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Let $k = F_p$, and let $k(x)$ be the rational function field in one variable over $k$. Define $φ : k(x) \to k(x)$ by $φ(x) = x+1$. Show that $φ$ has finite order in $Gal(k(x)/k)$. Determine this order, find a $u$ so that $k(u)$ is the fixed field of $φ$, determine the minimal polynomial over $k(u)$ of $x$, and find all the roots of this minimal polynomial.

Now order of $\phi$ is $p$ but what about the rest? I can see that if I consider $u=x^p -x $ then $\phi (x^p -x)=(x+1)^p-(x+1)= x^p -x$ So, $\phi$ fixes this $k(u)$. Now, is this the fixed field? If it is then why? and what about the rest?

Ri-Li
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    Not $\Bbb Q(u)$, but $k(u)$. – Lubin Aug 24 '15 at 15:08
  • Prof. Lubin's answer (+1) is, of course, what you need. If you are still wondering about irreducibility of $X^p-X-u$, then you can take a look at this thread. I shamelessly link to my answer, but the other answers there are good as well. I picked mine, because some of the arguments in the other answers depend on properties of the ground field that is finite in that question, but $k(u)$ in your case. AFAICT my answer has nothing that would not work in any field of characteristic $p$. – Jyrki Lahtonen Aug 25 '15 at 07:29

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I know from experience that this can get incredibly confusing. You have correctly identified the fixed field of $\varphi:x\mapsto x+1$ as $k(x^p-x)$, and called your generating element $u$. If you determine that $\bigl[k(x)\colon k(u)\bigr]=p$, you will have identified the complete fixed field as $k(u)$ because the degrees are right.

So all you need to do is find the $k(u)$-polynomial $\text{Irr}\bigl(x,k(u)[X]\bigr)$: an irreducible polynomial in $X$ with coefficients in $k(u)$ having $x$ as a root. In fact, the polynomial I’m going to show you has coefficients in the UFD $k[u]$.

The polynomial is $X^p-X-u$. Notice that when you plug in $x$ for $X$, you get zero. You see at a glance what the other roots are, and it’s not hard to see why the polynomial is irreducible.

Lubin
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  • Sir, later I have found that the polynomial is $X^p-X-u$ but why this polynomial is irred over $k(u)$ and what are the roots? – Ri-Li Aug 24 '15 at 16:56
  • I pointed out a root to you in my answer; you should see at a glance what the others are. Since $u$ is in the fixed field, you certainly have $[k(x)\colon k(u)]\ge p$; but you have a polynomial with $x$ for a root that’s of degree $\le p$ (it’s a factor of the polynomial I showed you). So the degree of $x$ over $k(u)$ is both at most and at least $p$. So equal, and the polynomial is irreducible. – Lubin Aug 24 '15 at 19:17
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    Yeah $x$ is a root, Okay then $x+1$ is a root then $x+2$ so on upto $x+(p-1)$. Thank you – Ri-Li Aug 25 '15 at 10:27