I am asked to say whether or not the following fields are splitting fields:

1) $ \mathbb{F_{3}}(t)(t^{\frac{1}{3}}) $ over $\mathbb{F_{3}}(t) $

2) $ \mathbb{F_{3}}(t)(t^{\frac{1}{4}}) $ over $\mathbb{F_{3}}(t) $

What I did was to notice that the irreducible polynomials $ X^{3}-t $ and $ X^{4}-t $ each have $ t^{\frac{1}{3}} $ and $ t^{\frac{1}{4}} $ as roots in the corresponding fields, but that the elements $ \omega t^{\frac{1}{3}},\omega^{2} t^{\frac{1}{3}} $ and $ it^{\frac{1}{4}} $ are not inside the corresponding fields, where $ \omega $ is a 3-rd root of unity, which shows that the extensions are not normal and since an extension is normal iff it is the splitting field of some polynomial with coefficients in the base field, we conclude that the two extensions in question are not splitting fields. Something doesn't seem to be right with this though.

I would appreciate any comments and help. Thank you!

  • 2
    $\begingroup$ Are you sure it doesn't contain $\omega$? How does $x^3-1$ factor over that field? $\endgroup$ – Wojowu Mar 17 '17 at 8:30
  • 1
    $\begingroup$ I see that the first one is actually the splitting field of $ X^{3}-t $ since in $ \mathbb{F_{3}}(t)(t^{\frac{1}{3}}) $ we have the factorization $ X^{3}-t=(X-t^{\frac{1}{3}})^{3} $ and therefore $ X^{3}-t $ has precisely one zero of multiplicity 3 in the extension, hence the first example is a splitting field. I can't use the same argument for the second one though. $\endgroup$ – Raizen Mar 17 '17 at 8:53

If by $\omega$ you denote a complex cubic root of unity, you're on the wrong path.

Note that, over a field of characteristic $3$, you can write $$ (a-b)^3=a^3-b^3 $$ so you have $$ X^3-t=(X-t^{1/3})^3 $$ On the other hand $$ a^4-b^4=(a^2+b^2)(a+b)(a-b) $$ and $-1$ is not a square in $\mathbb{F}_3$ nor in $\mathbb{F}_3(t)$.


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