If there exists a cyclic extension of degree $p$ of $K$, then there exists a cyclic extension of degree $p^n$ for every $n\ge 1$?

Question

Let $K$ be a field with characteristic $p\neq0$.

how to show that

if there exists a cyclic extension of degree $p$ of $K$, then there exists a cyclic extension of degree $p^n$ for every $n\ge 1$?

Answer 1

A possible solution is to construct inductively such a cyclic extension of degree $p^n$ using Artin-Schreier polynomials. Precisely, let $K$ be of characteristic $p \neq 0$, and suppose we have already a cyclic extension $L/K$ of degree $p^{n-1}, n \ge 2$, with Galois group $G = <\sigma>$. Let us construct a cyclic extension $M/L$ of degree $p$ such that $M/K$ is cyclic of degree $p^n$. By Artin-Schreier theory, $M/L$ is obtained by adding the roots of a polynomial $X^p - X - a$, with $a \in L$, whose roots are $\alpha , \alpha + 1, ... , \alpha + p - 1$. The extension $M/K$ is normal if and only if the extensions of $\sigma$ to embeddings of $M$ into an algebraic closure stabilize the set of roots described above, and the extension $M/K$ is cyclic if and only if morever it is not split, i.e. is not the composite of $L$ and of a cyclic extension of $K$ of degree $p$.

Let us introduce the Artin-Schreier operator $\mathcal P$ defined by $\mathcal P(x)=x^p -x$ in order to express these properties more conveniently. The extension $M/L$ is determined by a class $[a] \in L/ \mathcal P(L)$ , $M/K$ is normal if and only if $[a] \in (L/ \mathcal P(L))^G$ (the $G$-invariants), and $M/K$ is cyclic if and only if moreover $[a]$ is not in the image of the natural map $K/\mathcal P(L)^G \to (L/ \mathcal P(L))^G$ (see remark $1$ below). The rest of the argument will consist in showing the existence of such a class $[a]$. First pick up an element $b \in L, \notin K$ such that $Tr_{L/K}(b) = 1$ (this is always possible since the trace is a $K$-linear map). Since $Tr_{L/K}(b)$ is the sum of the conjugates of $b$ and char $K$ = $p$, we have $Tr_{L/K}(b^p)= (Tr_{L/K}(b))^p = 1$ , hence $b^{p}- b$ has trace $0$. It follows from Hilbert's theorem 90 (additive version) that there exists $a \in L$ such that $b^{p}- b = \sigma (a) - a$. This gives the desired class $[a]\in (L/ \mathcal P(L))^G$ , $\notin$ the image of $K/\mathcal P(L)^G \to (L/ \mathcal P(L))^G$.

Remarks: (1) The argument above is just a parallel of the one in characteristic zero, using Kummer theory. For more details, see e.g. https://math.stackexchange.com/a/1691332/300700 (2) Actually, in characteristic $p$, the theory of "Witt vectors" of length $n$ gives a much more satisfying description of the cyclic extensions of $K$ of degree $p^n$.