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Let $K$ be a field of characteristic $0$. Let $C/K$ be a be a quasi-projective conic defined over $K$. Let $L/K$ be a finite dimensional field extension of odd dimension. Assume that $C(L)$ is not empty.

Q: Then is true that $C(K)$ is not empty? If so, then how does one prove it?

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    $\begingroup$ The conic $C$ has a divisor of odd degree over $K$; since it has a divisor of degree 2 over $K$, it has a divisor of degree 1 over $K$. Use the divisor of degree 1 to show that $C$ embeds in $\mathbb{P}^1_K$. $\endgroup$
    – M P
    Commented Aug 6, 2011 at 15:52
  • $\begingroup$ Ok so your argument is kind of non trivial since here you are using Riemann-Roch over the field $K$ isn't it ? $\endgroup$
    – Hugo Chapdelaine
    Commented Aug 6, 2011 at 16:20
  • $\begingroup$ In any case, thanks MP for your solution, I don't mind to use Riemann-Roch. $\endgroup$
    – Hugo Chapdelaine
    Commented Aug 6, 2011 at 16:35
  • $\begingroup$ Addendum to MP: also use the fact that since $K$ has characteristic $0$, it is infinite, and therefore the finitely many $K$-rational points that you may lose from being quasi-projective instead of projective do not exhaust all the $K$-rational points on $C$. $\endgroup$
    – Pete L. Clark
    Commented Aug 6, 2011 at 20:52
  • $\begingroup$ I think that once you translate your claim into a statement about quaternion algebras it becomes an exercise in T.Y. Lam's quadratic forms over a field. $\endgroup$
    – stankewicz
    Commented Aug 6, 2011 at 21:07

2 Answers 2

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You can find an elementary proof in Un théorème arithmétique sur les coniques by Trygve Nagell in Ark. f. Mat. 2 (1952). This is a quadratic analogue of what Birch called Heegner's Lemma, which deals with the solvability of certain quartic equations.

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So if we use Riemann-Roch for smooth projective curves over $K$ the problem becomes easy. So without lost of generality we may assume that $C/K$ is smooth and projective since a conic admits a point over $K$ iff $C(K)$ is infinite (this is because the existence of a parametrization over $K$).

If $C(K)$ is not empty we are done. So now assume that $C(K)$ is empty. We will try to reach a contradiction. Let $Q$ be a point in $C(L)$ with minimal field of definition $L$ and let $[L:K]=m\equiv 1\pmod{2}$. Choose a point $P_1$ in $C(\overline{K})$ that lives in a quadratic extension of $K$. Such a point exists since we have a conic and $C(K)$ is empty. Let $P:=P_1+P_1^{\sigma}\in Div_K(C)$. Thus we have $deg(P)=2$ and $deg(Q)=m\geq 3$. We way thus write $m=2a+1$ for some positive integer $a$. Now let $$ D:=[Q]-a[P]\in Div_K(C) $$
We have $deg(D)=1$. Now let us consider the line bundle $L_D$ on $C$ where $$ L_D=\{f\in K(C):div(f)\geq -D\} $$ By Riemann-Roch, we have that $dim_K(L_D)=2$ and thus there exists a non-constant function $f\in L_D$. Note that the map $$ C(\overline{K})\rightarrow P^1(\overline{K}) $$ given by $x\mapsto [f(x),1]$ has degree $deg(div(f)_{\infty})$. So in general, it is not an embedding. Now let us work over $\overline{K}$ so that $[Q]=[Q_1]+[Q_2]\ldots+[Q_m]$ and $[P]=[P_1]+[P_2]$. Since $$ div(f)\geq -D, $$ $f\in K(C)$ (so $deg(div(f))=0$ and $div(f)$ is $G_K$-invariant) we must have that over $\overline{K}$ $$ div(f)=a[P_1]+a[P_2]+[P_3]-[Q_1]-[Q_2]-\ldots -[Q_m] $$ where $deg([P_3])=1$. This forces $P_3$ to be defined over $K$. This contradicts the fact that $C(K)$ was empty.

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  • $\begingroup$ @Hugo: I think you are making this a bit more complicated than is necessary. A smooth projective genus zero curve $C_{/K}$ has a $K$-rational divisor of degree $2$: the anti-canonical divisor. If it also has an $L$-rational point with $[L:K] = 2k+1$, then the trace from $L$ down to $K$ is a $K$-rational divisor of degree $2k+1$. Since $2$ and $2k+1$ are relatively prime, we get a $K$-rational divisor $D$ of degree $1$. Since the genus is zero, Riemann-Roch implies that $D$ is equivalent to an effective divisor of degree $1$. $\endgroup$
    – Pete L. Clark
    Commented Aug 6, 2011 at 21:53
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    $\begingroup$ Hi Pete, I guess that I had to reprove for myself that if $D$ (a divisor defined over $K$) has degree $1$ on a rational curve then necessarily it is equivalent to an effective divisor of degree $1$ defined over $K$ was not completely obvious to me, but I agree that it is fairly straightforward once you think about it. $\endgroup$
    – Hugo Chapdelaine
    Commented Aug 7, 2011 at 1:00
  • $\begingroup$ @Hugo: okay, sounds good. At this point I have written the last sentence in my above comment so many times in my own work that I have stopped thinking about it. Sorry that I didn't recognize your proof of that above. $\endgroup$
    – Pete L. Clark
    Commented Aug 7, 2011 at 21:35

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