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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 ?

Dylan Moreland
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Belgi
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  • In English, they are called "polynomials", not "polynoms". – Arturo Magidin Mar 28 '12 at 20:17
  • @Belgi : Actually two of the roots are NOT in $\mathbb R$ and one is, and that is because roots of real polynomials come either real or in complex conjugates. It makes no sense to notice that you have only one complex root, because then its conjugate would also be one. – Patrick Da Silva Mar 28 '12 at 20:35
  • @PatrickDaSilva - right, this is a mistake. but I still have to show that the root isn't in $\mathbb{Q}(\sqrt3)$... – Belgi Mar 28 '12 at 20:36

4 Answers4

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

Martin Brandenburg
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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.

Arturo Magidin
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  • But how do I show that it's irreducible ove $\mathbb{Q}(\sqrt3)$ ? – Belgi Mar 28 '12 at 20:44
  • @Belgi: You don't want that Polynomial irreducible over $\mathbb{Q}(\sqrt{3})$. This is a polynomial over $\mathbb{Q}$ that is satisfied by your element; since it is monic and irreducible, it is the monic irreducible of $\sqrt[3]{7-\sqrt{3}}$ over $\mathbb{Q}$. Since it is degree $6$, that proves directly that $[\mathbb{Q}(\sqrt[3]{7-\sqrt{3}}):\mathbb{Q}]=6$. No need to go through $\mathbb{Q}(\sqrt{3})$ if you do it this way. – Arturo Magidin Mar 28 '12 at 20:50
  • @Arturo : What is Schonemann's criterion? I've never heard such a name. – Patrick Da Silva Mar 28 '12 at 21:39
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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 – Arturo Magidin Mar 28 '12 at 21:42
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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 $\mathbb{Q}(\sqrt{3})$.

user8268
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0

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

davidlowryduda
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Bill Dubuque
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