Geometric genus of a singular curve

Assume the Axiom of Choice (The Axiom of Choice) and let $k$ be a perfect field (Perfect fields: every irreducible polynomial is separable). Let $X$ be a curve over $k$ (Curves over a field) that is proper over $k$. Write $\nu :X^{\mathrm{n}\mathrm{u}}\to X$ for its normalization from Normalization of an integral finite-type curve by gluing affine integral closures. We first verify that this normalization is itself a proper geometrically integral curve of chain dimension one, and then verify its smoothness before using the curve-genus definition.

First, $X$ is integral, separated, finite type, and has chain dimension one because it is a curve. Choose a finite affine cover $U_i=\mathrm{Spec}{A}_i$ of $X$. Since $\nu$ is finite, ${\nu }^{-1}(U_i)=\mathrm{Spec}{B}_i$ and $B_i$ is module-finite over $A_i$ (Finite morphisms of schemes, Finite is affine and local on its target). Each $A_i$ is a finite-type $k$-domain, so a finite set of algebra generators for $A_i$ together with a finite $A_i$-module generating set for $B_i$ generates $B_i$ as a $k$-algebra (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras). Thus $X^{\mathrm{n}\mathrm{u}}$ is finite type over $k$.

Its chain dimension is one as well. The chain-dimension hypothesis on $X$ gives a strict chain $Z_0⊊Z_1$ of nonempty irreducible closed subsets. Since $X$ is irreducible, $Z_1=X$: otherwise adjoining $X$ would give a chain of length two. Choose a point of $Z_0$ and an affine neighborhood $U=\mathrm{Spec}A$ of it. The nonempty open $U$ contains the generic point of $X$, so $Z_0\cap U$ is a nonempty proper irreducible closed subset of $U$. No chain in $U$ can have length two: if two closed subsets of $U$ had equal closures in $X$, intersecting the common closure with $U$ would make the original subsets equal (For $A\subseteq S\subseteq X$ the closure of $A$ in $S$ is ${\bar{A}}^X\cap S$, while the interior only contains ${\mathrm{int}}^X(A)\cap S$, with equality when $S$ is open; and a dense subset of $X$ traces to a dense subset of every open $S$). The chain $Z_0\cap U⊊U$ therefore shows that $U$ has dimension one. Closed irreducible subsets $V(\mathfrak{p})$ of $\mathrm{Spec}A$ correspond in reverse order to prime ideals (The prime spectrum and vanishing sets), so this is the ring dimension used in Affine-domain dimension equals transcendence degree. Since $\mathrm{Frac}(A)=k(X)$, that theorem gives ${\mathrm{trdeg}}_{k}k(X)=1$. Every nonempty affine chart $V=\mathrm{Spec}B$ of $X^{\mathrm{n}\mathrm{u}}$ is a finite-type domain with $\mathrm{Frac}(B)=k(X)$ by the normalization theorem and the function field lemma Function field of an integral finite-type scheme. Hence $\dim B=1$. To compare with chain dimension, any chain of irreducible closed subsets of $X^{\mathrm{n}\mathrm{u}}$ can be restricted to an affine neighborhood meeting its smallest member. Each trace is a nonempty irreducible closed subset of that affine open. Strictness is preserved: for nested irreducible closed subsets $Z⊊Z^{\mathrm{\prime }}$, both the chosen affine neighborhood's intersection with $Z^{\mathrm{\prime }}$ and $Z^{\mathrm{\prime }}\setminus Z$ are nonempty open subsets of the irreducible space $Z^{\mathrm{\prime }}$, so they intersect (Irreducibility via nonempty open subsets, connectedness and open subspaces). Conversely, a strict chain of closed subsets in an affine open remains strict after taking closures in the whole scheme, by For $A\subseteq S\subseteq X$ the closure of $A$ in $S$ is ${\bar{A}}^X\cap S$, while the interior only contains ${\mathrm{int}}^X(A)\cap S$, with equality when $S$ is open; and a dense subset of $X$ traces to a dense subset of every open $S$. Thus the affine-chart dimensions give chain dimension one for $X^{\mathrm{n}\mathrm{u}}$.

The map $\nu$ is finite, hence proper (Finite morphisms are proper); composing it with the proper structure map $X\to \mathrm{Spec}k$ shows that $X^{\mathrm{n}\mathrm{u}}$ is proper (Composite of a finite morphism and a proper morphism is proper). In particular its structure map is separated, since proper means separated, finite type and universally closed (Proper morphisms).

