the minimal polynomial of this number over $\mathbb Q$ has degree $6$ without too much annoying computations? I have "reduced" it to showing that both $\sqrt{2}$ and $\sqrt[3]{3}$ are in $\mathbb Q(\sqrt{2}+\sqrt[3]{3})$...

share|cite|improve this question
up vote 1 down vote accepted

Reference: Primitive Root Theorem

Just follow the proof, and you can show that the number $f$ in the proof, can be chosen to be $1$. Then $\mathbb{Q}(\sqrt 2, \sqrt[3]{3}) = \mathbb{Q}(\sqrt 2 + \sqrt[3]{3})$.

As $\alpha_1 = \sqrt 2$, and $\alpha_2 = -\sqrt 2$,

$\beta_1=\sqrt[3]{3}$, $\beta_2=\sqrt[3]{3} \omega$, $\beta_3= \sqrt[3]{3}\omega^2$, where $\omega=\exp(2\pi/3)$.

You just need the following equations are not true (which is obvious because one is real while the other is not),

$$-2\sqrt 2 = \sqrt[3]{3}-\sqrt[3]{3}\omega,$$ and $$-2\sqrt 2 = \sqrt[3]{3}-\sqrt[3]{3}\omega^2.$$

share|cite|improve this answer

There is a standard procedure for computing the minimal polynomial for the sum of two numbers from their minimal polynomials.

Let $f(X)$ and $g(Y)$ be the minimal polynomials of the algebraic numbers $\alpha$ and $\beta$. Set $Z=X+Y$. Now consider the polynomial $f(Z-Y)$ and eliminate from this the variable $Y$ using the relation $g(Y)=0$. Then thhe resulting polynomial involving just $Z$ is the minimal polynomial for $\alpha+\beta$.

Here it boils down to eliminating $Y$ from $ (Z-Y)^2-2$ using $Y^3-3=0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.