A genus-one curve with a rational point embeds as a plane cubic

(Degree-one divisor.) A rational point $p_0$ gives the divisor $[p_0]$ of degree one. Conversely, let $D$ be any divisor of degree one, not assumed effective. Since $1>2g-2=0$, [F1] gives $h^0(C,{\mathcal{O}}_C(D))=l(D)=1$. A nonzero section has an effective divisor $E$ linearly equivalent to $D$; the degree conventions in [F4] give ${\mathrm{deg}}_k(E)=1$. In ${\mathrm{deg}}_k(E)={\sum }_x{n}_x[\kappa (x):k]$, each $n_x$ is a nonnegative integer and each residue degree is a positive integer, so $E=[p]$ for one $k$-rational point $p$. Thus the parenthetical equivalence in the Statement holds also for signed divisors. The same degree convention gives $\mathrm{deg}({\mathcal{O}}_C(3p_0))=3$.

(Cohomology.) Since $\mathrm{deg}(L)=3>2g-2=0$, [F1] gives $H^1(C,L)=0$ and $h^0(C,L)=3+1-1=3$.

(Very ampleness.) The degree satisfies $\mathrm{deg}(L)=3=2g+1$, so [F2] applies. The complete linear system morphism of [F3] is a closed immersion ${\varphi }_L:C\to {\mathbf{P}}_k^{h^0(C,L)-1}={\mathbf{P}}_k^2$ and ${\varphi }_L^{\ast }\mathcal{O}(1)\cong L$.

(The homogeneous ideal of the image.) Put $Y={\varphi }_L(C)$, $R=k[X_0,X_1,X_2]$, and let $J\subset R$ be the unique saturated homogeneous ideal with $Y=V_{+}(J)$, given by [F5]. The scheme $Y$ is integral because it is isomorphic to $C$. For every degree, restriction to $Y$ embeds $S=R\mathrm{/}J$ into the section ring ${⨁}_{n\ge 0}{H}^0(Y,{\mathcal{O}}_Y(n))$: a homogeneous form maps to its restricted section, and if that section is zero then it vanishes on every standard projective chart, so the saturation criterion in [F5] puts the form in $J$. The section ring is a domain: a nonzero section of an invertible sheaf on an integral scheme remains nonzero at the generic point, since on a trivializing affine open its coefficient belongs to a domain embedded in the function field. At the generic point, the product of two nonzero homogeneous sections is the tensor product of two nonzero vectors over $k(Y)$, hence is nonzero; the grading then shows that products of arbitrary nonzero sums are nonzero by considering their least nonzero degrees. Thus $S$ is a domain and $J$ is prime. Each coordinate section is a nonzero member of the basis defining the complete linear system, so $X_i\mathrm{\notin }J$.

(The homogeneous ideal has height one.) Fix an index $i$ and write $A=(S_{X_i}{)}_0$. This is the coordinate ring of the nonempty affine chart $Y\cap D_{+}(X_i)$; it is a finite-type domain of dimension one because $Y$ is an integral curve. Since $S$ is standard graded and $X_i$ has degree one, $S_{X_i}\cong A[X_i,X_i^{-1}]$: a homogeneous fraction is its degree-zero part times the corresponding power of $X_i$, and the grading makes this decomposition direct. Therefore ${\mathrm{trdeg}}_k\mathrm{Frac}(S)=2$ by [F6]. Apply the dimension formula of [F6] to $J\subset R$; it gives $\mathrm{ht}(J)+2=3$, so $J$ has height one. By [F5], $J=(F)$ for an irreducible element $F$. Because $J$ is homogeneous, each homogeneous component of $F$ belongs to $(F)$. If $F$ had more than one degree, a nonzero component of degree smaller than $\mathrm{deg}F$ would equal $FG$ for some polynomial $G$, which is impossible by additivity of total degree in the polynomial domain. Thus $F$ is homogeneous. Therefore $Y=V_{+}(F)$ scheme-theoretically, and $F$ is square-free.

(The degree.) The irreducible homogeneous form $F$ has positive degree, and [F5] identifies the degree of its reduced hypersurface with $\mathrm{deg}F$. Choose a linear form $\mathrm{\ell }$ whose restriction to $Y$ is nonzero; when $\mathrm{deg}F=1$, choose one not proportional to $F$, and when $\mathrm{deg}F>1$ any nonzero linear form is coprime to $F$. The zero scheme of $s=\mathrm{\ell }{\mathrm{\mid }}_C\in H^0(C,L)$ is scheme-theoretically $Y\cap V_{+}(\mathrm{\ell })$ under ${\varphi }_L$. For each closed point $x$ in that intersection, the local ring of $C$ is a DVR by [F4]. If the local equation of $s$ is $u{t}^n$ with $u$ a unit and $t$ a uniformizer, then ${\mathrm{length}}_{{\mathcal{O}}_{C,x}}{\mathcal{O}}_{C,x}\mathrm{/}(s)=n$, exactly the coefficient of the zero divisor of $s$. Consequently the weighted sum of the intersection lengths in [F7] is $\mathrm{deg}(L)=3$ by [F4]. The forms $F$ and $\mathrm{\ell }$ are coprime, so the same sum is $\mathrm{deg}(F)$ by [F7]. Hence $\mathrm{deg}(F)=3$, and $Y$ is a plane cubic as claimed.

The very ample sheaf $L$ embeds every given genus-one curve into ${\mathbf{P}}_k^2$ as a scheme-theoretic cubic by steps 1.2, 2.1, and 3.1--5.1. Taking $L={\mathcal{O}}_C(3p_0)$ when a rational point is given proves that every such curve is isomorphic to a plane cubic. The Axiom of Choice is inherited through the cohomology, projective-space, affine-dimension, and Bezout suppliers cited above; no unmentioned rational intersection points are chosen.