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    Weil and Cartier

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    Weil and Cartier divisors, and when they agree

    Let \(X\) be a noetherian integral separated scheme regular in codimension one, and let \(K(X)\) be its function field. A prime divisor is a codimension-one integral closed subscheme of \(X\). A Weil divisor is a finite formal \(\ZZ\)-linear combination of prime divisors; these form the group \(\operatorname{Div}(X)\). For a prime divisor \(Y\) with generic point \(\eta_Y\), let \(v_Y\) be the normalized valuation of the DVR \(\OO_{X,\eta_Y}\). For \(f\in K(X)^\times\), the principal Weil divisor is \[ \operatorname{div}_X(f)\coloneqq\sum_{\operatorname{codim}(Y,X)=1}v_Y(f)Y. \] Only finitely many coefficients are nonzero [@Har10a, Lemma II.6.1]. The divisor class group \(\Cl(X)\) is \(\operatorname{Div}(X)\) modulo principal Weil divisors.

    (Cartier)

    Let \(X\) be a scheme, and let \(\mck\) be its sheaf of total quotient rings. Explicitly, for an open set \(U\), let \(S(U)\subseteq\Gamma(U,\OO_X)\) consist of sections whose germ at every point of \(U\) is a non-zero-divisor; \(\mck\) is the sheafification of \(U\mapsto S(U)^{-1}\Gamma(U,\OO_X)\). A Cartier divisor is a global section of \(\mck^\times/\OO_X^\times\). It is represented by an open cover \((U_i)\) and elements \(f_i\in\mck^\times(U_i)\) such that \(f_i/f_j\) is a unit on each overlap. For integral \(X\), the sheaf \(\mck\) is the constant sheaf of its function field and the \(f_i\) are nonzero rational functions [@Har10a, Chapter II, §6].

    (Principal, effective, support)

    A Cartier divisor \(D\) is principal if it is represented by a single element of \(\Gamma(X,\mck^\times)\). The Cartier divisor class group \(\CaCl(X)\) is the group of Cartier divisors modulo principal Cartier divisors. The divisor \(D\) is effective if it has local equations \(f_i\in\OO_X(U_i)\), necessarily non-zero-divisors in their stalks. Its support is the complement of the largest open subset on which all its local equations are units [@Har10a, Chapter II, §6].

    For a Cartier divisor \(D\) on a scheme \(X\), the sheaf \(\OO_X(D)\subseteq\mck\) is locally generated by \(f_i^{-1}\) when \(f_i\) is a local equation of \(D\). The rule \([D]\mapsto[\OO_X(D)]\) gives an injective homomorphism \(\CaCl(X)\to\Pic(X)\). It is an isomorphism when \(X\) is integral [@Har10a, Corollary II.6.14 and Proposition II.6.15]. If \(D\) is effective, \(\OO_X(-D)\) is the ideal sheaf locally generated by the \(f_i\) [@Har10a, Proposition II.6.18].

    (Cartier and Weil comparison)

    Let \(X\) be a noetherian integral separated normal scheme. Sending a Cartier divisor with local equations \(f_i\) to the Weil divisor with coefficient \(v_Y(f_i)\) at a prime divisor \(Y\) meeting \(U_i\) identifies Cartier divisors with locally principal Weil divisors. It induces an injection \(\CaCl(X)\to\Cl(X)\). For such \(X\), the support of a Cartier divisor is the union of the prime divisors with nonzero coefficient in its associated Weil divisor [@Har10a, Remark II.6.11.2 and the proof of Proposition II.6.11]. If \(X\) is locally factorial, the comparison is an isomorphism, and \(\Cl(X)\cong\Pic(X)\) [@Har10a, Proposition II.6.11 and Corollary II.6.16]. In particular, \(\CaCl(X)\cong\Cl(X)\cong\Pic(X)\) for every regular integral separated noetherian scheme \(X\), since regular local rings are factorial [@Har10a, Remark II.6.11.1A].

