This is inspired by 


[Does "all points rational" imply "constant" for this "cubic" curve over an arbitrary field?](https://mathoverflow.net/questions/67126/does-all-points-rational-imply-constant-for-this-cubic-curve-over-an-arbit) .


Say $K/F$ is a finite separable extension of fields. Assume $F$ is infinite (or else there are trivial counterexamples to the question below). Say 


$$t:\mathbf{P}^1\_K\to\mathbf{P}^1\_K$$


is a non-constant morphism defined over $K$, with the property that 


$$t(\mathbf{P}^1(K))\subseteq\mathbf{P}^1(F).$$


Does this imply that $K=F$?


Comments: (1) the question linked to above is a special case of this, written in a slightly more hands-on way. (2) There are counterexamples in the inseparable case of the form $t(x)=x^p$ with $K/F$ purely inseparable of degree $p$. (3) If $K$ is algebraically closed then $F=K$ trivially because $t$ is surjective on points. (4) If $F$ is a number field then I guess one can use the theory of "thin sets" (the ideas used in the proof of Hilbert irreducibility) to prove that $F=K$. (5) If you generalise to other projective curves then again the result fails, because e.g. the curves could have no $K$-points at all and $t$ could be the identity. (6) On the other hand one could ask about projective $n$-space or even affine 1-space -- one just needs a lot of $K$-points to make the question interesting... 



Image of projective 1-space contained in projective 1-space over a smaller field?
This proof was inspired by an idea of Francois Brunault and appears piecemeal in the comments to the question; it's probably not materially different from David's proof. Choose some $F$-basis $\{a\_1,\ldots,a\_n\}$ for $K$ with $a\_1=1$, and let $p:\mathbf{A}^n\_K\to\mathbf{A}^1\_K$ be the map $(X\_1,X\_2,\ldots,X\_n)\mapsto \sum\_ia\_iX\_i$. 


**Lemma** Suppose that $X$ and $Y$ are varieties over a field $F$ such that $X(F)$ is dense in $X$, that $K/F$ is a separable extension and that $f:X\_K\to Y\_K$ is a morphism such that $f(X(F))\subset Y(F)$. Then $f$ is defined over $F$.


**Proof:** Indeed, we can replace $K$ by its Galois closure and assume that $K/F$ is Galois. For every automorphism $g$ of $K/F$, $gfg^{-1}\vert\_{X(F)}=f\vert\_{X(F)}$; since $X(F)$ is dense in $X$, this implies that $gfg^{-1}=f$, and we conclude by Galois descent.


**Corollary 1** $t$ is defined over $F$.


**Corollary 2** $t\circ p$ is defined over $F$.


Let $L/F$ be the Galois closure of $K/F$. 


**Claim** The base changed map $p:\mathbf{A}^n\_L\to\mathbf{A}^1\_L$ is Galois-equivariant.


**End of proof assuming claim:** For any automorphism $g$ of $L/F$ and any $1\leq i\leq n$, $$ga\_i=g(p(e\_i))=p(g(e\_i)=p(e\_i)=a\_i,$$ 
where $e\_i$ is the $i^{\text{th}}$ standard basis vector for $\mathbf{A}^n\_L$. So all the $a\_i$ lie in $F$ (and, in particular, $n=1$).


**Proof of claim:** For any automorphism $g$ of $L/F$, we have
$$t\circ p=g(t\circ p)g^{-1}=gtg^{-1}\circ gpg^{-1}=t\circ gpg^{-1}.$$
So the pair $(p,gpg^{-1})$ defines a map $f:\mathbf{A}^n\_L\to Z\_L$, where $Z$ is the self fiber-product of $\mathbf{P}^1\_F$ over $\mathbf{P}^1\_F$ via $t$. $Z$ is a curve (not necessarily reduced or irreducible), and $f$ maps onto an irreducible component of $Z$. One of the irreducible components of $Z$ is the diagonal $\Delta$. Showing that $p=gpg^{-1}$ amounts to showing that $f$ maps onto $\Delta$. But $f$ maps the $F$-points of the first co-ordinate axis of $\mathbf{A}^n$ into the diagonal:
$$f(x,0,\ldots,0)=(x,gx)=(x,x),$$
for all $x\in\mathbf{A}^1(F)$. So it must map all of $\mathbf{A}^n\_K$ into the diagonal, and we are done.


EDIT: Here is one way to view this proof in a more general framework. Let $f:X\to Y$ be a map of $F$-varieties, with $X$ non-empty. Suppose that:


1. $f$ is finite.
2. $f(X(K))\subset Y(F)$.
3. $X(F)$ is dense in $X$.


We also suppose that the Weil restriction $X'=\text{Res}\_{K/F}X\_K$ is representable. Let $p:X'\_K\to X\_K$ be the natural adjunction map. The hypotheses imply (see Lemma above) that both $f$ and $f\circ p$ are defined over $F$. The same proof as in the claim above shows that $p$ is also defined over $F$. This shows that $K=F$.


