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Bezout's theorem states that the number of common points of two curves is at most equal to the product of their degrees, and equality holds if one counts points at infinity and points with complex coordinates (or more generally, coordinates from the algebraic closure of the ground field), and if each point is counted with its intersection multiplicity.

If we consider a elliptic curve $E(K)$ defined over a field $K$, the point addition is defined using a line (that has always $3$ points in common with the curve $E(K)$ according to Bezout's equality). But if $K$ is not algebraically closed, given $P$ and $Q$ on $E(K)$ with coordinates in $K$, why is there a third point $P + Q \in E(K)$ (and on the line $PQ$) with coordinates in $K$? In this case we cannot apply the Bezout's equality because $K$ is not algebraically closed

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    $\begingroup$ If an equation of degree 3 has two roots in $K$, then its third root is also in $K$. $\endgroup$ – Gerry Myerson Jun 14 '18 at 10:22
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Garry Myerson's comment is enough, but there is another easy way to see it:

Let $P_1=(x_1,y_1)$ and $P_2=(x_2,y_2)$ be points on $E(K)$ and $P_3:=P_1+P_2 \neq \infty$.

If $P_3=(x_3,y_3)$, then $x_3=m^2-x_1-x_2$ and $y_3=m(x_1-x_2)-y_1$, where $m$ is the slope of the line $L$ through $P_1$ and $P_2$ (or the tangent line if $P_1=P_2$).

Note that $m, x_1,x_2,y_1 \in K$, thus $x_3,y_3 \in K$ because $K$ is a field.

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