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I'm looking for the splitting fields of

(a) $x^3-3$

(b) $x^5-1$.


EDIT:

(a) Thanks to all the hints and suggestions, the three roots are

$x_1=3^{\frac{1}{3}}$, $x_2=e^{\frac{2 \pi i}{3}}3^{\frac{1}{3}}$, $x_3=e^{\frac{4 \pi i}{3}}3^{\frac{1}{3}}$

Now, the question doesn't specify the field over which these polynomials are defined, I'll take a guess and say $Q$. Now, all the roots can be generated from $x_2=e^{\frac{2 \pi i}{3}}3^{\frac{1}{3}}$ over the rationals, so is the answer $Q(e^{\frac{2 \pi i}{3}}3^{\frac{1}{3}})$ correct?


(b) Again, the roots are the 5 complex roots of unity, all of which can be generated by the root $x_1=e^{\frac{2 \pi i}{5}}$. So would the correct answer now be $Q(e^{\frac{2 \pi i}{5}})$

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  • $\begingroup$ Splitting fields over which ground field? $\endgroup$ – Torsten Schoeneberg Dec 2 '18 at 23:45
  • $\begingroup$ @TorstenSchoeneberg The question doesn't specify, perhaps there is an obvious choice? Most of the relevant section is concerned with extensions over the rationals, so my safe assumption is the rationals. $\endgroup$ – Mike Dec 2 '18 at 23:46
  • $\begingroup$ Also, be very careful with writing a negative number to the power of a fractional exponent. That is an ill-defined expression and that probably caused part of the problem here. Cf. math.stackexchange.com/q/317528/96384 $\endgroup$ – Torsten Schoeneberg Dec 2 '18 at 23:49
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    $\begingroup$ Craig, you should have a look first at what Mathematica means by $(-3)^{1/3}$. $\endgroup$ – Jean-Claude Arbaut Dec 2 '18 at 23:54
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    $\begingroup$ See math.stackexchange.com/questions/1597326/… $\endgroup$ – Jean-Claude Arbaut Dec 3 '18 at 0:14
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Hints:

  • A real number has $3$ cube roots in $\mathbf C$. One is the standard real cube root, he other two are this real cube root, multiplied by one of the complex cube roots of unity.
  • For $x^5-1$, solve it in the form $\mathrm e^{i\theta}$.
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  • $\begingroup$ That does not address the (correct) remark from the OP that the cube root of a negative number is "usually" (and at least in high school) defined to be a real negative number. That's a quirk of Mathematica. $\endgroup$ – Jean-Claude Arbaut Dec 2 '18 at 23:57
  • $\begingroup$ @Jean-ClaudeArbaut: I only gave hints for the implicit questions in the title. $\endgroup$ – Bernard Dec 2 '18 at 23:59
  • $\begingroup$ Correct. ${}{}{}$ $\endgroup$ – Jean-Claude Arbaut Dec 3 '18 at 0:02
  • $\begingroup$ @Bernard Thanks for the hint, I edited my question to re-write the roots as n-th roots of unity, hope I'm getting warmer. $\endgroup$ – Mike Dec 3 '18 at 0:08

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