    (A Weil divisor that is not Cartier)

    Let \(k\) be a field, put \(A=k[x,y,z]/(xy-z^2)\), and let \(X=\Spec A\). This is a normal domain: the map \(x\mapsto s^2\), \(y\mapsto t^2\), \(z\mapsto st\) identifies \(A\) with the second Veronese subring of \(k[s,t]\), which is normal by the graded-domain argument in P-AGH2514PROJNORM. The identification is injective because \(A=k[x,y]\oplus z k[x,y]\), whose two summands map to distinct even-even and odd-odd monomials in \(s,t\). The ruling \(D=V(x,z)\) is a prime divisor, and its class generates \(\Cl(X)\cong\ZZ/2\ZZ\) [@Har10a, Example II.6.5.2].

    At the vertex with maximal ideal \(\mathfrak m=(x,y,z)\), the ideal of \(D\) is \((x,z)A_{\mathfrak m}\). Its image in \(\mathfrak m A_{\mathfrak m}/\mathfrak m^2A_{\mathfrak m}\) has dimension two: the defining relation is quadratic, so the classes of \(x,y,z\) are linearly independent. A principal ideal contained in \(\mathfrak m A_{\mathfrak m}\) has image of dimension at most one there. Thus the ideal of \(D\) is not principal at the vertex, and \(D\) is not Cartier. Since every nonzero divisor class is represented by \(D\), the Cartier-to-Weil injection has zero image, so \(\Pic(X)=0\).

    On a smooth integral curve over a field, the local ring at every closed point is a DVR and the generic local ring is its function field. The curve is regular and locally factorial, so its Weil and Cartier divisor classes both identify with its Picard group.

    Describe Weil divisors and Cartier divisors on a curve.
    How do you get a Weil divisor from $f \in K^*$?
    When do the two notions agree?
    What are principal and effective Cartier divisors, and what is the support of a Cartier divisor?

    Zeros and poles of a rational function

    Codimension-one valuations detect only finitely many zeros and poles; on a normal scheme, no poles means regular.Suggest a slogan

    Let \(X\) be an integral Noetherian scheme with function field \(K\). For a prime divisor \(Y \subseteq X\) with generic point \(\eta_Y\) such that \(\OO_{X, \eta_Y}\) is a discrete valuation ring with valuation \(v_Y\), a function \(t \in K^\times\) has a zero along \(Y\) if \(v_Y(t) > 0\) and a pole along \(Y\) if \(v_Y(t) < 0\).

    1. A Noetherian topological space has finitely many irreducible components [@Har10a, Proposition I.1.5].

    2. For \(t \in K^\times\), there are only finitely many codimension-one points \(y \in X\) with \(t \notin \OO_{X,y}^\times\).

    3. If \(X\) is regular in codimension one, then \(\div t = \sum_Y v_Y(t)\, Y\) is a finite sum, hence a Weil divisor [@Har10a, Lemma II.6.1].

    4. If \(X\) is normal, then \(t \in K\) has no poles if and only if \(t \in \OO_X(X)\); on an affine open \(\Spec A\) this is \(A = \bigcap_{\operatorname{ht} \mathfrak{p} = 1} A_{\mathfrak{p}}\) for a Noetherian normal domain \(A\) [@Har10a, Proposition II.6.3A].

    Why does a rational function on an integral Noetherian scheme have only finitely many zeros and poles?
    Why is a rational function with no poles on a normal Noetherian scheme regular?

    A Weil divisor is a formal integer combination of prime divisors; a Cartier divisor is given by local equations \(f_i\in K^\times\) on an open cover \(\{U_i\}\) with \(f_i/f_j\in\OO^\times(U_i\cap U_j)\). If every local ring is a unique factorization domain, every prime divisor has a local equation, and the two groups agree. On the affine quadric cone \(xy=z^2\), the line \(x=z=0\) is a prime divisor with no local equation at the vertex; there \(\Cl\cong\ZZ/2\), generated by that line, while every Cartier divisor is principal. On a smooth curve every local ring is a discrete valuation ring, so codimension-one subvarieties and local equations are related by the valuation at each point.

    The degree of a divisor on a projective curve

    On a smooth projective curve, zeros and poles balance: every principal divisor has degree zero.Suggest a slogan

    For \(D = \sum n_i P_i\) on a curve, \(\deg D = \sum n_i [k(P_i) : k]\), which is \(\sum n_i\) when \(k = \bar{k}\).

    On a smooth projective curve, \(\deg \div(f) = 0\) for every \(f \in K(X)^*\). Degree therefore descends to a homomorphism \(\Pic(X) \to \ZZ\), and \(\Pic^0(X)\) is its kernel.

