# Artin-Schreier Extensions

This question is out-dated, a follow-up question can be found here.

Let $$K$$ be a field of characteristic $$p > 0$$, and denote by $$L_c$$ the splitting field of $$f_c := X^p-x+c \in K[X]$$ for some $$c \in K$$. It is not difficult to check that $$f_c$$ is reducible over $$K$$ if and only if $$c = b^p-b$$ for some $$b \in K$$. (The main idea is that $$f_c(\alpha) = 0$$ implies $$f(\alpha+u)=0$$ for every $$u \in \mathbb{F_p} \subseteq K$$. Hence all irreducible factors of $$f_c$$ need to have the same degree.)

However, how can I prove that $$L_c$$ and $$L_{c'}$$ are $$K$$-isomorphic (for some $$c,c' \in K$$) if and only if $$c-c' = b^p-b$$ for some $$b \in K$$?

The if-part is clear. And if we let $$\alpha \in L_c$$ be a root of $$f_c$$ and $$\beta \in L_{c'}$$ be a root of $$f_{c'}$$, then the only-if-part boils down to proving that $$\alpha-\beta \in K$$. However, I do not see how this could be done. I am grateful for any help!

• at the beginning: "It is not difficult to check that $f_c$ is irreducible ..." should be "reducible" – user8268 Nov 17 '18 at 7:50
• @user8268: You are right, I changed it! – Algebrus Nov 17 '18 at 7:52
• if $p\neq 2$ then $\beta=2\alpha$ is a root of $x^p-x-2c=f_{2c}$, and yet $\alpha-\beta\notin K$ (and $2c-c=c\neq b^p-b$ for any $b$ unless $f_c$ is reducible) – user8268 Nov 17 '18 at 8:05
• @user8268: This is interesting. I will ask a new question about that. – Algebrus Nov 17 '18 at 9:37
• The new question can now be found here: math.stackexchange.com/questions/3002150/… – Algebrus Nov 17 '18 at 9:53

For $$K$$ a finite field.

Let $$\phi(x) = x^p$$ the Frobenius. Then $$a^p-a = c$$ means $$\phi^{n+1}(a) = \phi^n(a) + \phi^n(c)= a + \sum_{l=0}^{n} \phi^l(c)$$.

Let $$m = [\mathbf{F}_p(c):\mathbf{F}_p]$$. Then $$\sum_{l=0}^{m-1} \phi^l(c) = Tr_{\mathbf{F}_p(c)/\mathbf{F}_p}(c)$$.

Since $$c \in \mathbf{F}_p(a)$$, there are two choices :

• Or $$Tr_{\mathbf{F}_p(c)/\mathbf{F}_p}(c) = 0$$ and the least integer with $$\phi^l(a) = a$$ is $$l=m$$ and $$[\mathbf{F}_p(a):\mathbf{F}_p]= m$$ and $$\mathbf{F}_p(a)=\mathbf{F}_p(c)$$.

• Or $$Tr_{\mathbf{F}_p(c)/\mathbf{F}_p}(c) \ne 0$$ and the least integer with $$\phi^l(a) = a$$ is $$l=mp$$ and $$[\mathbf{F}_p(a):\mathbf{F}_p]= mp$$. Only in that case $$X^p-X + c$$ is irreducible over $$\mathbf{F}_p(c)$$.

From this comparing the degree of the extension we see when $$\mathbf{F}_p(a) \simeq\mathbf{F}_p(b)$$ with $$a^p-a = c,b^p-b = d$$.