A smooth conic is a projective line once it has a rational point

Geometric integrality, normal form, and scheme dimension. Since $C$ is smooth, [F2] says it remains regular after every field extension, in particular over an algebraic closure $\bar{k}$. If $F_{\bar{k}}$ were reducible, a quadratic factorization would be either two distinct lines, singular at their intersection, or a repeated line, singular along that line; both contradict regularity. Thus $F_{\bar{k}}$ is irreducible and $C$ is geometrically integral. Choose homogeneous coordinates with $p=[0:1:0]$ and tangent line $T_{p}C=V(Z)$. Writing $F=a{X}^2+b{Y}^2+c{Z}^2+dXY+eXZ+fYZ$, the point condition gives $b=0$; the tangent condition gives $d=0$ and $f\mathrm{\ne }0$, so $F=a{X}^2+eXZ+fYZ+c{Z}^2$. If $a=0$, then $F=Z(eX+fY+cZ)$, contradicting geometric integrality; hence $a\mathrm{\ne }0$. The charts $D(Z)$ and $D(Y)$ cover $C$, since the only projective point with $Y=Z=0$ would be $[1:0:0]$, where $F=a\mathrm{\ne }0$. On $D(Z)$, the equation $a{x}^2+ex+fy+c=0$ is linear in $y$ with coefficient $f\mathrm{\ne }0$, so the coordinate ring is $k[x]$. On $D(Y)$, the ring is the domain $R=k[x,z]\mathrm{/}(a{x}^2+exz+fz+c{z}^2)$. Its defining polynomial has positive degree in $z$ because $f\mathrm{\ne }0$. The degree-in-$z$ product rule shows $k[x]\hookrightarrow R$: a nonzero polynomial in $x$ cannot be a multiple of a polynomial of positive $z$-degree. The equation also makes $z$ algebraic over $k(x)$, so ${\mathrm{trdeg}}_k\mathrm{Frac}(R)=1$. Both finite-type chart rings are Noetherian by [F12]; the same fact makes $C$ Noetherian, and [F12] gives chart dimension one and chain dimension one for $C$ by the finite open-cover lemma. Thus [F1] makes $C$ a curve; it is proper by [F13] and smooth by assumption.

The model computation on the projective line. Let ${\mathbb{P}}_k^1$ have coordinate $t$ on $U_0$ and point at infinity $\mathrm{\infty }=[0:1]$, the pole of $t=x_1\mathrm{/}{x}_0$ [F3]. For a monic irreducible $g\in k[t]$ of degree $d$, [F4] gives $\mathrm{div}(g)=[V(g)]-d[\mathrm{\infty }]$; if $P$ is a nonzero polynomial with factorization $P=c{\prod }_i{g}_i^{e_i}$, then [F5] gives $\mathrm{div}(P)={\sum }_i{e}_i[V(g_i)]-(\mathrm{deg}P)[\mathrm{\infty }]$. Let $Q\in {\mathbb{P}}^1(k)$. If $Q=\mathrm{\infty }$, write a nonzero $f\in L([\mathrm{\infty }])$ as $P\mathrm{/}{Q}_0$ with coprime $P,Q_0\in k[t]$. Any irreducible factor $g^e$ of a nonconstant denominator contributes coefficient $-e$ at the finite point $V(g)$ in $\mathrm{div}(f)+[\mathrm{\infty }]$, since $P$ and $Q_0$ are coprime. Thus $Q_0$ is constant and $f=P$; then $\mathrm{div}(f)+[\mathrm{\infty }]$ is effective exactly when $\mathrm{deg}P\le 1$, so $L([\mathrm{\infty }])=k\cdot 1\oplus k\cdot t$ has dimension two. If $Q=[1:c]$ for $c\in k$, then $\mathrm{div}(t-c)=[Q]-[\mathrm{\infty }]$ [F4]. For $f=P\mathrm{/}{Q}_0$ in lowest terms, every irreducible denominator factor other than $t-c$ would contribute a negative coefficient at its finite point, so $Q_0=(t-c{)}^m$. Coprimeness gives $t-c\nmid P$, and effectivity at $Q$ requires $1-m\ge 0$, hence $m\in \{0,1\}$. At infinity the coefficient is $m-\mathrm{deg}P$, so $\mathrm{deg}P\le m$. If $m=0$, $P$ is constant; if $m=1$, write $P=\alpha (t-c)+\beta$, giving $f=\alpha +\beta \mathrm{/}(t-c)$. Therefore $L([Q])=k\cdot 1\oplus k\cdot \frac{1}{t-c}$ has dimension two.

Arithmetic genus and normalization. By [F1] the conic $C$ is an integral curve in ${\mathbb{P}}_k^2$, so [F9] gives $H^0(C,{\mathcal{O}}_C)=k$ and $p_a(C)=\frac{(2-1)(2-2)}{2}=0$. By [F2] the local rings of the smooth curve $C$ are regular, hence integrally closed, so $C$ is normal; then the identity morphism $C\to C$ is a normal integral scheme, finite and birational over $C$, so by initiality of the normalization [F10] the normalization $\nu :C^{\mathrm{n}\mathrm{u}}\to C$ is an isomorphism, i.e. $C$ agrees with its normalization.

