Bézout's theorem tells:
Suppose that $X$ and $Y$ are two plane projective curves defined over a field $K$ that do not have a common component, Then the total number of intersection points of $X$ and $Y$ with coordinates in an algebraically closed field $E$ which contains $K$, counted with their multiplicities, is equal to the product of the degrees of $X$ and $Y$.
My question is: if $X$ and $Y$ with coordinates in field $K$, and there are $deg(X)deg(Y)-1$ intersection points lie in $K$, will the last intersection point(must in $E$) lies in $K$?
More generally, when all intersection points can lie into a field(not needed to be algebraically closed). I have know from Wikipedia's Examples(two cycles). It is wrong when there are only $deg(X)deg(Y)-2$ intersection points lie in $K$
Thanks for your help.
Because the action of $\operatorname{Gal}(E/K)$ fixes the polynomials defining your scheme, it also on all of your schemes as well: the conjugate of a point in your scheme is also in the scheme.
In particular, it acts on the scheme consisting of the $\operatorname{deg}(X)\operatorname{deg}(Y)$. And since the Galois group fixes your first $\operatorname{deg}(X)\operatorname{deg}(Y) - 1$ points; any element of the Galois group cannot map the last intersection point to any of the first $\operatorname{deg}(X)\operatorname{deg}(Y)-1$ points, and thus it must also fix the last point.
Since the last point is fixed by $\operatorname{Gal}(E/K)$, it too must have coordinates ni $K$.