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In other words, show that for $a,b,c\in \mathbb{Q}$, $a\sqrt{2}+b \sqrt[3]{2}+c \sqrt[5]{2}=0$ implies that $a=b=c=0$.

There might be some questions on this forum that are similar to mine (e.g. Linear independence of $\sqrt{2}$, $\sqrt[3]{2}$, $\sqrt[4]{2}$, . . .). However, I was given the following hint:

"Note: If organized smartly, this comes out as first the third order radical, then the second order is added to form an extension of degree 6 (since it can't belong to the first), and then, the radical of order 5 cannot be in the extension since it would violate the degree."

So I would love to see a solution to this using the field extensions, and how the hint makes sense. Huge thanks in advance!

Alex
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1 Answers1

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  • $x^6-2$ and $x^5-2$ both are irreducible over $\mathbb{Q}$ thus $\mathbb{Q}(\sqrt[6]{2}) \cong \mathbb{Q}[x]/(x^6-2)$ and $\mathbb{Q}(\sqrt[5]{2})\cong \mathbb{Q}[x]/(x^5-2)$ are $\mathbb{Q}$ vector spaces of dimension $6$ and $5$.

  • Let $K = \mathbb{Q}(\sqrt[6]{2},\sqrt[5]{2})$ a $\mathbb{Q}$ vector space of dimension $n$.

    By the multiplicativity of degree of extensions we know that $5|n$ and $6|n$ thus $30| n$.

  • Assume that $a\sqrt{2}+b \sqrt[3]{2}= \sqrt[5]{2}$ with $a,b \in \mathbb{Q}$.

    Then $\sqrt[5]{2} \in \mathbb{Q}(\sqrt[6]{2})$ and $n= 6$ which is a contradiction.

  • The same idea applies to show that $\sqrt[3]{2} \not \in \mathbb{Q}(\sqrt{2})$ so $a\sqrt{2}+b\sqrt[3]{2} =0$ implies $a=b=0$.

  • An alternative is to develop a similar contradiction over some finite field $\mathbb{F}_p$ instead of $\mathbb{Q}$ and then to lift back the contradiction to $\mathbb{Q}$.

reuns
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  • Thanks very much for your answer! It made a lot more sense to me. So the idea is first to show that $\sqrt[3]{2}\not\in \mathbb{Q}(\sqrt{2})$, showing that $\sqrt{2}$ and $\sqrt[3]{2}$ are linearly independent. Then by the same idea, show that $\sqrt[5]{2}\not\in \mathbb{Q}(\sqrt[6]{2})$, so that $\sqrt[5]{2}\not\in \mathbb{Q}(\sqrt[3]{2}), \mathbb{Q}(\sqrt[3]{2})$, since $\mathbb{Q}(\sqrt[3]{2}),\mathbb{Q}(\sqrt{2})\subseteq \mathbb{Q}(\sqrt[6]{2})$. – Alex Dec 29 '18 at 02:55
  • @Alex And to show that $x^6-2,x^5-2$ are irreducible ! Factorize them, let $f_6,f_5$ be the minimal polynomials of $\sqrt[6]{2},\sqrt[5]{2}$ then look at $f_6(0),f_5(0)$ – reuns Dec 29 '18 at 05:08
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    @renus Aren't they irreducible by Eisenstein's Criterion? – Alex Dec 29 '18 at 05:10