TITLE: Image of projective 1-space contained in projective 1-space over a smaller field?
QUESTION [12 upvotes]: This is inspired by 
Does "all points rational" imply "constant" for this  "cubic" curve over an arbitrary field? .
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...

REPLY [7 votes]: 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:

$f$ is finite.
$f(X(K))\subset Y(F)$. 
$X(F)$ 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 $K=F$.