Vrftjkry

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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!

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 $mathbf{F}_p(a) simeqmathbf{F}_p(b)$ with $ a^p-a = c,b^p-b = d$.