Let $K$ be a field of characteristic different from two. Conics over $K$ (that is curves of degree two in $\mathbb{P}^2_K$) are parametrized by $\mathbb{P}(k[x,y,z]_2) = \mathbb{P}^5_K$.

Conisider the set $R = \{C \in \mathbb{P}(k[x,y,z]_2) \: | \: C \text{ has a $K$-point}\}\subseteq\mathbb{P}^5$.

If $K$ is algebraically closed then $R = \mathbb{P}^5$ but if not there are conics without points.

When $K$ is not algebraically closed does $R$ have any meaningful geometric structure?

  • 2
    $\begingroup$ It is an orbit of $\operatorname{PGL}(3,K) $ acting on $\mathbb{P}^5_K$. $\endgroup$
    – abx
    Jan 12 at 19:53
  • $\begingroup$ If I understand correctly you take a conic with a point and look at its orbit under $PGL(3,K)$. But what is the dimension of this orbit? $\endgroup$
    – Arty
    Jan 12 at 20:03
  • 1
    $\begingroup$ It is a set; it has no geometric structure. In any case the orbit is Zariski dense. $\endgroup$ Jan 12 at 20:16
  • $\begingroup$ Since the OP does not exclude singular conics, it is not true that $\mathrm{PGL}(3,K)$ acts transitively on $R$. There is one orbit consisting of smooth conics (rank 3), one for double lines (rank 1), and a whole family for rank 2, parametrized by $K^\times/K^{\times2}$, with $d\in K^\times$ coding for the conic $x^2-dy^2$. $\endgroup$ Jan 13 at 12:41
  • $\begingroup$ @DanielLoughran: the orbit is not Zariski-dense if (and only if) $K$ is finite. But in this case, of course, $R=\mathbb{P}^5(K)$. And also, what we mean by Zariski-dense is debatable. $\endgroup$ Jan 15 at 9:37

1 Answer 1


Let $$X: \quad a_{0,0}x^2 + a_{1,1}y^2 + a_{2,2}z^2 + a_{0,1}xy + a_{0,2}xz + a_{1,2}yz = 0 \quad \subset \mathbb{P}^2 \times \mathbb{P}^5$$ be the total space of the family of all plane conics. Then your set $R$ is the image $\pi_2(X(K)) \subset \mathbb{P}^5(K)$ of the $K$-rational points on $X$ with respect to the second projection.


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