# Splitting fields for $x^3-3$ and $x^5-1$

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 $$Q(e^{\frac{2 \pi i}{5}})$$

• Splitting fields over which ground field? – Torsten Schoeneberg Dec 2 '18 at 23:45
• @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. – Mike Dec 2 '18 at 23:46
• 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 – Torsten Schoeneberg Dec 2 '18 at 23:49
• Craig, you should have a look first at what Mathematica means by $(-3)^{1/3}$. – Jean-Claude Arbaut Dec 2 '18 at 23:54
• – Jean-Claude Arbaut Dec 3 '18 at 0:14

• 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 $$\mathrm e^{i\theta}$$.
• Correct. ${}{}{}$ – Jean-Claude Arbaut Dec 3 '18 at 0:02