TITLE: What exactly is the fixed field of the map $t\mapsto t+1$ in $k(t)$?
QUESTION [15 upvotes]: Suppose $k$ is a field, and $k(t)$ is the rational function field. If $f(t)=P(t)/Q(t)$ for some polynomials $P(t)$ and $Q(t)\neq 0$, then the map $t\mapsto t+1$ sends $f(t)$ to $f(t+1)$. 
So the fixed field for this map the set of rational functions $P(t)/Q(t)$ such that
$$
\frac{P(t)}{Q(t)}=\frac{P(t+1)}{Q(t+1)}\text{ or }P(t)Q(t+1)=P(t+1)Q(t).
$$
This does not seem awfully descriptive. Is there a more explicit description of what the fixed field looks like, (perhaps in terms of the form of $P$ and $Q$?), in order to get a better handle on it? 
Later: With the wonderful help of Professor Suárez-Alvarez, I understand that the $\sigma$-invariant polynomials in $k[t]$ have all factors of form $\phi_\alpha$ where $\phi_\alpha=\prod_{i=1}^{p-1} (t-\alpha-i)$ for $\alpha$ a root of $f$ in $k$ when $k$ is algebraically closed.
What happens when $k$ is not algebraically closed? I think you can still factor out polynomials of form $\phi_\alpha$ for $\alpha$ roots of $f$ in $k$. But if $f$ does not split completely then there will be some $\sigma$-invariant factor with not roots in $k$ in the factorization of $f$. Is there anything more one can say to describe this remaining part? Or is that about the best we can do? Thank you for your time.

REPLY [4 votes]: In characteristic $p$ case we can also proceed as follows. We get from Professor Suárez-Alvarez' answer that $u=t^p-t$ is contained in the fixed field $L=k(t)^\sigma$, so $k(u)\subseteq L$. Galois theory tells us that $[k(t):L]=\mathop{ord}(\sigma)=p$. OTOH clearly $[k(t):k(u)]=[k(t,u):k(u)]\le p$, because
the minimal polynomial $m(x)$ of $t$ over $k(u)$ must be a factor of $x^p-x-u$.
Therefore we can conclude that $L=k(u)$, and also that $m(x)=x^p-x-u$.
We can also tell right away that the fixed field of a finite group of automorphisms of $k(t)$ will be of the form $k(v)$ for some rational function $v$.  This follows from Lüroth's theorem.