I need to find a basis of $\mathbb{Q}(\sqrt{2}, \sqrt[3]{2})$ over $\mathbb{Q}$. I think I have it, but I wanted to check to see if what I did was correct:

Let $F = \mathbb{Q}$, $w = \sqrt{2}+\sqrt[3]{2}$.

The element $w$ is algebraic over $\mathbb{Q}$ because it belongs to a finitely generated extension $\mathbb{Q}(\sqrt{2}, \sqrt[3]{2})$ of $\mathbb{Q}$ by two algebraic elements $\sqrt{2}$ and $\sqrt[3]{2}$.

We consider $G = \mathbb{Q}(\sqrt{2}, \sqrt[3]{2})$ as two subsequent simple extensions of $\mathbb{Q}$.

First, $\mathbb{Q}(\sqrt{2})$ is an extension of $\mathbb{Q}$ by a root of the irreducible polynomial $x^{2} - 2$. The normal form of an element of $\mathbb{Q}(\sqrt{2})$ is $$ a + b\sqrt{2}, \qquad a,b \in \mathbb{Q}$$

Now, $G = \mathbb{Q}(\sqrt{2})(\sqrt[3]{2})$.

The element $\sqrt[3]{2}$ is a root of the polynomial $x^{3}-2$.

Since this polynomial is of degree $3$, it is irreducible iff it has no roots in $\mathbb{Q}(\sqrt{2})$ - i.e., if and only if $\sqrt[3]{2} \in \mathbb{Q}(\sqrt{2})$.

Suppose $\sqrt[3]{2} = a + b\sqrt{2}, \qquad a,b \in \mathbb{Q}$.

Cubing this, we obtain $2 = (a^{3}+6ab^{2}) + (3a^{2}b+2b^{3})\sqrt{2}$.

If $3a^{2}b+2b^{3} \neq 0$, then we can solve this equation for $\sqrt{2}$ in terms of $a$ and $b$ which implies that $\sqrt{2}$ is rational, a contradiction.

If $b = 0$, then $\sqrt[3]{2} = a$, which implies that $\sqrt[3]{2}$ is rational, which it is not (right?). Contradiction.

If $a = 0$, then $\sqrt[3]{2} = b\sqrt{2} \, \implies \, \left( \frac{\sqrt[3]{2}}{2}\right)\sqrt{2} = b\, \implies\, $ $\sqrt{2}$ is rational. Contradiction.

So, $b \neq 0$, $a \neq 0$, but $3a^{2}b + 2b^{3} = 0$, so we must have that $3a^{2}b = -2b^{3} \, \implies -\frac{3}{2} = \frac{a^{2}}{b^{2}}\, \implies \, \frac{\sqrt{3}}{\sqrt{2}}i = \frac{a}{b}$, but since $a,b \in \mathbb{Q}$, their ratio cannot be a complex number. Contradiction.

This shows that $\sqrt[3]{2} \notin \mathbb{Q}(\sqrt{2})$. Therefore, $[\mathbb{Q}(\sqrt{2}, \sqrt[3]{2}): \mathbb{Q}]=4$.

So, we need at most $5$ powers, $1, w, \cdots, w^{4}$ of $w$ to obtain a linear dependency over $\mathbb{Q}$.

Note that any element of $G = \mathbb{Q}(\sqrt{2}, \sqrt[3]{2})$ can be written uniquely in the form $$ (a+b\sqrt{2}) + (c+d\sqrt{2})\sqrt[3]{2}, \qquad a,b,c,d \in \mathbb{Q}$$

or equivalently, $$a + b\sqrt{2} + c\sqrt[3]{2}+d\sqrt{2}\sqrt[3]{2}\, \qquad a,b,c,d \in \mathbb{Q}$$

So, the set $\{1, \sqrt{2}, \sqrt[3]{2}, \sqrt{2}\sqrt[3]{2} \}$ spans $G$ and is not linearly dependent over $\mathbb{Q}$, so it is a basis of $G$ over $\mathbb{Q}$.

Did I do this correctly? If not, what is missing and/or wrong?

Thank you ahead of time for your time and patience.

  • $\begingroup$ You're missing $\sqrt[3]{2}^2$ and $\sqrt2\sqrt[3]2^2$. Alternatively, since $\sqrt2\sqrt[3]2^2=2\sqrt[6]2$, you can express all your basis elements as $\sqrt[6]2^n$. $\endgroup$ – Arthur Aug 11 '17 at 20:00
  • $\begingroup$ @Arthur that didn't come out of where I wrote " note that any element of $G = \mathbb{Q}(\sqrt{2}, \sqrt[3]{2})$ can be written uniquely in the form$(a + b \sqrt{2}) + (c+ d\sqrt{2})\sqrt[3]{2}$..." I suppose I need another term with unknowns $e$ and $f$ $\in \mathbb{Q}$ in order to get the $\sqrt[3]{2}^{2}$ and $\sqrt{2}\sqrt[3]{2}^{2}$. What should those terms look like so that I get those additional basis elements? $\endgroup$ – ALannister Aug 11 '17 at 20:03
  • $\begingroup$ @Arthur Also, does that mean that $[\mathbb{Q}(\sqrt{2}, \sqrt[3]{2}): \mathbb{Q}]=6$?? How do I show that part? $\endgroup$ – ALannister Aug 11 '17 at 20:05
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    $\begingroup$ Well, $\sqrt[3]{2}\not\in{\Bbb Q}$ but that doesn't mean $[\Bbb Q(\sqrt[3]{2}):\Bbb Q]=2$ does it? $\endgroup$ – anon Aug 11 '17 at 20:22
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    $\begingroup$ Start with a basis for $\Bbb{Q}[\sqrt{2}]$ and a basis for $\Bbb{Q}[\sqrt[3]{2}]$. Since those fields are linearly disjoint, the pair-wise product of the bases will give you a basis for the composite field. $\endgroup$ – sharding4 Aug 11 '17 at 20:22

Hint for simple answer: $\mathbb{Q}(2^{1/2},2^{1/3})=\mathbb{Q}(2^{1/6})$.

  • $\begingroup$ a result that itself needs to be proven. $\endgroup$ – ALannister Aug 11 '17 at 20:21
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    $\begingroup$ Indeed. You must figure out how to get $2^{1/2}$ and $2^{1/3}$ from $2^{1/6}$ and vice-versa. $\endgroup$ – anon Aug 11 '17 at 20:22
  • $\begingroup$ I'll go ahead and give you the hint of how to get $2^{1/2}$ and $2^{1/3}$ from $2^{1/6}$. hint: take second and third powers of $2^{1/6}$. $\endgroup$ – mdave16 Aug 11 '17 at 23:04

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