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The question reads: Find a basis of $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$ over $\mathbb{Q}$. Describe the elements of $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$.

My original thought on the approach was to find the minimum polynomial (which would have degree 6), and then just take create a basis where every element is $(\sqrt{2}+\sqrt[3]{4})^n$ for $n=0,\ldots,5$.

The hint in the back recommends adjoining $\sqrt[3]{4}$ first, but I'm not sure where to go once I've done that. Can anyone help guide me through the process the book wants me to use?

Thank you very much in advance.

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    While the process suggested by the book and fleshed out by Arturo is probably the nicest way to solve the problem, there is nothing wrong with your original thought. Find a polynomial of degree 6 satisfied by $\sqrt2+\root3\of4$, prove that it's irreducible, then those powers you wrote down form a basis.2012-04-14
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    There is another question which deals with exactly the same field extensions $\mathbb Q(\sqrt2+\sqrt[3]4)$: [Finding a basis for the field extension ${\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})}$](http://math.stackexchange.com/questions/124425/finding-a-basis-for-the-field-extension-mathbbq-sqrt2-sqrt34/)2012-05-08

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Dedekind's Product Theorem (which a friend of mine calls the "Royal Dutch Airlines Theorem") states that if $K\subseteq L\subseteq M$ are fields, then $$[M:K] = [M:L][L:K].$$ The proof of the theorem is constructive: if $\{m_i\}_{i\in I}$ is a basis for $M$ as an $L$-vector space, and $\{\ell_j\}_{j\in J}$ is a basis for $L$ as a $K$-vector space, then one shows that $\{m_i\ell_j\}_{(i,j)\in I\times J}$ is a basis for $M$ as a $K$-vector space.

The Hint in the book is suggesting that you do this. Since $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4}) = \mathbb{Q}(\sqrt{2},\sqrt[3]{4})$, you can find a basis for $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$ by considering the tower of extensions $$\mathbb{Q}\subseteq \mathbb{Q}(\sqrt[3]{4}) \subseteq \mathbb{Q}(\sqrt[3]{4})(\sqrt{2}).$$ It is easy to find a minimal polynomial for $\sqrt[3]{4}$ over $\mathbb{Q}$, from which you easily get a basis for the first extension; and it is easy to find a minimal polynomial for $\sqrt{2}$ over $\mathbb{Q}(\sqrt[3]{4})$, from which you get a basis for the second extension. Now you can combine them, using the argument of the Dedekind Product Theorem, to get a basis for $\mathbb{Q}(\sqrt{2},\sqrt[3]{4}) = \mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$ over $\mathbb{Q}$.

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    Ah yes, that makes sense. Now, my minimum polynomial for $\sqrt[3]{4}$ over $\mathbb{Q}$ is $x^3 - 4$, and yields the basis $\{1, 4^{1/3}, 4^{2/3} \}$. But what is my minimum polynomial for $\sqrt{2}$ over $\mathbb{Q}(\sqrt[3]{3})$?2012-04-14
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    @JoeDub: Well... $\sqrt{2}$ certainly satisfies $x^2-2$. So if $\sqrt{2}\notin \mathbb{Q}(\sqrt[3]{4})$, what could the polynomial be? It has to divide *any* polynomial that is satisfied by $\sqrt{2}$, so...2012-04-14
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    Since every element in $\mathbb{Q}(\sqrt[3]{4})$ is of the form $p+q\sqrt[3]{4}$, then it would have to satisfy $(x-\sqrt[3]{4})^2 - 2$? Am I thinking through that correctly?2012-04-14
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    @JoeDub: No. Again: in **any** fields $F\subseteq K$, if $a\in K$ satisfies a polynomial $g(x)\in F[x]$, then the minimal polynomial of $a$ over $F$ must divide $g(x)$. Here, $F=\mathbb{Q}(\sqrt[3]{4})$; and $a=\sqrt{2}$ satisfies the polynomial $x^2-2$. So whatever the minimal polynomial of $\sqrt{2}$ is, it must divide $x^2-2$; it must be monic. If $\sqrt{2}$ is **not** in $\mathbb{Q}(\sqrt[3]{4})$, then it cannot be degree $1$. How many degree 2 monic polynomials divide $x^2-2$?2012-04-14
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    @JoeDub: Also, you are incorrect about the elements of $\mathbb{Q}(\sqrt[3]{4})$. The extension is of degree $3$, not $2$, so the elements of $\mathbb{Q}(\sqrt[3]{4})$ are of the form $p+q\sqrt[3]{4} + r\sqrt[3]{16}$, with $p,q,r\in\mathbb{Q}$.2012-04-14
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    There aren't any degree 2 monic polynomials that divide $x^2 - 2$ other than $x^2 -2$ itself, so then that means that $x^2 - 2$ is the minimum polynomial in $\mathbb{Q}(\sqrt[3]{4},\sqrt{2})$. Okay. So then that yields the basis $\{1,\sqrt{2}\}$ of $\mathbb{Q}(\sqrt[3]{4},\sqrt{2})$ over $\mathbb{Q}(\sqrt[3]{4})$. And then from there I can just multiply the bases to determine my basis for $\mathbb{Q}(\sqrt[3]{4},\sqrt{2})$ over $\mathbb{Q}$. Did I get it right this time? I'm sorry, I don't know why this stuff isn't sticking with me.2012-04-14
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    @JoeDub: Note: it's the minimal polynomial **over $\mathbb{Q}(\sqrt[3]{4})$** (not "in $\mathbb{Q}(\sqrt[3]{4},\sqrt{2})$"). Otherwise, yes, it's fine.2012-04-14
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    Thank you for pointing that out. I'll have to go back through and fix the grammar. Thank you again for all of your help.2012-04-14
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    @ArturoMagidin I always wonder how you type up these answers so quickly!!2012-04-14
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    @Benjamin, "Arturo" is actually the collective name for a sweatshop of over a dozen mathematicians, chained to their computers and forced to answer questions on m.se.2012-04-14
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    @ArturoMagidin how do we know that $\mathbb{Q}(\sqrt{2},\sqrt[3]{4}) = \mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$?2017-03-29
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    @GuachoPerez: former clearly contains the latter; the former has degree $6$, so the question is just whether the latter can be degree $2$ or degree $3$. If it were of degree $2$, then $\sqrt{2}\sqrt[3]{4}$ would be the root of a quadratic. If it were of degree $3$, then it would be the root of a cubic. It can't be either.2017-03-29