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


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


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...

  • $\begingroup$ To me, (4) seems most significant: it seems that one should look to easy examples of non-Hilbertian fields for counterexamples. E.g. what happens over $\mathbb{Q}_p$? $\endgroup$ Jun 8, 2011 at 20:47
  • $\begingroup$ I also took the liberty of adding the (new!) tag field-arithmetic. This is kind of based on a hunch, but I hope it will be helpful. $\endgroup$ Jun 8, 2011 at 20:49
  • $\begingroup$ Isn't it the case that $t$ is automatically defined over $F$ ? (at least if $t$ is a polynomial, by interpolation). If so I have an argument (I think) for polynomials, but I'm not sure whether it extends for rational functions. $\endgroup$ Jun 8, 2011 at 21:47
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    $\begingroup$ Yes, $t$ must be defined over $F$: We can assume $K/F$ is Galois; then, for all automorphisms $g$ of $K/F$, $gtg^{-1}\vert_{\mathbf{P}(F)}=t\vert_{\mathbf{P}(F)}$. This shows that $gtg^{-1}=t$, and we conclude by Galois descent. $\endgroup$ Jun 8, 2011 at 23:41
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    $\begingroup$ An easy comment: The claim is true for polynomial maps in characteristic zero. Proof: Let $t(x) = \sum_{k=0}^d a_k t^k$ with $a_d \neq 0$. For any $f: F \to F$, set $(\Delta f)(x) = f(x+1)-f(x)$. Then $(\Delta^{d-1} t)(x) = d! a_k x + b$ for some constant $b$. We see that, for any $x \in K$, the quantity $d! a_k x+b$ is in $F$. Plugging in $x=0$ and $x=1$, we have $b \in F$ and $d! a_k \in F$. So $x \in K$ implies $((d! a_k x+b)-b)/(d! a_k)$ is in $F$; that is to say, $x \in F$. (We used characteristic zero to make sure that $d! a_k$ is nonzero.) $\endgroup$ Jun 9, 2011 at 14:16

2 Answers 2


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 $K=F$.

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    $\begingroup$ That looks pretty different to me! You never use the fact that $\mathbb{P}^1_K$ has a lot of endomorphisms, which is pretty crucial for me. I think your argument should generalize to other maps $X \to Y$ with $X(F)$ infinite much better than mine. $\endgroup$ Jun 10, 2011 at 18:37
  • $\begingroup$ So I get home from work and find I have two proofs to choose from :-/ Which one do I accept?? $\endgroup$ Jun 10, 2011 at 19:51
  • $\begingroup$ Is the Weil restriction ever not representable? $\endgroup$ Jun 10, 2011 at 20:41
  • $\begingroup$ The best statement I can find is in Bosch, et. al.'s 'Neron models', Theorem 4 on p. 194, which implies that the Weil restriction of anything quasi-projective is representable (it's the same issue as with existence of quotients by finite groups). There are probably some pathological beasts hidden somewhere in the work of Raynaud. In any case, we can probably get away in this case with requiring $X'$ to be something like the stack quotient of a scheme by a finite group, which is fine. $\endgroup$ Jun 10, 2011 at 20:54
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    $\begingroup$ Max--This is where we use the finiteness of $t$ (or $f$ in the more general statement). The point is that every component of $Z$ has dimension $1$, and so each component is a curve. The diagonal is obviously a curve inside $Z$, and you can't have a curve be a proper sub-variety of another curve (any non-trivial equation will cut out only finitely many points). In general, every component of $Z$ will have dimension equal to $dim X$, and a similar statement holds: no variety can have a proper sub-variety of the same dimension, and in particular, the diagonal will be a component. $\endgroup$ Jun 24, 2011 at 16:28

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.

  • $\begingroup$ David -- I wonder to what extent Luroth is necessary. It had occurred to me that Luroth might be relevant -- this is one of the reasons I mentioned projective $n$-space in the question. Thanks for the answer. I've not read it yet (or Keerthi's) so I'll refrain from accepting anything. $\endgroup$ Jun 10, 2011 at 19:52
  • $\begingroup$ It isn't actually. I thought about explaining this, but I decided my answer was already too long. What I really need to know is that $L$ is the fraction field of some curve $Y$ over $F$, and that I get a separable map $u: \mathbb{P}^1_F \to Y$. The fact that $Y \cong \mathbb{P}^1_F$ isn't necessary, it just makes the argument much more readable. I think I wrote the above in such a way that it works unaltered with the weaker level of knowledge about the target of $u$. $\endgroup$ Jun 10, 2011 at 19:59
  • $\begingroup$ OK I finally read it. This seems fine to me. You seem to assume $t$ separable though. $\endgroup$ Jun 11, 2011 at 8:38
  • $\begingroup$ PS thanks! This question was bugging me :-) $\endgroup$ Jun 11, 2011 at 8:38
  • $\begingroup$ OK so finally the impossible problem of which of two distinct correct proofs to accept. I tossed a coin and Keerthi won. Thanks to both of you. $\endgroup$ Jun 11, 2011 at 8:54

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