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It is not hard to construct such curves explicitly, e.g. my favourite example is a curve $U$ singular at $(3:4:5i)$ and also passing through $(1:0:0)$, $(0:1:0)$, $(0:0:1)$, $(1:1:0)$, $(1:0:1)$, $(0:1:1)$. (Amusingly, one more real point on $U$ is also rational, namely $(3^4:4^4:5^4)$; two more are quadratic irrational).

Is there an easier argument demonstrating that such curves exist? A reference would be good enough, too.

Edit: It is immediate from Bezout theorem that a curve $W$ with more than 9 real points must contain a real oval (i.e. a curve in $\mathbb{RP^2}$), just look at the intersection of $W$ and its complex conjugate $W^*$. A similar argument (see Noam's answer) shows that if it has 8 real points it must also have 9 of them (counting multiplicities, as usual).

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Let $P_1,\ldots,P_8$ be "random" real points (which could even be rational). Then the space of cubic polynomials vanishing on all $P_i$ has dimension $10 - 8 = 2$. Let $(C_1, C_2)$ be a basis of this real vector space. Then the cubic curves $C_1=0$ and $C_2=0$ meet at $P_1,\ldots,P_8$ and at some ninth point $P_9$, which is automatically real (and also rational if $P_1,\ldots,P_8$ were). The cubic $C_1 + i C_2 = 0$ then passes through $P_1,\ldots,P_9$ and has no other real point.

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  • $\begingroup$ Well, this is the intuition (one can probably even replace "random" by "no 4 on a line, no 7 on a conic"). But I don't know how one does make this into a proof. $\endgroup$
    – Dima Pasechnik
    Commented Jun 13, 2018 at 16:37
  • $\begingroup$ OK --- indeed, all you need is that the space of real cubics vanishing on the 8 points is 2-dimensional, and $C_1$ and $C_2$ do not share a component. $\endgroup$
    – Dima Pasechnik
    Commented Jun 13, 2018 at 17:25

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