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?  
 A: 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.