    Projectivity enters because a nonconstant \(f\) defines a finite morphism \(X\to\PP^1\), and \(\div(f)=f^*(0)-f^*(\infty)\), where both fibres have degree \(\deg f\) counted with multiplicity. Zeros and poles balance because the map has a well-defined degree, and the map has one because \(X\) is projective.

    On \(\AA^1\) the statement is false — \(\div(x) = [0]\) has degree one — and this is the cleanest illustration that divisor theory is a theory about complete curves.

    What is the degree of a divisor?
    Why does a principal divisor have degree zero?

    Degrees of points, finite morphisms, line bundles and coherent sheaves on curves

    Let \(k\) be a field.

    (Degree of a closed point)

    Let \(X\) be a scheme locally of finite type over \(k\) and \(p \in X\) a closed point. The degree of \(p\) is \(\deg p = [\kappa(p) : k]\), which is finite by the Nullstellensatz.

    (Degree of a finite morphism)

    Let \(\pi \colon X \to Y\) be a finite morphism and \(y \in Y\). The degree of \(\pi\) at \(y\) is \(\dim_{\kappa(y)} (\pi_* \OO_X \otimes_{\OO_Y} \kappa(y))\). Since \(\pi\) is affine, \(\pi^{-1}(y) = \Spec (\pi_* \OO_X \otimes \kappa(y))\), so this is the dimension of the ring of functions on the fibre. If \(X\) is a curve without embedded points, \(Y\) is a regular curve and \(\pi\) is finite and surjective, then \(\pi_* \OO_X\) is locally free of finite rank, and its rank, the degree at every point, is the degree \(\deg \pi\).

    (Degree of a line bundle and of a coherent sheaf)

    Let \(C\) be a projective curve over \(k\). The degree of an invertible sheaf \(\mathcal{L}\) on \(C\) is \(\deg_C \mathcal{L} = \chi(C, \mathcal{L}) - \chi(C, \OO_C)\). If \(C\) is integral, the degree of a coherent sheaf \(\mathcal{F}\) on \(C\) is \(\deg \mathcal{F} = \chi(C, \mathcal{F}) - (\operatorname{rank} \mathcal{F}) \chi(C, \OO_C)\), where \(\operatorname{rank} \mathcal{F}\) is the dimension of the stalk of \(\mathcal{F}\) at the generic point over the function field. For rank \(1\) locally free \(\mathcal{F}\) the two definitions agree.

    Let \(C\) be a regular projective integral curve over \(k\).

    1. For a divisor \(D = \sum_i n_i p_i\), \(\deg \OO_C(D) = \deg D = \sum_i n_i \deg p_i\) (PR-Y5S7V); so if \(s\) is a nonzero rational section of \(\mathcal{L}\), then \(\deg \mathcal{L} = \deg \div s\).

    2. For a locally free sheaf \(\mathcal{E}\) of rank \(r\) and an invertible sheaf \(\mathcal{L}\), \(\chi(\mathcal{E} \otimes \mathcal{L}) = \chi(\mathcal{E}) + r \deg \mathcal{L}\).

    3. If \(\pi \colon C' \to C\) is a finite surjective morphism from a projective integral curve, then \(\deg_{C'} \pi^* \mathcal{L} = (\deg \pi)(\deg_C \mathcal{L})\) [@Har10a, Proposition II.6.9].

    What is the degree of a closed point?
    What is the degree of a morphism at a point?
    How is the degree of a line bundle or a coherent sheaf on a projective curve defined, and how does it behave under pullback?

    The moving lemma

    Move within rational equivalence until intersections are proper; that is what makes Chow products well defined.Suggest a slogan

    Let \(X\) be a quasiprojective variety over an algebraically closed field, \(D\) a Cartier divisor on \(X\), and \(S \subseteq X\) a finite set of points. Then \(D\) is linearly equivalent to a Cartier divisor \(D'\) whose support contains no point of \(S\).

    (Chow's moving lemma)

    Let \(X\) be a smooth quasiprojective variety, and let \(\alpha\) and \(\beta\) be algebraic cycles on \(X\). Then \(\alpha\) is rationally equivalent to a cycle \(\alpha'\) that meets \(\beta\) properly: every irreducible component of \(\supp \alpha' \cap \supp \beta\) has codimension \(\codim \alpha + \codim \beta\).

