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Let $k$ be a field of characteristic $p>0$ ,and let $a∈k$. Let $h(x)= x^p − a^{p−1}x ∈ k[x]$. Show that $h$ is fixed by the automorphism $φ$ of $k(x)$ defined by $φ(f (x)/g(x)) = f (x + a)/g(x + a)$ for any $f(x),g(x) ∈ k[x]$. Show that $k(h)$ is the fixed field of $φ$.

Here also I am not getting any clue, as well as I find anything I can update. Till then anyone please try to help.

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Eoin (+1) already showed that $h$ is invariant under $\phi$.

By induction on $n$ we see that $\phi^n(r(x))=r(x+na)$ for all $n\in\Bbb{N}$ and all $r(x)\in k(x)$. From this it follows that $\phi$ is of order $p$. So $\phi$ generates a subgroup $G\le Aut(k(x))$ of order $p$. By basic facts of Galois theory the fixed field $K=\operatorname{Inv}(G)$ satisfies $[k(x):K]=|G|=p$.

We know that $k(h)\subseteq K$, so it suffices to show that $[k(x):k(h)]=p$. Because $[k(x):k(h)]=[k(x):K]\cdot [K:k(h)]$ we see that $[k(x):k(h)]$ is a multiple of $p$, in particular it is $\ge p$.

OTOH $x$ is a zero of the polynomial $$ m(T)=T^p-a^{p-1}T-h\in k(h)[T]. $$ Therefore $[k(h)(x):k(h)]=[k(x):k(h)]\le \deg m(T)=p$.

The claim $K=k(h)$ follows from this. As a bonus we get that $m(T)$ is the minimal polynomial of $x$ over $k(h)$, and hence irreducible in $k(h)[T]$.

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  • $\begingroup$ I was a bit surprised not to find a duplicate of this question. If proven wrong, let me know and I will delete my answer. $\endgroup$ – Jyrki Lahtonen Aug 30 '15 at 7:27
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We can show that $h$ is fixed by this mapping by direct computation:

$$\varphi(h)=h(x+a)=(x+a)^p-a^{p-1}(x+a)=x^p+a^p-a^{p-1}x-a^p=h$$

Since $k\subset k(x)$ is fixed (the map $f(x)=c=f(x+a)$ for every $a$), we now have a fixed field $k(h)\subset k(x)$ by the above computation and noting that all elements of $k(h)$ are rational functions in the "variable" $h$ with coefficients in $k$.

Lastly, we should show that this is all that is fixed by $\varphi$. But I can't think of how to do this right now so maybe someone else will answer this part.

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By unique factorization, the fixed field of the automorphism is the field of fractions of the fixed ring, since assuming the denominator and numerator are coprime your automorphism permutes the irreducible factors appearing in them. To compute the fixed ring use induction on the degree and polynomial division.

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    $\begingroup$ It may be worth emphasizing that your first claim holds because $\phi$ maps polynomials to polynomials. If in place of $\phi$ we had an automorphism involving fractional linear transformations (aka Möbius group) then this method of identifying the fixed field would not work. +1 of course. $\endgroup$ – Jyrki Lahtonen Aug 30 '15 at 7:24
  • $\begingroup$ @Jyrki of course, more complicated automorphisms don't fall to this technique. $\endgroup$ – Pedro Tamaroff Aug 30 '15 at 12:25

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