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I am practicing writing down random algebraic numbers and finding their degrees over $\mathbb Q$ and have fumbled when coming to $\sqrt[3]{2} + \sqrt[3]{3}$. Mathematica tells me the minimal polynomial is $x^9-27 x^6-135 x^3-729$, which is a polynomial in $x^3$. I figured by cubing the element I could show that the cube has degree 3, which is some partial progress, but I have failed to even do that. If anyone has any neat tricks up their sleeves about these kinds of results, I would be forever indebted to you.

user369210
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    Mathematica (10.4.1) gives $-125 - 87 x^3 - 15 x^6 + x^9$ as the minimal polynomial. Factoring your polynomial also gives $\sqrt[3]{7}$s mixed in, e.g., $3+\sqrt[3]{28} + \sqrt[3]{98}$. – Eric Towers Jun 05 '16 at 04:41
  • There must have been an error in entering into Mathematica the irrationality whose minimal polynomial was to be found. – Lubin Jun 05 '16 at 04:53

2 Answers2

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Here's a way to do it:

This element is certainly in the field $K=\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3})$. Choose a basis for this field as a vector space over $\mathbb{Q}$; I'd probably choose $1,\sqrt[3]{2},\sqrt[3]{3},\sqrt[3]{6},\sqrt[3]{4},\sqrt[3]{9},\sqrt[3]{12},\sqrt[3]{18},\sqrt[3]{24}$. Kind of an extensive list.

Now, multiplication by $\sqrt[3]{2}+\sqrt[3]{3}$ acts on $K$ as a $\mathbb{Q}$-linear endomorphism. With the basis chosen above, it isn't too dificult to write out a matrix for this endomorphism. Taking the characteristic polynomial of it will give you the minimal polynomial (well, a power of the minimal polynomial - of course this field could be a finite extension of the splitting field. It's not but you'd have to show the characteristic polynomial is irreducible - which it is).

If you can use a computer, then a $9\times 9$ determinant isn't too bad.

Eoin
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  • Actually, the comment above is probably an even better way to take care of the situation. – Eoin Jun 05 '16 at 04:04
  • maybe you could explain how the minimal polynomial is related to the $\mathbb{Q}$-conjugates – reuns Jun 05 '16 at 04:09
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First, Mathematica is wrong. The minimal polynomial is $-125-87X^3-15X^6+X^9$.

The irreducible equation for $\sqrt[3]2$ over $\Bbb Q$ is $X^3-2$, and the irreducible equation for $\sqrt[3]2+\sqrt[3]3$ over $\Bbb Q(\sqrt[3]3\,)$would apparently be $F(X)=(X-\sqrt[3]3\,)^3-2=X^3-3\sqrt[3]3X^2+3\sqrt[3]9X-5$. Now, to get a $\Bbb Q$-polunomial out of $F$, you need to multiply it by its two conjugates, gotten by (1) replacing $\sqrt[3]3$ by $\omega\sqrt[3]3$, and (2) replacing $\sqrt[3]3$ by $\omega^2\sqrt[3]3$, where $\omega=\frac12(-1+\sqrt{-3}\,)$ is a chosen primitive cube root of $1$. The conjugate polynomials are \begin{align} F_1(X)&=X^3-3\omega\sqrt[3]3X^2+3\omega^2\sqrt[3]9X-5\>,\>\text{and}\\ F_2(X)&=X^3-3\omega^2\sqrt[3]3X^2+3\omega\sqrt[3]9X-5\,. \end{align} I’ll leave it to you to do the multiplication $FF_1F_2$. It’s tedious by hand, but far easier than computing a nine-by-nine determinant!

Lubin
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