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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.

Joe Wells
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1 Answers1

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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 $\mathbb{Q}$.

Arturo Magidin
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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 $\mathbb{Q}(\sqrt[3]{3})$? – Joe Wells Apr 14 '12 at 03:33
  • @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 03:40
  • 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? – Joe Wells Apr 14 '12 at 03:46
  • @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 03:48
  • @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 $p,q,r\in\mathbb{Q}$. – Arturo Magidin Apr 14 '12 at 03:49
  • 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. – Joe Wells Apr 14 '12 at 03:59
  • @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. – Arturo Magidin Apr 14 '12 at 04:02
  • 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. – Joe Wells Apr 14 '12 at 04:05
  • @ArturoMagidin I always wonder how you type up these answers so quickly!! –  Apr 14 '12 at 04:58
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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 05:44
  • @ArturoMagidin how do we know that $\mathbb{Q}(\sqrt{2},\sqrt[3]{4}) = \mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$? – GuPe Mar 29 '17 at 02:06
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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. – Arturo Magidin Mar 29 '17 at 06:06
  • @ArturoMagidin why u consider only degree 2 and 3 for later? – Meet Patel Jan 18 '23 at 00:20
  • @MeetPatel Please type out complete words. Comments almost six years later are hard enough to understand without having to wonder if "u" is supposed to be a variable/function, or just (not-at-all) cute and lazy textspeak for "you". You aren't charged by the letter. The reason is that the degree must divide $6$, by Dedekind's product theorem, and it is plainly not equal to $1$. – Arturo Magidin Jan 18 '23 at 02:08
  • @ArturoMagidin i will from next time . how you express the later in terms of two extension to use Dedekind's product theorem? – Meet Patel Jan 18 '23 at 10:58
  • @MeetPatel What are you talking about? It is sitting between $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{2},\sqrt[3]{4})$. That's your two extensions. – Arturo Magidin Jan 18 '23 at 12:36