Fixed field and automorphisms. Galois Theory.

Question

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

Notation. If $S$ is a subset of Aut$(K)$, the fixed field of $S$ is denoted by $\mathcal{F}(S)$.

• For the first part:

$$\varphi(f(x))=f(x+a)/1=\underbrace{(x+a)^{p}}_{char(k)=p}-a^{p-1}(x+a)=x^{p}+a^{p}-a^{p-1}x-a^p=x^{p}-a^{p-1}x = f(x).$$

• For the second part:

If $c \in k$, $c=c(x)$ is a constant polynomial, so $\varphi(c(x)/1) = c(x+a)=c$. Then $\varphi$ fixed $k$ and $f$. Therefore, $k(f) \subset \mathcal{F}(\varphi)$.

Now, how I show that $\boldsymbol{\mathcal{F}(\varphi) \subset k(f)}$? Thanks for any hint!

Answer 1

The problem should say that $a\neq 0$, because for $a=0$, $\phi$ is the identity and the fixed field would be $k(x)$, instead of $k(x^p)$.

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Assume that $\phi(m)=m$, for some $m\in k[x]$.

Write $m=fq+r$, with $q,r\in k[x]$ and $\deg(r)<\deg(f)$.

Applying $\phi$, we get $m=\phi(m)=\phi(f)\phi(q)+\phi(r)=f\phi(q)+\phi(r)$.

Since $\deg(\phi(r))=\deg(r)<\deg(f)$, and the uniqueness of Euclidean division, we must have $\phi(r)=r$. Since $\deg(r)<\deg(f)$, this can only happen if $r$ is constant. Here is where $a\neq0$ is used. If $r$ were not constant, then it has a root $t$ in some algebraically closed extension of $k$. But then $t,t+a,t+2a,...,t+(p-1)a$ are also roots of $r$, and they are different, and there are $p$ of them. Therefore, $\deg(r)$ would be $\geq p=\deg(f)$. Contradiction.

Therefore $q=(m-r)/f$ is again invariant with respect to $\phi$ and $\deg(q)<\deg(m)$. Dividing $q$ by $f$ and applying the same reasoning several times, we get that $m$ is of the form $f(...f(f+r_1)+r_2)+...)+r_n$, for some constants $r_1,...,r_n$. This means that $m\in k[f]$.

If $m/n\in k(x)$ such that $\phi(m/n)=m/n$, then we must have $\phi(m)=m$ and $\phi(n)=n$, since $\deg(m)=\deg(\phi(m))$, $\deg(n)=\deg(\phi(n))$, and $\phi$ doesn't change the coefficient of the leading term. Applying the argument above, for $m$ and $n$, we get that $m/n\in k(f)$.

Answer 2

Observe that $k(f)\le \mathcal{F}(\varphi)\le k(x)$. Moreover, we can show that the minimal polynomial of $x$ over $k(f)$ is $$h(y) = y^p - a^{p-1}y -(x^p - a^{p-1}x).$$ (Note that $h(y)=f(y)-f(x)$. $h$ is irreducible over $k(f(x))(y)$ because it is isomorphic to $k(z)(y)$ and $f(y)-z$ is irreducible over $k(z)(y)$.)

Therefore $[k(x):k(f)]=p$ as $k(x) = k(f)(x)$. Hence either $[k(x):\mathcal{F}(\varphi)]=1$ or $[\mathcal{F}(\varphi):k(f)]=1$. If $[k(x):\mathcal{F}(\varphi)]=1$ then every polynomial $g\in k(x)$ must be fixed by $\varphi$, however we know this is not the case. Hence $[\mathcal{F}(\varphi):k(f)]=1$ and we have $\mathcal{F}(\varphi)=k(f)$.