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.

  • 2
    $\begingroup$ The bijection is rather between the Galois extensions of $K$ of order $p$ and $\bigl(K\setminus\{a^p-a|a\in K\}\bigr)/\Bbb F_p^*$. (If $\lambda\in\Bbb F_p^*$ then $L_{\lambda c}=L_c$.) $\endgroup$ – user8268 Nov 17 '18 at 20:45
  • $\begingroup$ @user8268: Yes, this seems to be the case, but why is the AS-correspondence always stated as I did it above? $\endgroup$ – Algebrus Nov 18 '18 at 0:12

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 $\psi:L\to L'$ such that $$\psi\circ\sigma=\sigma'\circ\psi.$$ Then the AS-correspondence is a bijection between the non-trivial cosets of the subgroup $A=\{x^p-x\mid x\in K\}\le(K,+)$ 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 $\sigma:L\to L$ 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 $\beta\mapsto \beta+1$ will map $2\beta\mapsto 2\beta+2$, 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 $K=\Bbb{Q}(\omega)$, $\omega=e^{2\pi i/3},$ the cyclic cubic extensions $K(\root3\of2)$ and $K(\root3\of4)$ 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 $K(\root3\of2)$ that map $\root3\of2\mapsto\omega\root3\of2$ and $\root3\of4\mapsto \omega\root3\of4$ 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 $b\in B$ 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.

  • $\begingroup$ Thank you for adding this explanation. My answer was missing the part about using characters of the Galois group. What I was trying to describe as the need to specify a preferred generator can equally well (=more naturally) be described with characters. $\endgroup$ – Jyrki Lahtonen Nov 24 '18 at 7:05

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