Krdytkyu

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 $2alpha + u$, where $u$ ranges through $mathbb{F}_p subseteq K$, hence $L_c = L_{2c}$. But $2c - c = c notin {a^p-a mid a in K}$.

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.

I should really check out a definite source, but I think the Artin-Schreier correspondence means the following. To summarize, the problem observed by user8268 can be resolved by insisting that the Galois groups should have a preferred generator.

So something like the following.

Let $L$ and $L'$ be two cyclic degree $p$ extensions of $K$, and let $sigma$ (resp. $sigma'$) be the respective preferred generators. We call $(L,sigma)$ and $(L',sigma')$ equivalent, if there exists a $K$-isomorphism $psi:Lto L'$ such that $$psicircsigma=sigma'circpsi.$$
Then the AS-correspondence is a bijection between the non-trivial cosets of the subgroup $A={x^p-xmid xin K}le(K,+)$ and the equivalence classes of pairs $(L,sigma)$. If $c+A$ is a non-trivial coset of $A$ then it corresponds to an extension $L=K(beta)$ with $beta^p-beta+c=0$ together with the preferred automorphism $sigma:Lto L$ uniquely determined by $sigma(beta)=beta+1$.

A consequence of this formulation is that while the splitting fields of $x^p-x+c$ and $x^p-x+2c$ are both equal to $L=K(beta)$, the above correspondence associates a different generator of the Galois group with the latter polynomial. The automorphism $sigma$ that maps $betamapsto beta+1$ will map $2betamapsto 2beta+2$, so we should associate $x^p-x+2c$ with the pair $(K(beta),sigma^2)$ instead of $(K(beta),sigma)$. Observe that those two pairs cannot be equivalent according to the above definition because $sigma'=sigma^2$ (and here every possible $psi$ commutes with $sigma$).

A few closing remarks:

• I may be out of my depth here, but I hazard a guess that the interpretation of this result in terms of Galois cohomology makes it necessary to specify the action of a fixed generator of the cyclic group.


• An analogous problem is present in Kummer theory also. For example, with $K=Bbb{Q}(omega)$, $omega=e^{2pi i/3},$ the cyclic cubic extensions $K(root3of2)$ and $K(root3of4)$ are clearly equal (as subsets of $Bbb{C}$) even though $2$ and $4$ belong to distinct (multiplicative) cosets of the subgroup of cubes of $K^*$. Here we also resolve the problem by observing that the automorphisms of $K(root3of2)$ that map $root3of2mapstoomegaroot3of2$ and $root3of4mapsto omegaroot3of4$ are similarly squares of each other, i.e. distinct generators of the Galois group.

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 $bin B$ 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 $chi:G to mathbf Z/pmathbf Z$, and the problem consists in the description of $Hom(G,mathbf Z/pmathbf Z$). The quickest and clearest proof uses the additive version of Hilbert's thm. 90. More precisely, consider the exact sequence of $G$-modules $0to mathbf Z/pmathbf Z to K_s to K_s to 0$, where the righmost map, defined by $P$, is surjective because the polynomial $P(X)-b$ is separable. The associated cohomology exact sequence gives $K to K to H^1(G, mathbf Z/pmathbf Z) to H^1(G, K_s)$. But $H^1(G, K_s)=0$ (Hilbert's 90) and $H^1(G, mathbf Z/pmathbf Z)= Hom (G, mathbf Z/pmathbf Z)$ because $G$ acts trivially on $mathbf Z/pmathbf Z$, hence $K/P(K)cong Hom (G, mathbf Z/pmathbf Z) $, and one can check that this isomorphism associates to $bin K$ 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 $pneq 2$.

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.