# How to understand the Artin-Schreier correspondence?

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.

• 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 $L_{\lambda c}=L_c$.) – user8268 Nov 17 '18 at 20:45
• @user8268: Yes, this seems to be the case, but why is the AS-correspondence always stated as I did it above? – 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.

• 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. – Jyrki Lahtonen Nov 24 '18 at 7:05