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Find the degree of $[F:\mathbb{Q}]$, where $F$ is the splitting field of $x^3-11$ over $\mathbb{Q}$

Roots are $\sqrt[3]{11}$ and the other $2$ are complex, I think. I don't think that finding the roots and then what field they generate is a good idea. I think it's better to prove this polynomial is irreducible.

By Eiseintein criterion, $p=11$ is prime, does divide every $a_i$ except $a_3=1$, $p^2$ does not divide $a_0$. So the polynomial is irreducible over $\mathbb{Q}$ so the degree is $3$. Is this right?

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There is a more general result:

For all prime numbers $p$ the splitting field $F$ of $f=x^3-p$ over $\mathbb{Q}$ has degree 6.

Reason: The polynomial is irreducible by Eisenstein criterion and has exactly one real root $z_1=\sqrt[3]{p}$, because the polynomial is strictly increasing. Now there must be a pair of non-real roots $z_2$ and $z_3$. Because the coefficients of $f$ are real, we have $\overline{z_2}=z_3$.

Now I use the following theorem

If $f$ is an irreducible polynomial of degree $n$ over $\mathbb{Q}$, there is a splitting field $F$ of $f$ with $[F:\mathbb{Q}]\leq n!$

Let $F$ be the splitting field of $x^3-p$. Remark that $F=\mathbb{Q}(z_1,z_2,z_3)$ is a splitting field of $f$ over $\mathbb{Q}$ up to isomorphism. The theorem says $[F:\mathbb{Q}]\leq 3!=6$. Note that $[\mathbb{Q}(z_1):\mathbb{Q}]=3$, because $f$ is the minimal polynomial of $z_1$. But both $z_2,z_3\notin\mathbb{Q}(z_1)$, thus $1<[\mathbb{Q}(z_1,z_2,z_3):\mathbb{Q}(z_1)]\leq 2$ and consequently \begin{align} [\mathbb{Q}(z_1,z_2,z_3):\mathbb{Q}]=[\mathbb{Q}(z_1,z_2,z_3):\mathbb{Q}(z_1)]\cdot [\mathbb{Q}(z_1):\mathbb{Q}]=2\cdot 3=6 \end{align}

If I am missing some details or there is a mistake, please let me know.

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    $\begingroup$ In fact, the claim is true for all integers $p$ that are not perfect cubes, as in that case, $x^3 - p$ has no linear factor and hence is irreducible. $\endgroup$ – Travis Oct 3 '17 at 19:14
  • $\begingroup$ @Travis Very good comment. Thank you! $\endgroup$ – Fakemistake Oct 6 '17 at 4:30
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HINT: The roots are: $$a_1 = \sqrt[3]{11} \quad \quad a_2 = \xi_{3}\sqrt[3]{11} \quad a_2 = \xi^2_{3}\sqrt[3]{11}$$

where $\xi_{3}$ is the third root of unity. Now you need an extension of $\mathbb{Q}$ that contains both $\xi_{3}$ and $\sqrt[3]{11}$

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The roots of $x^3-11$ are $11^{\frac{1}{3}}$ ,$11^{\frac{1}{3}}\omega$ ,$11^{\frac{1}{3}} \omega^2$

Hence $F=\Bbb Q(11^{\frac{1}{3}},w)$

Hence $[F:\Bbb Q]=[\Bbb Q(11^{\frac{1}{3}},w):\Bbb Q(11^{\frac{1}{3}})][\Bbb Q(11^{\frac{1}{3}}):\Bbb Q]=2\times 3=6$

[Since the minimal polynomial of $\omega$ is $x^2+x+1$ and minimal polynomial of $11^{\frac{1}{3}}$ over $\Bbb Q$ is $x^3-11$.]

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Your argument that $x^3 - 11$ is irreducible over $\Bbb Q$ is correct, but this does not imply that the degree of the extension is $3$. (It does, however, imply that the degree is at least $3$; we can thus conclude that the only possible degrees of a splitting field of an irreducible cubic are $3$ and $6$.)

Hint Since the polynomial has nonreal roots, (the restriction to $F$ of) complex conjugation is an automorphism of $F / K$ of order $2$.

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  • $\begingroup$ I thought the degree of the splitting field was the degree of the irreducible polynomial that has all the roots. What am I confusing here? $\endgroup$ – Guerlando OCs Oct 3 '17 at 17:15
  • $\begingroup$ It's perhaps unfortunate that the term "degree" is used for two related by distinct concepts here. For any field extension $F / K$, $F$ is a vector space over $K$, and the degree of the extension is just the dimension $\dim_K F$ of this vector space. In this case we may take the basis $\{1, \omega, \alpha, \alpha \omega, \alpha^2, \alpha^2 \omega\}$, where $\omega$ is a primitive third root of unity and $\alpha = \sqrt[3]{11}$. $\endgroup$ – Travis Oct 3 '17 at 18:31

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