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Find the splitting field of $x^6-2x^3-1$ over $\mathbb{Q}$.

I know that the real roots of $x^6-2x^3-1$ are $\sqrt[3]{1+\sqrt{2}}$ and $-\sqrt[3]{-1+\sqrt{2}}$. And I know $\zeta_{6}$ is included somehow, but I'm honestly a little lost on how to approach this polynomial. Is there a general way to go about finding splitting fields for a polynomial?

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    $\begingroup$ Splitting field over what base field ? $\endgroup$ – Belgi May 1 '15 at 6:55
  • $\begingroup$ Over $\mathbb{Q}$. Sorry for not specifying. $\endgroup$ – user3566523 May 1 '15 at 6:56
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Let $K$ be the splitting field of $f(x) = x^6-2x^3-1$ over $\mathbb Q$, and $L$ be the splitting field of $y^2-2y-1$ over $\mathbb Q$. We know the following facts:
- $L = \mathbb Q(\sqrt{2})$;
- $K$ contains $L$;
- $f(x)$ factorizes over $L$ as $(x^3-y_0)(x^3 - y_1)$, where $y_0, y_1$ are the two roots of $y^2 - 2y - 1$ in $L$.
Therefore, $K$ is the compositum of the splitting fields $K_i$ of $x^3 - y_i$ over $L$. Since $L \subset \mathbb R$, it does not contain any nontrivial cube root of $1$, therefore $K_i = L(\sqrt[3]{y_i}, j)$ where $j = \exp (2i\pi/3)$. This means that $$ K = \mathbb Q (x_0 = \sqrt[3]{y_0}, x_1 = \sqrt[3]{y_1}, j),$$ where $f$ factors as $$ f(x) = (x-x_0)(x-jx_0)(x-j^2 x_0)(x-x_1)(x-jx_1)(x-j^2x_1).$$

This answer is, however, possible only since $f$ is soluble by radicals over $\mathbb Q$. Another, more general answer, would be: factorize $f$ as a product of irreducible polynomials $p_i$; its splitting field is then the compositum of the splitting fields of the $p_i$. Since $f$ is itself irreducible over $\mathbb Q$, we know that $K = \mathbb Q(x)/f(x)$ as a field extension of $\mathbb Q$. But this is a bit of a cheat...

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