[proofplan] We prove that $O$ is the identity directly from the [chord-and-tangent law](/page/Chord-and-Tangent%20Law), used here as the defining reference for the elliptic curve operation on a [projective Weierstrass cubic](/page/Projective%20Weierstrass%20Cubic). For an affine point $P=(x,y)$, the line through $P$ and $O$ is the vertical projective line $X=xZ$, whose three intersections with $E$ are $P$, its reflection $\iota(P)=(x,-y)$, and $O$, counted with [intersection multiplicity](/page/Intersection%20Multiplicity). Thus $P*O=\iota(P)$, and applying the reflection once more gives $P+O=P$. The case $P=O$ is handled separately by checking that the tangent line at $O$ is the line at infinity and meets $E$ only at $O$, with multiplicity three. [/proofplan]

[step:Record the reflection map and the chord-and-tangent convention] Define the reflection morphism $\iota: E \to E$ by sending each projective point $[X:Y:Z] \in E$ to $[X:-Y:Z] \in E$. This map is well-defined on $E$ because substituting $-Y$ for $Y$ leaves the equation \begin{align*} Y^2Z = X^3+aXZ^2+bZ^3 \end{align*} unchanged. The nonsingularity hypothesis says precisely that the projective cubic $E \subset \mathbb{P}^2_k$ is smooth, so each point of $E$ has a well-defined tangent line and every line used in the chord-and-tangent construction meets $E$ in a degree-three intersection cycle counted with [intersection multiplicity](/page/Intersection%20Multiplicity). The line-intersection convention in the [Chord-and-Tangent Law](/page/Chord-and-Tangent%20Law) is therefore applicable: for distinct inputs one uses the projective line through them, and for equal inputs one uses the tangent line at that point. Since $\operatorname{char}(k) \ne 2$, the affine formula is genuinely the reflection sending an affine point $(x,y)$ to the affine point $(x,-y)$. Also, $O=[0:1:0]$ is fixed by $\iota$, since \begin{align*} [0:-1:0]=[0:1:0] \end{align*} in projective space. By the defining convention fixed in the [Chord-and-Tangent Law](/page/Chord-and-Tangent%20Law), if $R,S \in E(k)$, then \begin{align*} R+S=\iota(R*S), \end{align*} where $R*S$ is the third intersection point of $E$ with the line determined by $R$ and $S$, counted with [intersection multiplicity](/page/Intersection%20Multiplicity). [/step]

[step:Compute the intersections of the vertical line through an affine point and $O$]Let $P \in E(k)$ be an affine point, and write \begin{align*} P=[x:y:1] \end{align*} with $x,y \in k$. Since $P \in E(k)$, we have \begin{align*} y^2=x^3+ax+b. \end{align*} Let $\mathbb{P}^2_k$ denote the projective plane over $k$ with homogeneous coordinates $[X:Y:Z]$. Let $L_P \subset \mathbb{P}^2_k$ be the projective line defined by \begin{align*} X=xZ. \end{align*} Then $P \in L_P$ and $O \in L_P$, because \begin{align*} x=x \cdot 1 \end{align*} for $P=[x:y:1]$, while \begin{align*} 0=x \cdot 0 \end{align*} for $O=[0:1:0]$. To compute $E \cap L_P$, use the homogeneous parametrization of $L_P$ by projective coordinates $[Y:Z]$ via \begin{align*} [Y:Z]\mapsto [xZ:Y:Z]. \end{align*} Pulling back the homogeneous cubic equation along this parametrization is exactly the substitution $X=xZ$. We obtain \begin{align*} Y^2Z=x^3Z^3+axZ^3+bZ^3=(x^3+ax+b)Z^3=y^2Z^3. \end{align*} Thus the restricted cubic on $L_P$ is the homogeneous degree-three equation \begin{align*} Z(Y^2-y^2Z^2)=0. \end{align*} The factor $Z=0$ gives the point $O=[0:1:0]$. On the affine chart $Z \ne 0$, write $t=Y/Z$. The remaining equation becomes \begin{align*} t^2-y^2=0, \end{align*} so the affine intersection points are \begin{align*} [x:y:1]=P \end{align*} and \begin{align*} [x:-y:1]=\iota(P), \end{align*} with multiplicity two at $P=\iota(P)$ when $y=0$. Therefore the third intersection point of $E$ with the line through $P$ and $O$ is \begin{align*} P*O=\iota(P). \end{align*}[/step]