It remains to check geometric integrality. Fix an algebraic closure $\bar{k}$ of $k$, and put $K=k(X)=k(X^{\mathrm{n}\mathrm{u}})$. Since $X$ is geometrically integral, Function field of an integral finite-type scheme gives that $K{\otimes }_k\bar{k}$ is a domain (under the stated Choice premise). For each nonempty affine chart $V_i=\mathrm{Spec}{B}_i$ of $X^{\mathrm{n}\mathrm{u}}$, the function-field identification embeds $B_i$ into $K$. The $k$-module $\bar{k}$ is flat by Modules over a field are projective, flat, and injective, so tensoring this injection gives $B_i{\otimes }_k\bar{k}\hookrightarrow K{\otimes }_k\bar{k}$. Thus each chart ring $B_i{\otimes }_k\bar{k}$ is a nonzero domain. For any pair of these charts their intersection is nonempty because it contains the generic point; it is affine because $X^{\mathrm{n}\mathrm{u}}$ is separated, by Affine-overlap criterion for separatedness. Write it as $W_{ij}=\mathrm{Spec}{D}_{ij}$. The same function-field identification embeds $D_{ij}$ into $K$, so flatness gives $D_{ij}{\otimes }_k\bar{k}\hookrightarrow K{\otimes }_k\bar{k}$. This is a nonzero domain. By Affine charts after extension of the ground field, these tensor-product charts and overlaps are exactly the charts and intersections after extension to $\bar{k}$. Nonzero affine rings have points under Choice, by In a nonzero commutative ring, every proper ideal is contained in a maximal ideal, so the base-changed charts cover a nonempty scheme and remain pairwise intersecting. A cover by integral affine opens with pairwise nonempty intersections is reduced and irreducible; therefore $X_{\bar{k}}^{\mathrm{n}\mathrm{u}}$ is integral. This proves geometric integrality by Geometric fibres and geometric points and Geometric properties of fibres.

We have now established that $X^{\mathrm{n}\mathrm{u}}$ is a proper geometrically integral separated finite-type curve of chain dimension one. Its finite affine cover above has Noetherian coordinate rings by the finite-type-over-a-field case of Every algebra of finite type over a principal ideal domain is a Noetherian ring, so $X^{\mathrm{n}\mathrm{u}}$ is Noetherian by Locally Noetherian and Noetherian schemes. Its normality means that its local rings are integrally closed domains (Weil divisor normal noetherian scheme). The localizations of the chart rings are Noetherian by Every quotient and every localisation of a Noetherian ring is Noetherian. At a non-generic point $x$, an affine chart $\mathrm{Spec}B$ identifies $x$ with a nonzero prime $\mathfrak{p}$. The chain $0⊊\mathfrak{p}$ and the dimension-one bound give $\dim B_{\mathfrak{p}}=1$. Thus ${\mathcal{O}}_{X^{\mathrm{n}\mathrm{u}},x}$ is a one-dimensional Noetherian local integrally closed domain, hence a discrete valuation ring by Equivalent characterizations of a DVR and therefore regular. The generic local ring is the field $K$, also regular. So $X^{\mathrm{n}\mathrm{u}}$ is regular; since $k$ is perfect, it is smooth by Regular equals smooth over a perfect field.

The geometric genus of $X$ is $g(X):=g(X^{\mathrm{n}\mathrm{u}})=h^1(X^{\mathrm{n}\mathrm{u}},{\mathcal{O}}_{X^{\mathrm{n}\mathrm{u}}}),$ the genus Genus and arithmetic genus of a curve of the smooth proper curve $X^{\mathrm{n}\mathrm{u}}$. It is a nonnegative integer, finite-dimensionality of the cohomology being part of that definition. If $X$ is itself smooth, then $X$ is regular by Regular equals smooth over a perfect field and therefore normal (regular local rings are normal). The normalization's initiality then identifies $\nu$ with an isomorphism, so $g(X)$ is the genus of $X$.

The definition is independent of all choices: the normalization is unique up to unique isomorphism over $X$ (Normalization of an integral finite-type curve by gluing affine integral closures), and the genus of a smooth proper curve is an isomorphism invariant of the curve together with its structure morphism to $k$. We do not call $g(X)$ the genus of $X$ without qualification, reserving the unqualified word for the smooth case; the geometric genus is an invariant of the singular curve $X$ and is insensitive to the singularities, in contrast with the arithmetic genus $p_a(X)=1-\chi ({\mathcal{O}}_X)$.