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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!

  • $\begingroup$ at the beginning: "It is not difficult to check that $f_c$ is irreducible ..." should be "reducible" $\endgroup$ – user8268 Nov 17 '18 at 7:50
  • $\begingroup$ @user8268: You are right, I changed it! $\endgroup$ – Algebrus Nov 17 '18 at 7:52
  • 1
    $\begingroup$ 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) $\endgroup$ – user8268 Nov 17 '18 at 8:05
  • $\begingroup$ @user8268: This is interesting. I will ask a new question about that. $\endgroup$ – Algebrus Nov 17 '18 at 9:37
  • $\begingroup$ The new question can now be found here: math.stackexchange.com/questions/3002150/… $\endgroup$ – 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$.


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