[guided]Let $P=[x:y:1]$ be an affine $k$-point of $E$. Let $\mathbb{P}^2_k$ denote the projective plane over $k$ with homogeneous coordinates $[X:Y:Z]$, and recall that the reflection map is the morphism $\iota: E \to E$ given by $\iota([X:Y:Z])=[X:-Y:Z]$. The first task is to identify the projective line determined by $P$ and $O$. Since $O=[0:1:0]$ is the point at infinity in the vertical direction, the line through $P$ and $O$ should be the vertical line over the affine $x$-coordinate of $P$. In homogeneous coordinates, this line is \begin{align*} L_P=\{[X:Y:Z]\in \mathbb{P}^2_k : X=xZ\}. \end{align*} It contains $P=[x:y:1]$ because $X=x$ and $Z=1$, and it contains $O=[0:1:0]$ because both sides of $X=xZ$ are zero. Now we compute the scheme-theoretic intersection in the elementary form needed for the [chord-and-tangent law](/page/Chord-and-Tangent%20Law). The line $L_P$ is parametrized by projective coordinates $[Y:Z]$ through the map \begin{align*} [Y:Z]\mapsto [xZ:Y:Z]. \end{align*} Therefore restricting the cubic equation to $L_P$ means substituting $X=xZ$. This gives \begin{align*} Y^2Z=(xZ)^3+a(xZ)Z^2+bZ^3=(x^3+ax+b)Z^3. \end{align*} Because $P \in E(k)$, its coordinates satisfy \begin{align*} y^2=x^3+ax+b. \end{align*} Hence the intersection equation along $L_P$ becomes \begin{align*} Y^2Z=y^2Z^3, \end{align*} or equivalently \begin{align*} Z(Y^2-y^2Z^2)=0. \end{align*} This factorization accounts for all three intersections, counted with multiplicity. The factor $Z=0$ gives the point at infinity \begin{align*} O=[0:1:0]. \end{align*} On the affine chart $Z \ne 0$, define the affine coordinate \begin{align*} t:=Y/Z. \end{align*} Then the remaining factor becomes \begin{align*} t^2-y^2=0. \end{align*} Since $\operatorname{char}(k)\ne 2$, this factors as \begin{align*} t^2-y^2=(t-y)(t+y). \end{align*} Thus the affine intersection points are \begin{align*} [x:y:1]=P \end{align*} and \begin{align*} [x:-y:1]=\iota(P). \end{align*} If $y=0$, these two points coincide, and the same factorization records the doubled intersection multiplicity at $P$. The chord-and-tangent definition says that $P*O$ is the third intersection point of $E$ with the line through $P$ and $O$, counted with multiplicity. We have just shown that the three intersections are $P$, $\iota(P)$, and $O$, with the expected multiplicity in the tangent case $y=0$. Therefore \begin{align*} P*O=\iota(P). \end{align*}[/guided]

[step:Reflect the third intersection point to obtain $P+O=P$]For the affine point $P=[x:y:1]$, the previous step gives \begin{align*} P*O=\iota(P). \end{align*} Let $\operatorname{id}_E: E \to E$ denote the identity morphism on $E$. Using the definition of addition and the fact that $\iota^2=\operatorname{id}_E$, we get \begin{align*} P+O=\iota(P*O)=\iota(\iota(P))=P. \end{align*}[/step]

[guided]The chord-and-tangent convention defines addition by first taking the third intersection point and then reflecting it across the $x$-axis. In the present case, the previous step proved that the third intersection point is \begin{align*} P*O=\iota(P). \end{align*} Let $\operatorname{id}_E: E \to E$ denote the identity morphism on $E$. Since $\iota$ sends $[X:Y:Z]$ to $[X:-Y:Z]$, applying $\iota$ twice gives \begin{align*} \iota(\iota([X:Y:Z]))=[X:Y:Z], \end{align*} so $\iota^2=\operatorname{id}_E$. Therefore the addition law gives \begin{align*} P+O=\iota(P*O)=\iota(\iota(P))=P. \end{align*}[/guided]

[step:Repeat the argument with the inputs reversed]For an affine point $P \in E(k)$, the projective line through $O$ and $P$ is the same line $L_P$ as the projective line through $P$ and $O$. Hence the third intersection point is the same: \begin{align*} O*P=P*O=\iota(P). \end{align*} Therefore \begin{align*} O+P=\iota(O*P)=\iota(\iota(P))=P. \end{align*}[/step]

[guided]The operation $*$ is defined from the line determined by the two inputs. Reversing the order from $(P,O)$ to $(O,P)$ does not change that projective line: both inputs determine the same vertical line $L_P$. Hence the third intersection point is the same one computed before, \begin{align*} O*P=P*O=\iota(P). \end{align*} Applying the definition of addition and the identity $\iota^2=\operatorname{id}_E$ gives \begin{align*} O+P=\iota(O*P)=\iota(\iota(P))=P. \end{align*}[/guided]

