MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 $\mathbb{P}^1_K$. – M P Aug 6 '11 at 15:52
Ok so your argument is kind of non trivial since here you are using Riemann-Roch over the field $K$ isn't it ? – Hugo Chapdelaine Aug 6 '11 at 16:20
In any case, thanks MP for your solution, I don't mind to use Riemann-Roch. – Hugo Chapdelaine Aug 6 '11 at 16:35
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 $C$. – Pete L. Clark Aug 6 '11 at 20:52
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. – stankewicz Aug 6 '11 at 21:07
up vote 5 down vote accepted

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.

share|cite|improve this answer
Thanks Franz for the reference – Hugo Chapdelaine Aug 7 '11 at 3:02

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.

share|cite|improve this answer
@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 $1$. – Pete L. Clark Aug 6 '11 at 21:53
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. – Hugo Chapdelaine Aug 7 '11 at 1:00
@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. – Pete L. Clark Aug 7 '11 at 21:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.