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By simple algebraic means I got that $P(x):=(x^3-7)^2-3$ is a polynomial s.t $P(\alpha)=0$ where $\alpha = \sqrt[3]{7-\sqrt{3}}$.

I wish to show that $P$ is of minimal degree, is this proof ok ?

Clearly $\mathbb{Q}(\sqrt3)$ is of degree 2 over $\mathbb{Q}$ so If I show that $F(x):=x^3-7+\sqrt3$ is irreducible over $\mathbb{Q}(\sqrt3)$ then the degree is 6 [Here I am missing an argument, not too sure why this will end things]

I know that $F(x)$ have exactly 3 roots, 2 are in $\mathbb{R}$ and one is not. So I also have to prove that the two real roots are not in $\mathbb{Q}(\sqrt3)$...

Any ideas ?

  • 0
    @PatrickDaSilva - right, this is a mistake. but I still have to show that the root isn't in $\mathbb{Q}(\sqrt3)$...2012-03-28

4 Answers 4

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Hint $\rm\ (i + j\sqrt{3})^3\:=\ k + 3\:m\: \sqrt{3}\:\neq\: n \pm \sqrt{3}$

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Eisenstein's Criterion (with $p=2$) implies that $P=x^6-14 x^3+46$ is irreducible.

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If $x^3-7+\sqrt{3}$ is irreducible over $\mathbb{Q}(\sqrt{3})$, then the degree of the extension $\mathbb{Q}(\sqrt{3},\sqrt[3]{7-\sqrt{3}})$ over $\mathbb{Q}(\sqrt{3})$ will be $3$. By Dedekind's Product Theorem, that will mean that $[\mathbb{Q}(\sqrt{3},\sqrt[3]{7-\sqrt{3}}):\mathbb{Q}] = [\mathbb{Q}(\sqrt{3},\sqrt[3]{7-\sqrt{3}}):\mathbb{Q}(\sqrt{3})][\mathbb{Q}(\sqrt{3}):\mathbb{Q}] = 3\times 2 = 6.$ Finally, you observe that $\sqrt{3}\in\mathbb{Q}(\sqrt[3]{7-\sqrt{3}})$, so that $\mathbb{Q}(\sqrt{3},\sqrt[3]{7-\sqrt{3}}) = \mathbb{Q}(\sqrt[3]{7-\sqrt{3}})$, and you would be done.

But you can show directly that your polynomial $(x^3-7)^2 -3 = x^6 - 14x^3 + 46$ is irreducible, e.g., by Eisenstein's or Schönemann's Criterion.

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    @Patrick: I hadn't either until recently, but it actually is a more general criterion than Eisenstein, and implies Eisenstein. See [this answer](http://math.stackexchange.com/a/107152/742)2012-03-28
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One way to finish: the norm $N_{\mathbb{Q}(\sqrt{3})/\mathbb{Q}}(7+\sqrt{3})=7^2-3=46$; it is not a cube, so $7+\sqrt{3}$ is not a cube in $\mathbb{Q}(\sqrt{3})$.