Image of projective 1-space contained in projective 1-space over a smaller field?

Question

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...

Answer 1

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 $K=F$.

Answer 2

Here is a proof. Let $t$, $K$ and $F$ be as stated.

Preliminary reduction: $t$ is defined over $F$.

Proof: Let $t(x) = p(x)/q(x)$, with $p=\sum_{i=0}^r p_i x^i$ and $q=\sum_{j=0}^s q_j x^j$. For every $x \in F$, let $y = t(x)$; then $\sum q_j x^j y - \sum p_i x^i$ is a linear constraint on the $p_i$ and $q_j$, with coefficients in $F$. Since there is a nonzero solution to these constraints in $K$, there is also a solution in $F$; call it $(p'_0, \ldots, p'_r, q'_0, \ldots, q'_s)$. So we have a rational function $p'/q'$ such that $p'(x)/q'(x) = p(x)/q(x)$ for all $x \in F$. Since $F$ is infinite, this means that $p/q = p'/q'$. QED

Theorem: Let $F$ be infinite. Let $t \in F(x)$ be nonconstant and separable. Let $L$ be the subfield of $F(x)$ generated by $t(g(x))$, as $g$ ranges over all nonconstant rational functions in $F(x)$. Then $L=F(x)$.

Proof: Since $F(t) \subseteq L \subseteq F(x)$, we see that $F(x)/L$ is a separable finite extension. By Luroth's theorem, $L = F(u)$ for some $u \in F(x)$, and this $u$ is a separable map $\mathbb{P}^1_F \to \mathbb{P}^1_F$. Suppose for the sake of contradiction that $u$ has degree $d>1$.

Since $F$ is infinite, we can find two points in $\mathbb{P}^1_F$ where $t$ takes two different values; using an automorphism of $\mathbb{P}^1_F$, we may assume that $t(\infty) \neq t(0)$.

Since $u$ is separable, for all but finitely many $y \in \mathbb{P}^1_F$, the preimage $u^{-1}(y)$ in $F^{\mathrm{alg}}$ has size $d$. Since $F$ is infinite, we can choose $\zeta \in \mathbb{P}^1_F$ so that $u^{-1}(u(\zeta))$ has size $d$ (with the preimage taken in $F^{\mathrm{alg}}$.)

Let $\alpha$ be an element of $u^{-1}(u(\zeta))$ other than $\zeta$. So $\alpha$ lies in some finite extension of $F$ (possibly $F$ itself). Let $p$ be the minimal polynomial of $\alpha$ and set $g(x) = p(x)/(x-\zeta)$. So $(t \circ g)(\alpha) = t(0) \neq t(\infty) = (t \circ g)(\zeta)$. So $t \circ g$ takes different values on $\alpha$ and on $\zeta$, and thus cannot be in $F(u)$. We have found an element of $L$ which is not in $F(u)$, a contradiction. The theorem is proved. QED

By the Theorem, $x$ can be written as $h(t \circ g_1, t \circ g_2, \ldots, t \circ g_N)$ for some rational functions $g_i \in F(x)$ and some $h \in F(y_1, \ldots, y_N)$. So, for any $x \in K$, we have $$x = h(t(g_1(x)), \ldots, t(g_N(x)) ).$$

Now, all of the $g_i(x)$ are in $K$. So, by hypothesis, all the $t(g_i(x))$ are in $F$. So $h(t(g_1(x)), \ldots, t(g_N(x)) )$ is in $F$. In short, we have shown that, for every $x \in K$, we have $x \in F$, which is what we wanted.