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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. –  Gerry Myerson Apr 14 '12 at 5:49
    
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 ${\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})}$](math.stackexchange.com/questions/124425/…) –  Martin Sleziak May 8 '12 at 9:41
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1 Answer

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 $\mathbb{Q}(\sqrt[3]{3})$? –  JoeDub Apr 14 '12 at 3:33
    
@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... –  Arturo Magidin Apr 14 '12 at 3:40
    
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? –  JoeDub Apr 14 '12 at 3:46
    
@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 $x^2-2$? –  Arturo Magidin Apr 14 '12 at 3:48
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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. –  Gerry Myerson Apr 14 '12 at 5:44
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