Let $k$ perfect field.

If we have a cubic non-singular projective curve $C(k)$ (over a field $k$), take two diferent points $P_1,P_2 \in C(k)$ and consider the line through the points, by Bezout theorem this line intersect to $C(\overline{k})$ in a unique third point $P_3$, now my question is:

Why the Galois Group $Gal(\overline{k}/k)$ allow us to say that $P_3 \in C(k)$?

  • 1
    $\begingroup$ It’s the same reasoning that tells you that if $f$ is a cubic polynomial over $k$ with two roots in $k$, then the third root will also be in $k$. $\endgroup$
    – Lubin
    Commented Jan 18, 2016 at 1:57

1 Answer 1


The equations which define $P_3$ have coefficients in $k$, so $Gal(\bar k/k)$ preserves the locus of these equations i.e it fixes the coordinates of $P_3$ since it is the unique point of this locus.


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