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This is exercise A4 in chapter 29 from Pinter's A Book of Abstract Algebra. It is not homework but hints/roadmap would be preferred to a full solution for now.

First some context

The book works out the example of $\mathbb{Q}(\sqrt{1+\sqrt[3]{2}})$. If $a = \sqrt{1+\sqrt[3]{2}}$ then $a^2-1 = \sqrt[3]{2}$, so that $\sqrt[3]{2} \in \mathbb{Q}(a)$, from which it follows that $\mathbb{Q}(a) = \mathbb{Q}(\sqrt[3]{2},a)$. We then adjoin $\sqrt[3]{2}$ and $a$ to $\mathbb{Q}$ in order.

The field $\mathbb{Q}(\sqrt[3]{2})$ is an extension of degree $3$ over $\mathbb{Q}$ with basis $\{1,2^{1/3},2^{2/3}\}$. Since $x^2-1-\sqrt[3]{2}$ is irreducible over $\mathbb{Q}(\sqrt[3]{2})$, the field $\mathbb{Q}(\sqrt[3]{2},a)$ is an extension of degree $2$ over $\mathbb{Q}(\sqrt[3]{2})$ with basis $\{1,a\}$. Thus $\mathbb{Q}(\sqrt[3]{2},a)$ is an extension of degree $6$ over $\mathbb{Q}$ with basis $\{1,2^{1/3},2^{2/3},a,a 2^{1/3},a 2^{2/3}\}$.

The question at hand

I have to find a basis for the field extension $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$.

A hint is given: This is similar to the case for $\mathbb{Q}(\sqrt{1+\sqrt[3]{2}})$. Adjoin first $\sqrt[3]{4}$, then $a$.

I'm not sure if the $a$ in the hint refers to the same $a$ as in the example or if it refers to $\sqrt{2}+\sqrt[3]{4}$.

Naively I computed

$$ \left(\sqrt{2}+\sqrt[3]{4}\right)^2 = 2 + 2 \sqrt{2} \sqrt[3]{4} + \left(\sqrt[3]{4}\right)^2 $$

and

$$ \left(\sqrt{2}+\sqrt[3]{4}\right)^3 = 4 + 2\sqrt{2} + 6 \sqrt[3]{4} + 3\sqrt{2}\left(\sqrt[3]{4}\right)^2, $$

but I'm not seeing how to combine these to show that $\sqrt[3]{4} \in \mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$, which would allow me to follow the steps of the example.

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It might be easier to see what's going on if you write everything as a fractional power of 2, e.g., $\sqrt2=2^{1/2}$, $\root3\of4=2^{2/3}$, $(\root3\of4)^2=2^{4/3}=2\cdot2^{1/3}$, $\sqrt2\root3\of4=2^{7/6}=2\cdot2^{1/6}$, etc. –  Gerry Myerson Mar 26 '12 at 0:13
    
There is another question which deals with exactly the same field extensions $\mathbb Q(\sqrt2+\sqrt[3]4)$: Finite Extensions and Bases –  Martin Sleziak May 8 '12 at 9:41

1 Answer 1

up vote 4 down vote accepted

Denote $u=\sqrt2+\sqrt[3]4$. Then you have $$\begin{align} u-\sqrt2&=\sqrt[3]4\\ u^3-3u^2\sqrt2+6u-2\sqrt2&=4\\ u^3+6u-4&=(3u^2+2)\sqrt2\\ \sqrt2&=\frac{u^3+6u-4}{3u^2+2}\in\mathbb Q(u) \end{align}$$ Note that $3u^2+2>0$, hence we can divide by this number. From $\sqrt2\in\mathbb Q(u)$ follows also $\sqrt[3]4\in\mathbb Q(u)$.

Note also that by squaring the equation $u^3+6u-4=(3u^2+2)\sqrt2$ we get $$\begin{align} u^6+12u^4-8u^3+36u^2-48u+16&=(9u^4+12u^2+4)2\\ u^6-6u^4-8u^3+12u^2-48u+8&=0 \end{align}$$ If we show that $[\mathbb Q(u):\mathbb Q]=6$, then $x^6-6x^4-8x^3+12x^2-48x+8$ must be the minimal polynomial.

This can also be checked at WolframAlpha.

Very similar approach was used in an answer to this question: Finding the minimal polynomial of $\sqrt 2 + \sqrt[3] 2$ over $\mathbb Q$.

Another possible approach would be expressing some powers of $u$ as linear combinations of $1$, $2^{1/6}$, $2^{1/3}$, $2^{1/2}$, $2^{2/3}$, $2^{5/6}$ and then try to find a non-trivial linear combination which yields zero. (Again, you can have a look at the linked question for this approach for $\sqrt2+\sqrt[3]2$.)

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