1

I came across this problem which I couldn't solve thus far:

Definition Let $F_0$ be a field with $char(F)=p>0$. A polynomial $f \in F_0[X]$ is said to be Schreier solvable if there exists a series of extensions $F_i/F_{i-1}, \, F_i=F_{i-1}[\alpha_i], \, i=1,...,r$, such that $\alpha_i$ is a root of a polynomial $X^p-X-a_i \in F_{i-1}[X]$ and $f$ splits in $F_r$.

Show that a Galois group of a Schreier solvable polynomial is a solvable group.

Any ideas?

Community
  • 1
Brassican
  • 577

1 Answers1

2

You have a tower of Artin-Schreier extensions (adjoining a root of $X^p-X-a$). An Artin-Schreier extension is Galois with a trivial or a cyclic order $p$ group. So a tower of Artin-Schreier extensions will have a Galois group a $p$-group. Each $p$-group is soluble...

Angina Seng
  • 158,341
  • I'm not familiar yet with Artin-Schreier extensions, only started to read about it following your comment. Could it be solved without it? Also, do you have a reference for the fact that an Artin-Schreier extension is Galois with a trivial or a cyclic order $p$-group? – Brassican Oct 02 '17 at 18:05
  • @Brassican In some sense the answer is yes, since Artin-Schreier extensions are a special case of the result you seek. Any big book on Galois theory will do Artin-Schreier extensions; I'm sure that Lang's *Algebra" has it. – Angina Seng Oct 02 '17 at 18:11
  • 1
    @Brassican If $\alpha_i\notin F_{i-1}$, then $x^p-x-a_i$ is irreducible over $F_{i-1}$, and the Galois group is generated by $\alpha_i\mapsto\alpha_i+1$. The fun fact is that any Galois extension of degree $p$ of char $p$ fields has that form. As Lord Shark (+1) said this is very much textbook stuff. Yet, these bits have been covered on our site as well. See here for the irreducibility result and here for the other result. – Jyrki Lahtonen Oct 02 '17 at 19:22