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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.

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Suppose first that $k$ is algebraically closed and of characteristic zero.

Let $P$, $Q$ be coprime. The equality $P(t)/Q(t)=P(t+1)/Q(t+1)$ implies that $P(t)$ and $P(t+1)$ have the same zeroes in $k$, and the same for $Q(t)$ and $Q(t+1)$. This is easily seen to be impossible, unless they are constant.

If now $k$ is not necessarily algebraically closed and $\overline k$ is its algebraic closure, the invariants in $k(t)$ are contained in the invariants of $\overline k(t)$. Therefore again in this case we only have constants.

Now, if $k$ is of positive characteristic $p$, things are more subtle. Let $\sigma:k[t]\to k[t]$ be the unique autmorphism such that $\sigma(t)=t+1$, and let $\sigma:k(t)\to k(t)$ be its natural extension. The difference now is that $\sigma $ has order $p$.

If $f/g\in k(t)$, then $$ \frac fg = \frac{f\sigma(g)\sigma^2(g)\cdots\sigma^{p-1}(g)}{g\sigma(g)\sigma^2(g)\cdots\sigma^{p-1}(g)} $$ and the denominator in the right is $\sigma$-invariant. We thus see that every element in $k(t)$ can be written with an invariant denominator, so it is itself invariant iff its numerator is invariant. This means that $k(t)^\sigma$ is the quotient field of the ring of $\sigma$-invariant polynomials $k[t]^\sigma$.

Can you describe the elements of $k[t]^\sigma$?

Later: Suppose $f$ is in $k[t]^\sigma$ and, at first, that $k$ is algebraically closed. If $\alpha$ is a root of $f$, then $\alpha+1$, $\dots$, $\alpha+p-1$ are also roots of $f$ because of invariant, and therefore the product $\phi_a=\prod_{i=0}^{p-1}(t-a-i)=t^p-t-a^p+a$ divides $f$. The quotient $f/\phi_a$ is also in $k[t]^\sigma$. By induction, we conclude that $f$ is a product of polynomials of the form $\phi_a$ for various $a$s in $k$.

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So it seems as if most of the question is about the positive-characteristic case. – Michael Hardy Nov 30 '11 at 4:09
See also my answer here. – Bill Dubuque Nov 30 '11 at 4:14
I'm trying to think of the elements of $k[t]^\sigma$. If $g(t)\in k[t]^\sigma$, then $g(t+1)-g(t)=0$. Does this imply that $g(t)$ is constant, else the nonzero polynomial $g(t+1)-g(t)$ has an infinite number of roots? Or have I gone astray? – Enrique Cartan Nov 30 '11 at 4:42
Thanks for the added later part, I forgot the roots would cycle if the field has characteristic $p$. If $k$ is not algebraically closed, then $f$ is the product of $\phi_\alpha$ for roots $\alpha$ of $f$ in $k$, and something else left over possibly, if not all roots of $f$ are in $k$, correct? Is there any way to determine what that left over part is? – Enrique Cartan Nov 30 '11 at 5:23

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.

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Hello Jyrki, I must be dense, but from $[k(t):k(u)]=[k(t,u):k(u)]\le p$ how do you conclude $L\subseteq k(u)$, to get the equality $L=k(u)$? – yunone Nov 30 '11 at 20:09
@yunone: $k(u)\subseteq L \subseteq k(t)$ gives that $$[k(t):k(u)]=[k(t):L]\cdot [L:k(u)].$$ We also know that $[k(t):L]=p$, so $[k(t):k(u)]=p[L:k(u)]$, and also $p\le [k(t):k(u)]$. Together with $[k(t):k(u)]\le p$ this implies $p=[k(t):k(u)]$, and hence also $[L:k(u)]=1$. – Jyrki Lahtonen Nov 30 '11 at 20:46
Makes sense, thanks! – yunone Nov 30 '11 at 22:29

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