[step:Handle the point at infinity]It remains to consider $P=O$. Let $F \in k[X,Y,Z]$ be the homogeneous cubic polynomial \begin{align*} F(X,Y,Z):=Y^2Z-X^3-aXZ^2-bZ^3. \end{align*} The tangent line to $E$ at $O=[0:1:0]$ is obtained from the linear part of $F$ at $O$. The partial derivative with respect to $X$ is \begin{align*} \frac{\partial F}{\partial X}=-3X^2-aZ^2. \end{align*} The partial derivative with respect to $Y$ is \begin{align*} \frac{\partial F}{\partial Y}=2YZ. \end{align*} The partial derivative with respect to $Z$ is \begin{align*} \frac{\partial F}{\partial Z}=Y^2-2aXZ-3bZ^2. \end{align*} At $O=[0:1:0]$, these evaluate to \begin{align*} \frac{\partial F}{\partial X}(O)=0. \end{align*} Also, \begin{align*} \frac{\partial F}{\partial Y}(O)=0. \end{align*} Finally, \begin{align*} \frac{\partial F}{\partial Z}(O)=1. \end{align*} Thus the tangent line at $O$ is \begin{align*} Z=0. \end{align*} Intersecting $E$ with this line gives \begin{align*} 0=X^3. \end{align*} On the projective line $Z=0$, the coordinate $Y$ is nonzero at every point of the intersection, so the affine coordinate $u:=X/Y$ is a local coordinate around $O=[0:1:0]$ on that line. Dividing the equation $X^3=0$ by $Y^3$ on the chart $Y \ne 0$ gives \begin{align*} u^3=0. \end{align*} Thus the local intersection algebra is $k[u]/(u^3)$, whose $k$-dimension is $3$. Therefore the only intersection point is $O$, and its intersection multiplicity with the line $Z=0$ is exactly three. Hence \begin{align*} O*O=O. \end{align*} Since $\iota(O)=O$, the addition law gives \begin{align*} O+O=\iota(O*O)=\iota(O)=O. \end{align*} Combining this with the affine case proves that for every $P \in E(k)$, \begin{align*} P+O=O+P=P. \end{align*}[/step]

[guided]When $P=O$, the line used in the chord-and-tangent definition is the tangent line at $O$. We compute this tangent line from the homogeneous cubic polynomial $F: k^3 \to k$ defined by \begin{align*} F(X,Y,Z):=Y^2Z-X^3-aXZ^2-bZ^3. \end{align*} The projective curve $E$ is cut out by $F=0$, and at the point $O=[0:1:0]$ the tangent line is obtained from the linear form determined by the gradient of $F$ at $O$. The partial derivative with respect to $X$ is \begin{align*} \frac{\partial F}{\partial X}=-3X^2-aZ^2. \end{align*} The partial derivative with respect to $Y$ is \begin{align*} \frac{\partial F}{\partial Y}=2YZ. \end{align*} The partial derivative with respect to $Z$ is \begin{align*} \frac{\partial F}{\partial Z}=Y^2-2aXZ-3bZ^2. \end{align*} Evaluating these derivatives at the representative $(0,1,0)$ of $O$ gives \begin{align*} \frac{\partial F}{\partial X}(O)=0, \qquad \frac{\partial F}{\partial Y}(O)=0, \qquad \frac{\partial F}{\partial Z}(O)=1. \end{align*} Therefore the tangent line is the projective line defined by \begin{align*} Z=0. \end{align*} This is the line at infinity. We now compute its intersection with $E$. Substituting $Z=0$ into the equation \begin{align*} Y^2Z=X^3+aXZ^2+bZ^3 \end{align*} gives \begin{align*} 0=X^3. \end{align*} Thus every intersection point on the projective line $Z=0$ has $X=0$. Since $[0:0:0]$ is not a projective point, the coordinate $Y$ must be nonzero, so the only point of the intersection is \begin{align*} [0:1:0]=O. \end{align*} On the chart $Y \ne 0$, define the affine coordinate $u:=X/Y$ on this projective line near $O=[0:1:0]$. Dividing the equation $X^3=0$ by $Y^3$ gives \begin{align*} u^3=0. \end{align*} Thus the local intersection algebra is $k[u]/(u^3)$, which has $k$-basis $1,u,u^2$ and therefore $k$-dimension $3$. Hence the tangent line meets $E$ only at $O$, with intersection multiplicity three, so \begin{align*} O*O=O. \end{align*} Because $\iota(O)=O$, the addition law gives \begin{align*} O+O=\iota(O*O)=\iota(O)=O. \end{align*} Together with the affine computations $P+O=P$ and $O+P=P$, this proves \begin{align*} P+O=O+P=P \end{align*} for every $P \in E(k)$.[/guided]