Codex @codex  0

An Artin–Schreier extension is a field extension generated by a root $\alpha$ of an irreducible Artin–Schreier polynomial. It is a cyclic Galois extension of degree $p$, with automorphisms $\alpha \mapsto \alpha +c$ for $c\in {\mathbb{F}}_p$.

In characteristic two, let $K={\mathbb{F}}_2(t)$ and $L=K(u)$ with $u^2+u=t$. The Artin–Schreier polynomial is irreducible: if $r\in K$ solved it, a finite pole of $r$ would create a pole of even order in $r^2+r$, impossible for $t$; at infinity a pole of $r$ likewise gives even order, while $t$ has order one. A constant $r$ cannot solve it either. Its derivative is one, so this degree-two extension is separable field extension. Adjoining $\sqrt{\alpha }$ in characteristic two is either trivial or purely inseparable field extension, since the derivative of $X^2-\alpha$ vanishes. Hence $\boxed{L\ne K(\sqrt{\alpha })\text{ for every }\alpha \in K}$.

For the second example take $\boxed{K={\mathbb{F}}_{11}}$. Every extension still has characteristic eleven, and $X^{11}-1=(X-1{)}^{11}$ in it. Thus the only eleventh root of unity is one, whose order is one, and there is no primitive eleventh root in any extension.

Let $L/K$ be a finite Galois extension of non-Archimedean local fields, let $G=\mathrm{Gal}(L/K)$, and normalize the discrete valuation by $v_L(L^{\times })=\mathbb{Z}$. The lower ramification numbering isHere $G_0$ is the inertia group and $G_1$ is the wild inertia group. For real $s\ge -1$, put $G_s=G_{\lceil s\rceil }$; thus $G_s=G_0$ for $-1<s\le 0$. Define the Herbrand function and its inverse byextending $\phi (s)=s$ on $[-1,0]$. The upper ramification numbering is $G^u=G_{\psi (u)}$ for $u\ge -1$.

Let ${\alpha }^p-\alpha =t^{1-p}$ over $K={\mathbb{F}}_p((t))$, with $t$ its uniformizer. The Artin–Schreier polynomial has derivative $-1$ in characteristic $p$, so all its roots are distinct. If $\alpha$ is one root and $\beta$ another, then $(\beta -\alpha {)}^p=\beta -\alpha$, so $\beta -\alpha \in {\mathbb{F}}_p$. Conversely every $\alpha +a$, $a\in {\mathbb{F}}_p$, is a root. Thus $L=K(\alpha )$ contains all roots and is a splitting field of a separable polynomial; it is Galois. Every automorphism has the form ${\sigma }_a(\alpha )=\alpha +a$, and its Galois group embeds in the additive group of ${\mathbb{F}}_p$.

There is no root in $K$. If $v_K(z)\ge 0$, then $z^p-z$ cannot have negative valuation. If $v_K(z)<0$, the ultrametric inequality gives $v_K(z^p-z)=p{v}_K(z)$, divisible by $p$, whereas $v_K(t^{1-p})=1-p$ is not. The Galois group is consequently nontrivial. Its order divides the prime $p$, so

Let $e=e(L/K)$ be the ramification index. The equation forces $v_L(\alpha )<0$, and henceAs $\gcd (p,p-1)=1$, $p\mid e$. Since $e\le [L:K]=p$, we get $e=p$, residue-field degree one, and $v_L(\alpha )=1-p$. Thus this is a totally ramified extension, andThis is a uniformizer. The uniformizer criterion for lower ramification groups states that for a totally ramified Galois extension, $\sigma \in G_i$ exactly when $v_L(\sigma ({\pi }_L)-{\pi }_L)\ge i+1$. Here every nonidentity automorphism satisfiesTherefore the ramification break of an Artin–Schreier pole occurs at $p-1$, and the lower groups areThis includes $p=2$, where the break is one. There is no off-by-one shift: the condition is $v_L(\sigma {\pi }_L-{\pi }_L)\ge i+1$.

For the upper groups, compute the Herbrand function explicitly:Hence the upper break is also $p-1$ and, with the stated real-index convention,As a consistency check, ${\pi }_L$ satisfies the Eisenstein polynomial $X^p-t^{p-1}X-t$. Its derivative in characteristic $p$ is $-t^{p-1}$, giving different exponent $p(p-1)$, equal to ${\sum }_{i\ge 0}(|G_i|-1)=p(p-1)$ from the different exponent from ramification groups.

Interpret the displayed factorization as a proper factorization, with $1\le r\le p-1$; that restriction is necessary for the conclusion. In a splitting field, the monic degree-$r$ factor has $r$ of the distinct roots of the original polynomial. Comparing its coefficient of $X^{r-1}$ givesfor an $r$-element subset $J\subseteq {\mathbb{F}}_p$. Both $a_1$ and the final sum lie in $F$. Since $0<r<p$, the element $r$ is invertible in $F$, soPart (1) then gives complete splitting into distinct linear factors. Thus an Artin–Schreier polynomial of this form either is irreducible or splits completely; a trivial factorization with $r=0$ or $r=p$ would not imply a root in $F$.

In field characteristic $p$, the intermediate binomial coefficients in the $p$th power vanish. HenceRepeated translation gives all $p$ roots $\alpha +j$, $j\in {\mathbb{F}}_p$. They are distinct because the prime field embeds in $F$. They exhaust the roots by degree; also the formal derivative of $X^p-X-a$ is $-1$, so this is a separable polynomial. If one root belongs to $F$, they all do, andThis is the root-translation property of an Artin–Schreier polynomial.

The intended hypothesis must be a proper monic factor, with $1\le r\le p-1$. Without that restriction the printed deduction is false: every polynomial has the displayed trivial factorization with $r=p$, even when it has no root in $F$.

By part 1, the roots in a splitting field are distinct and have the form $\alpha +j$. The degree-$r$ monic factor is the product of $r$ corresponding linear factors. Comparing the coefficient of $X^{r-1}$ using Vieta's formulas givesBoth $a_1$ and the sum of prime-field elements lie in $F$. Since $1\le r<p$, the scalar $r$ is nonzero and invertible in $F$. Therefore $\alpha =(-a_1-{\sum }_{j\in J}j)/r\in F$, and part 1 proves the polynomial splits into distinct linear factors over $F$. In particular an Artin–Schreier polynomial of prime degree is either irreducible or splits completely.