0

The question: if $F$ is a field, and $K = F(x^{5}/(x^{3}+x^{2}+x+1))$, find the degree of the extension $[F(x) : K]$.

You can brute-force this problem by getting in the weeds and finding a basis, but that feels crude (and maybe time-consuming). You can also note $$[F(x) : K] = [F(x) : F(x^{5})] [F(x^{5}) : K],$$ which makes the problem slightly easier (I think), but it still just pushes the problem down the road to computing $[F(x^{5}) : K]$.

Alternatively, you can compute the minimal polynomial $f(t)$ of $x$ over $K[t]$, so that $[F(x) : K] = \text{deg}(f)$, but this feels similarly difficult.

There are many ways to attack this problem. I'm not necessarily asking for a step-by-step solution. But am I missing a "nice" way to approach it?

Qiaochu Yuan
  • 468,795
roblich mandervach
  • 133
  • In order to use the tower law $K$ has to be a subfield of $F(x^5)$. – Qiaochu Yuan Jul 26 '24 at 20:05
  • Well, by inspection, you have $x^5 - \alpha x^3 - \alpha x^2 - \alpha x - \alpha = 0$, where $\alpha = \frac{x^5}{x^3+x^2+x+1}$, so to conclude the degree $[F(x):K] = [K(x):K]$ is 5 it would suffice to show that polynomial is irreducible in $K[x]$. But since $\alpha$ is transcendental over $F$, I'd think it should be just a matter of applying Eisenstein's criterion to get it's irreducible in $(F[\alpha])[x]$, and then Gauss' lemma to get it's irreducible in $K[x]$. – Daniel Schepler Jul 26 '24 at 20:33
  • 1
    See here – leoli1 Jul 26 '24 at 20:51

0 Answers0