Let $K$ be a field of characteristic $p > 0$. Then it is due to Artin and Schreier that the assignment
$$c \in K \mapsto \text{Splitting field } L_c \text{ of } X^p-X+c$$
induces a bijection between the non-trivial elements in $K/\{a^p-a \mid a \in K\}$ and the $K$-isomorphism classes of Galois extensions of degree $p$ over $K$.
In particular, this should imply that if $c, c' \in K$ are such that $L_c$ and $L_{c'}$ are $K$-isomorphic, then there is $k \in K$ such that $k^p-k = c-c'$.
However, what about the following example (which was suggested by user8268 in this question): Let $p > 2$ and $c \in K \setminus \{a^p-a \mid a \in K\}$ and let $\alpha \in L_c$ be a root of $x^p-x+c$. Then the roots of $x^p-x+2c$ are given by $2\alpha + u$, where $u$ ranges through $\mathbb{F}_p \subseteq K$, hence $L_c = L_{2c}$. But $2c – c = c \not\in \{a^p-a \mid a \in K\}$.
How is this compatible with the Artin-Schreier correspondence? I am grateful for any help!
EDIT 1: Note that the Artin-Schreier correspondence is usually proved by constructing an inverse map, as it was done e.g. in this answer.
Best Answer
Perhaps you should define precisely what you mean by a $K$-isomorphism class of Galois extensions of degree $p$ of $K$, and also give a reference for your formulation of the AS theorem. Because the classical formulation (as in Lang's "Algebra") reads : if $K$ is of characteristic $p$, the operator $P$ defined by $P(x)=x^p-x$ is an additive homomorphism of $K$ into itself; if $B$ is a subgroup of $(K,+)$ containing $P(K)$, the map $B \to K_B=$ the splitting field of all the polynomials $P(X)-b$ for $b\in B$ gives a bijection between all such groups $B$ and all the abelian extensions of $K$ of exponent $p$. This can be shown as follows :
If $K_s$ be a separable closure of $K$ and $G=Gal(K_s/K)$, a cyclic extension of degree $p$ of $K$ is obviously determined by the kernel of a (continuous) character $\chi:G \to \mathbf Z/p\mathbf Z$, and the problem consists in the description of $Hom(G,\mathbf Z/p\mathbf Z$). The quickest and clearest proof uses the additive version of Hilbert's thm. 90. More precisely, consider the exact sequence of $G$-modules $0\to \mathbf Z/p\mathbf Z \to K_s \to K_s \to 0$, where the righmost map, defined by $P$, is surjective because the polynomial $P(X)-b$ is separable. The associated cohomology exact sequence gives $K \to K \to H^1(G, \mathbf Z/p\mathbf Z) \to H^1(G, K_s)$. But $H^1(G, K_s)=0$ (Hilbert's 90) and $H^1(G, \mathbf Z/p\mathbf Z)= Hom (G, \mathbf Z/p\mathbf Z)$ because $G$ acts trivially on $\mathbf Z/p\mathbf Z$, hence $K/P(K)\cong Hom (G, \mathbf Z/p\mathbf Z) $, and one can check that this isomorphism associates to $b\in K$ the character $\chi_b$ defined by $\chi_b(g)=g(x)-x$, where $x$ is a root of $P(x)=b$.
In your example involving $X^p -X - c$ and $X^p -X - 2c$, the AS extenions coincide because $c$ and $2c$ generate the same (additive) group of order $p$ when $p\neq 2$.
NB: in the kummerian case, the same arguments work for the multiplicative version of Hilbert 's 90 and any integer $n$ s.t. $K$ contains a primitive $n$-th root of unityin place of the prime $p$, and the same remark applies to the example given by @Jirki Lahtonen.