[proofplan]
We use the defining geometry of the chord-and-tangent group law. The vertical projective line through an affine point $P=(x_0,y_0)$ meets the cubic at $P$, at its reflected point $(x_0,-y_0)$, and at the point at infinity $O$, with multiplicities included. The group law says that the sum of two intersection points on a line is the reflection of the third one; since the reflection of $O$ is $O$, this gives $P+(-P)=O$. The final assertion follows algebraically from the equation $y^2=x^3+ax+b$.
[/proofplan]
[step:Identify the third intersection of the vertical line through $P$]
Let $P=(x_0,y_0) \in E(\mathbb{Q})$ be an affine point, so
\begin{align*}
y_0^2 = x_0^3 + ax_0 + b.
\end{align*}
Let $L \subset \mathbb{P}^2_{\mathbb{Q}}$ be the projective line
\begin{align*}
L := \{[X:Y:Z] \in \mathbb{P}^2_{\mathbb{Q}} : X - x_0 Z = 0\}.
\end{align*}
This line contains $P=[x_0:y_0:1]$, contains $-P=[x_0:-y_0:1]$, and contains $O=[0:1:0]$.
On the affine chart $Z \neq 0$, write $x=X/Z$ and $y=Y/Z$. The equation of $L$ becomes $x=x_0$, and the equation of $E$ becomes
\begin{align*}
y^2 = x_0^3 + ax_0 + b = y_0^2.
\end{align*}
Thus the affine intersections are precisely the solutions of
\begin{align*}
(y-y_0)(y+y_0)=0,
\end{align*}
namely $(x_0,y_0)$ and $(x_0,-y_0)$, counted with multiplicity when $y_0=0$. On the line at infinity $Z=0$, the equation $X-x_0Z=0$ gives $X=0$, so the only point of $L$ at infinity is $[0:1:0]=O$. To record the intersection multiplicity at this point, use the projective equation
\begin{align*}
Y^2Z=X^3+aXZ^2+bZ^3.
\end{align*}
Substituting $X=x_0Z$ on $L$ gives
\begin{align*}
Y^2Z=(x_0^3+ax_0+b)Z^3=y_0^2Z^3,
\end{align*}
so the restriction of the cubic equation to $L$ is
\begin{align*}
Z(Y^2-y_0^2Z^2)=0.
\end{align*}
At $O=[0:1:0]$ we have $Y\neq 0$, and in the affine coordinate $u=Z/Y$ on the chart $Y\neq 0$ this equation becomes
\begin{align*}
u(1-y_0^2u^2)=0.
\end{align*}
The factor $1-y_0^2u^2$ has value $1$ at $u=0$, so the intersection at $O$ has multiplicity one. Hence $L$ intersects $E$ in $P$, $-P$, and $O$, with multiplicity counted.
[guided]
We want to understand exactly which three points the vertical line contributes to the chord-and-tangent construction. Define the projective line
\begin{align*}
L := \{[X:Y:Z] \in \mathbb{P}^2_{\mathbb{Q}} : X - x_0 Z = 0\}.
\end{align*}
This is the projective closure of the affine vertical line $x=x_0$. It contains $P=[x_0:y_0:1]$ because $x_0-x_0=0$, and it contains $-P=[x_0:-y_0:1]$ for the same reason. It also contains $O=[0:1:0]$, since $X-x_0Z=0-x_0\cdot 0=0$.
Now restrict to the affine chart $Z\neq 0$, where $x=X/Z$ and $y=Y/Z$. The line equation becomes $x=x_0$. Substituting this into the affine Weierstrass equation gives
\begin{align*}
y^2 = x_0^3 + ax_0 + b.
\end{align*}
Because $P=(x_0,y_0)$ lies on $E$, we already know
\begin{align*}
y_0^2 = x_0^3 + ax_0 + b.
\end{align*}
Therefore the affine intersection equation is
\begin{align*}
y^2=y_0^2,
\end{align*}
or equivalently
\begin{align*}
(y-y_0)(y+y_0)=0.
\end{align*}
Thus the affine intersections are $(x_0,y_0)$ and $(x_0,-y_0)$. If $y_0=0$, these two points coincide, and the factorization shows that the affine root is repeated.
It remains to check the point at infinity on this line and its multiplicity. Setting $Z=0$ in the line equation gives $X=0$, so every point of $L$ at infinity has the form $[0:Y:0]$ with $Y\neq 0$. In projective space all such points are equal to $[0:1:0]=O$.
For multiplicity, use the projective Weierstrass equation
\begin{align*}
Y^2Z=X^3+aXZ^2+bZ^3.
\end{align*}
On the line $L$ we have $X=x_0Z$, so substitution gives
\begin{align*}
Y^2Z=(x_0^3+ax_0+b)Z^3=y_0^2Z^3.
\end{align*}
Equivalently, the restricted equation is
\begin{align*}
Z(Y^2-y_0^2Z^2)=0.
\end{align*}
Near $O$ we work in the affine chart $Y\neq 0$ and set $u=Z/Y$. Dividing by the nonzero factor $Y^3$ rewrites the restricted equation as
\begin{align*}
u(1-y_0^2u^2)=0.
\end{align*}
Since $1-y_0^2u^2$ equals $1$ at $u=0$, the zero at $u=0$ is simple. Thus the point at infinity contributes exactly one intersection point, namely $O$, counted with multiplicity one. Hence the vertical line meets $E$ at $P$, $-P$, and $O$, with multiplicity included.
[/guided]
[/step]
[step:Apply the chord-and-tangent group law to the vertical line]
By the [chord-and-tangent definition of the group law](/page/Elliptic%20Curve%20Group%20Law) on a nonsingular plane cubic, if a line meets $E$ in three points $R,S,T$, counted with multiplicity, then
\begin{align*}
R+S+T=O.
\end{align*}
Applying this rule to the line $L$ from the previous step gives
\begin{align*}
P+(-P)+O=O.
\end{align*}
Since $O$ is the identity element for the group law on $E(\mathbb{Q})$, this simplifies to
\begin{align*}
P+(-P)=O.
\end{align*}
[/step]
[step:Show that two distinct affine points with the same $x$-coordinate are negatives]
Let $P=(x_1,y_1)$ and $Q=(x_2,y_2)$ be affine points of $E(\mathbb{Q})$ with $x_1=x_2$ and $P\neq Q$. Since both points lie on $E$,
\begin{align*}
y_1^2 = x_1^3 + ax_1 + b.
\end{align*}
Also,
\begin{align*}
y_2^2 = x_2^3 + ax_2 + b.
\end{align*}
Using $x_1=x_2$, we obtain
\begin{align*}
y_1^2=y_2^2.
\end{align*}
Therefore
\begin{align*}
(y_2-y_1)(y_2+y_1)=0.
\end{align*}
If $y_2-y_1=0$, then $y_2=y_1$, and together with $x_2=x_1$ this gives $Q=P$, contradicting $P\neq Q$. Hence $y_2+y_1=0$, so $y_2=-y_1$. Therefore
\begin{align*}
Q=(x_1,-y_1)=-P.
\end{align*}
This proves the second assertion and completes the proof.
[/step]
