Smooth plane quartic has genus three

Dimension of the hypersurface. The nonzero quartic $F$ does not vanish identically on the irreducible surface ${\mathbb{P}}_k^2$. By [F7], $X=V_{+}(F)$ is nonempty and each of its irreducible components has dimension one. Thus every closed point used below has local dimension one by [F8].

No repeated factor. Factor $F$ into irreducible homogeneous forms in the UFD $k[x_0,x_1,x_2]$ [F5]; homogeneous factors can be taken homogeneous because the lowest and highest graded degrees of a product add. Suppose an irreducible factor $G$ occurs with multiplicity at least two, so $F=G^{2}H$. By [F7], $V_{+}(G)$ is nonempty; choose a closed point $P\in V_{+}(G)$. In a standard affine chart through $P$, the actual equation is $f=g^{2}h$, so its first partial derivatives all vanish at $P$. This remains true in every characteristic because each derivative is divisible by $g$. By [F8] the local dimension is one, so [F4] says the zero Jacobian row makes the local ring nonregular. This contradicts smoothness. Thus $F$ is square-free.

No reducible square-free factorization. If square-free $F$ were reducible, write $F=GH$ with coprime homogeneous forms $G,H$ of positive degree. By [F6], their projective intersection is nonempty; choose a closed point $P$ in it. On a standard chart through $P$, the equation is $f=gh$ with $g(P)=h(P)=0$, so every first partial derivative $h\mathrm{\partial }g+g\mathrm{\partial }h$ vanishes at $P$. Its local ring has dimension one [F8] and is nonregular by [F4], contradicting smoothness again. Therefore $F$ is irreducible.

The curve hypothesis. By [F5], irreducible $F$ is prime, so the homogeneous coordinate ring $k[x_0,x_1,x_2]\mathrm{/}(F)$ is a domain. The nonempty standard projective charts are spectra of domains, so [F12] makes its Proj reduced and irreducible; it is nonempty and one-dimensional by [F7]. Since $k$ is algebraically closed, its algebraic-closure fibre is itself, so $X$ is geometrically integral. As a closed subscheme of projective space, $X$ is separated and finite type. Hence $X$ is a curve over $k$ by [F11].

The arithmetic genus and delta invariants. Smoothness makes every local ring regular [F4], so every delta invariant is zero [F2] and the sum in [F3] is empty. Applying [F1] with $d=4$ gives $H^0(X,{\mathcal{O}}_X)=k$ and $p_a(X)=\frac{(4-1)(4-2)}{2}=3$. The Axiom of Choice [A1] supplies the DC needed in [F3] through [A2].

The geometric genus. Applying [F3] to the integral quartic and using step 5.1 gives $g(X^{\mathrm{n}\mathrm{u}})=3$. By [F9], smoothness makes $X$ normal, so its normalization is isomorphic to $X$; therefore $g(X)=g(X^{\mathrm{n}\mathrm{u}})=3$, agreeing with $p_a(X)$. More generally, the same square-free and irreducibility argument of steps 2.1 and 3.1 applies to any smooth plane curve of degree $d\ge 1$ over this algebraically closed field, so it is integral. Then [F1] gives $p_a=(d-1)(d-2)\mathrm{/}2$, smoothness makes every delta invariant zero by [F2], and [F3] and [F9] give $g=p_a$. Thus the displayed quartic is the $d=4$ case of the triangular-number formula, without using a separate general-genus supplier.