# Artin-Schreier tower

Let $$K$$ be an algebraically closed field with positive characteristic $$p > 0$$.

Let $$L$$ be a Galois extension of $$K(T)$$ with degree $$[L:K(T)] = p^2$$

I know that there exist a normal sub-extension $$E$$ such that $$[L:E] = [E:K(T)] = p$$ and $$L/E,E/K(T)$$ are Artin-Schreier extension.

I want to find $$Gal(L/K(T))$$ .

By trying some simple examples in Magma i can guess it isomorphic to $$Z/(pZ)^2$$

I want to prove this is always the case. Any idea how?

• The Galois group is $\cong C_p\times C_p$ whenever $L$ is the compositum of two linearly disjoint, degree $p$ Artin-Schreier extension of $K(T)$. It is possible to get extensions with Galois group $C_{p^2}$. Those can be classified with the aid of Witt vector arithmetic. See chapter 8 of Jacobson's Basic Algebra II. – Jyrki Lahtonen Aug 6 at 7:39
• Here I jotted down a few details of what it looks like in the simplest case $p=2$. Some care is required in the choice of $f_1$ and $f_2$ for the construction to play out. Basically you need to make sure that both components of the Witt vector equation force a proper extension, first of $k(T)$ then of its extension. In that thread a finite field is used in place of your $K$, but that is irrelevant. – Jyrki Lahtonen Aug 6 at 7:41

I am puzzled by the condition you put on the base field, which should be of the form $$C(T)$$, where $$C$$ is algebraically closed with characteristic $$\neq 0$$. Actually, any field $$k$$ with characteristic $$p\neq 0$$ s.t. $$P(k)\neq k$$, where $$P$$ is the Artin-Schreier operator, admits a cyclic extension $$L/k$$ of degree $$p^2$$. This can be shown elementarily, without resorting to Witt vectors, just by using Artin-Schreier equations. For clarity, let me sketch a parallel between the Kummer and the Artin-Schreier situations. Throughout, $$K/k$$ will be a cyclic extension of degree $$p$$, with Galois group $$<\sigma>$$, and we aim to construct a tower $$L/K/k$$ s.t. $$L/k$$ is cyclic of degree $$p^2$$.
1) In the Kummer case, $$k$$ has characteristic $$\neq p$$ and contains a primitive $$p$$-th root $$\zeta$$ of unity. By Kummer theory, a cyclic $$L/K$$ has the form $$L=K(\sqrt [p]{b})$$, with $$b\in K^*/{K^*}^p$$ (obvious notations). Moreover, $$L/k$$ will be galois iff, for all extensions $$\tau$$ of $$\sigma$$ to a normal closure of $$L,\tau(\sqrt [p]{b})\in L$$, or equivalently $$\sigma(b)/b \in {K^*}^p$$, say $$\sigma(b)/b=x^p, x\in K^*$$, or equivalently, by the (multiplicative) Hilbert thm.90, the norm of $$x$$ has the form $$N(x)=\zeta^i$$. All galois extensions $$L/k$$ of degree $$p^2$$ have Galois group $$\cong (\mathbf Z/p)^2$$ or $$\mathbf Z/p^2$$ and, obviously, the first case occurs iff $$b\in k^*{K^*}^p/{K^*}^p$$, or equivalently $$N(x)=1$$. Summarizing, a cyclic $$L/K/k$$ of degree $$p^2$$ exists iff $$\zeta\in N(K^*)$$, in which case $$L$$ can be explicitly constructed.
2) In the Artin-Schreier case, $$k$$ has characteristic $$p$$ and we use the A-S. operator $$P$$ defined by $$P(x)=x^p-x$$. The arguments in 1) can be repeated word for word, just replacing $$\sqrt [p]{b}$$ by a root of $$P(x)=b$$, the norm of $$K/k$$ by the trace, and the multiplicative Hilbert thm.90 by its additive counterpart. The existence of a cyclic $$L/K/k$$ boils down to the existence of $$b\in K$$ s.t. $$\sigma(b)-b \in P(K)$$. Here we use the A-S. description of $$K$$ as $$K=k(\alpha)$$, where $$P(\alpha)=a\in k$$. Taking traces in $$K/k$$ we obtain $$Tr(P(\alpha))=Tr(a)$$ $$=pa=0$$, and Hilbert's 90 guarantees the existence of $$b\in K$$ s.t. $$\sigma(b)-b=P(\alpha)$$. Summarizing, a cyclic explicit $$L/K/k$$ of degree $$p^2$$ always exists in the A-S. case.
• This is all fine (+1). But I think the OP is looking for concrete choices of $b$ and $a$ to have something they can test in Magma or some such CAS. – Jyrki Lahtonen Aug 11 at 6:27
• So, given $b_1\in C(T)$ such that $E/C(T)$ is the splitting field of $x^p-x=b_1$, how to find $b_2\in E$ such that the splitting field $L$ of $x^p-x=b_2$ over $E$ is A) Galois over $C(T)$, and B) cyclic? – Jyrki Lahtonen Aug 11 at 6:47