The parametrization and the projection. Keep the normal form of step 1.1 and let $[X:Z]$ be homogeneous coordinates on ${\mathbb{P}}_k^1$. Define $\mathrm{\Phi }:{\mathbb{P}}_k^1⟶{\mathbb{P}}_k^2,[X:Z]⟼[fXZ:-(a{X}^2+eXZ+c{Z}^2):f{Z}^2],$ whose components are homogeneous of degree two; they do not all vanish, because $Z=0$ forces $X\mathrm{\ne }0$ and then the image is $[0:-a{X}^2:0]=[0:1:0]$ since $a\mathrm{\ne }0$, so $\mathrm{\Phi }$ is a $k$-morphism [F3]; and $\mathrm{\Phi }$ lands in $C$: substituting gives $a(fXZ{)}^2+e(fXZ)(f{Z}^2)+f(-(a{X}^2+eXZ+c{Z}^2))(f{Z}^2)+c(f{Z}^2{)}^2=f{Z}^2\cdot 0$. In the other direction the projection $\mathrm{\Pi }:C\setminus \{p\}⟶{\mathbb{P}}_k^1,[X:Y:Z]⟼[X:Z],$ is a $k$-morphism: on $C$ the equations $X=Z=0$ define the single point $p=[0:1:0]$, as $F(0,Y,0)=0$ for all $Y$, so off $p$ at least one of $X,Z$ is nonzero.

The two maps are mutually inverse on dense opens. For $[X:Z]\in {\mathbb{P}}_k^1$ with $Z\mathrm{\ne }0$ one has $\mathrm{\Pi }(\mathrm{\Phi }([X:Z]))=\mathrm{\Pi }([fXZ:-(a{X}^2+eXZ+c{Z}^2):f{Z}^2])=[fXZ:f{Z}^2]=[X:Z]$, so $\mathrm{\Pi }\circ \mathrm{\Phi }$ is the identity on the dense open $\{Z\mathrm{\ne }0\}\subseteq {\mathbb{P}}_k^1$. On the open chart $Z\mathrm{\ne }0$ of $C$, with homogeneous coordinates $[X:Y:Z]$, one has $Z\mathrm{\ne }0$ (if $Z=0$ then $0=F(X,Y,0)=a{X}^2$ forces $X=0$, hence $[X:Y:Z]=p$), and the conic equation $YfZ=-(a{X}^2+eXZ+c{Z}^2)$ gives $\mathrm{\Phi }(\mathrm{\Pi }([X:Y:Z]))=[fXZ:-(a{X}^2+eXZ+c{Z}^2):f{Z}^2]=[X(fZ):Y(fZ):Z(fZ)]=[X:Y:Z]$, since $fZ\mathrm{\ne }0$; so $\mathrm{\Phi }\circ \mathrm{\Pi }$ is the identity on the dense open $C\setminus \{p\}$.

The isomorphism. The morphism $\mathrm{\Pi }$ of step 2.1 represents a rational map $C⇢{\mathbb{P}}_k^1$ [F7]; the curve $C$ is smooth and ${\mathbb{P}}_k^1$ is proper over $k$, so under Choice [F11] the extension lemma [F7] represents this rational map by a morphism $\bar{\mathrm{\Pi }}:C\to {\mathbb{P}}_k^1$ extending $\mathrm{\Pi }$. The morphisms $\mathrm{\Phi }\circ \bar{\mathrm{\Pi }}$ and ${\mathrm{i}\mathrm{d}}_C$ from the reduced scheme $C$ to the separated $k$-scheme $C$ agree on the dense open $C\setminus \{p\}$ by step 3.1, so they are equal by [F8]; similarly $\bar{\mathrm{\Pi }}\circ \mathrm{\Phi }$ and ${\mathrm{i}\mathrm{d}}_{{\mathbb{P}}^1}$ agree on the dense open $\{Z\mathrm{\ne }0\}\subseteq {\mathbb{P}}^1$ by step 3.1, so $\bar{\mathrm{\Pi }}\circ \mathrm{\Phi }={\mathrm{i}\mathrm{d}}_{{\mathbb{P}}^1}$. Hence $\bar{\mathrm{\Pi }}$ is an isomorphism of $k$-schemes with inverse $\mathrm{\Phi }$, which is the first assertion: projection from $p$ exhibits $C\cong {\mathbb{P}}_k^1$.

The Riemann-Roch space of a rational point. Let $D=[P]$ with $P\in C(k)$ and put $Q=\bar{\mathrm{\Pi }}(P)\in {\mathbb{P}}_k^1(k)$. An isomorphism of $k$-schemes induces a $k$-isomorphism of function fields and a bijection of closed points preserving residue fields, hence a degree-preserving bijection of divisor groups intertwining $\mathrm{div}$ and $\mathrm{ord}$ by [F5]; under the isomorphism $\bar{\mathrm{\Pi }}:C\to {\mathbb{P}}_k^1$ the pullback of $[Q]$ is $[P]=D$, and pullback of rational functions $f\mapsto f\circ \bar{\mathrm{\Pi }}$ carries $L([Q])$ onto $L(D)$ [F6]. By step 1.2 the space $L([Q])$ is $2$-dimensional over $k$, with basis $\{1,t\}$ for $Q=\mathrm{\infty }$ and $\{1,\frac{1}{t-c}\}$ for the finite point $Q=[1:c]$ with $t=x_1\mathrm{/}{x}_0$; hence ${\dim }_{k}L(D)=2$.

Conclusion. For a smooth conic $C\subseteq {\mathbb{P}}_k^2$ with a $k$-rational point $p$, step 4.1 exhibits an explicit isomorphism $C\cong {\mathbb{P}}_k^1$ from the projection at $p$ and its residual-intersection parametrization, step 5.1 computes ${\dim }_{k}L([P])=2$ for every $k$-rational point $P$, and step 1.3 gives $p_a(C)=\frac{(2-1)(2-2)}{2}=0$ together with the agreement of $C$ with its normalization. The Axiom of Choice is assumed for the geometric-regularity characterization and normalization in steps 1.1 and 1.3, as well as the rational-map extension and dense-open uniqueness in step 4.1 [F2, F10, F7, F8]. The divisor calculation in step 1.2 uses the verified clauses 1 and 2 of the current draft supplier [F4].