    The moving lemma is what makes the intersection product on the Chow ring \(A^*(X)\) of a smooth quasiprojective variety well defined: move one cycle to meet the other properly, intersect with multiplicities, and the rational equivalence class of the result does not depend on the choice of \(\alpha'\). For divisors on a smooth projective surface, it lets the intersection number \(C \cdot D\) be computed as a sum of local intersection multiplicities after replacing \(D\) by a linearly equivalent divisor with no common component with \(C\).

    What is the moving lemma?

    Computing a Picard group

    The Picard group of a nodal and of a cuspidal curve

    Let \(C\) be a singular curve with normalization \(\nu : \tilde{C} \to C\). Comparing units gives the exact sequence \[ 0 \to \OO_C^* \to \nu_* \OO_{\tilde{C}}^* \to \mcs \to 0 \] with \(\mcs\) a skyscraper at the singular points, and its cohomology sequence computes \(\Pic(C)\) from \(\Pic(\tilde{C})\) and the local unit groups.

    For the cusp \(A = k[t^2,t^3] \subseteq k[t]\), the conductor square gives \(\Pic(\Spec A) \cong k^+ = \GG_a\). For the node \(k[x,y]/(y^2 - x^2(x+1))\), the same computation gives \(\GG_m\).

    At a node \(p\) with \(\nu^{-1}(p)=\ts{p_1,p_2}\), the stalk \(\mcs_p\) is \((k^\times\times k^\times)/k^\times\cong k^\times\), the ratio of the values of a unit of \(\OO_{\tilde C}\) at \(p_1\) and \(p_2\); so the node contributes \(\GG_m\). At the cusp of \(\Spec k[t^2,t^3]\), with \(\nu^{-1}(p)=\ts{\tilde p}\) and \(\OO_{C,p}\) the functions whose derivative in \(t\) vanishes at \(\tilde p\), the stalk \(\mcs_p\) is \((k[t]/(t^2))^\times/k^\times\cong k^+\), via \(a+bt\mapsto b/a\); so the cusp contributes \(\GG_a\).

    For the projective cubics, \(\Pic^0\) of the nodal cubic is \(\GG_m\) and \(\Pic^0\) of the cuspidal cubic is \(\GG_a\); in each case the smooth locus, with the chord-tangent group law, is isomorphic to \(\Pic^0\).

    Compute $\Pic(k[t^2,t^3])$.
    What is the Picard group of a nodal cubic?

    For an integral projective curve \(X\) over an algebraically closed field with normalization \(\tilde X\), the comparison with \(\tilde X\) is the exact sequence \[ 0\to\bigoplus_{p\in X}\tilde\OO_p^\times/\OO_{X,p}^\times\to\Pic(X)\to\Pic(\tilde X)\to0 , \] where \(\tilde\OO_p\) is the integral closure of \(\OO_{X,p}\). For an open subset \(U = X \sm Z\) of a smooth variety, the comparison is the excision sequence \[ \ZZ^{\ts{\text{components of } Z \text{ of codimension } 1}} \to \Cl(X) \to \Cl(U) \to 0 . \] It gives \(\Cl(\AA^n)=0\) from \(\Cl(\PP^n)=\ZZ\), and \(\Cl(\PP^n\setminus V(f))\cong\ZZ/d\) for an irreducible form \(f\) of degree \(d\).

    The class group of projective space and of a hypersurface complement

    Let \(k\) be a field and let \(n\ge1\). Then \(\Cl(\PP_k^n)\cong\ZZ\), generated by the hyperplane \(H=V(x_0)\), with the isomorphism given by degree. For a prime divisor \(D=V(f)\) defined by an irreducible homogeneous polynomial of degree \(d\), the rational function \(g=f/x_0^d\) satisfies \(D-dH=\operatorname{div}(g)\). Every divisor is therefore equivalent to a multiple of \(H\), and the degree of a principal divisor is zero, so that multiple is unique [@Har10a, Proposition II.6.4]. For \(n=0\), projective space is \(\Spec k\) and its class group is zero. For every \(n\ge0\), \(\Cl(\AA_k^n)=0\), because \(k[x_1,\ldots,x_n]\) is a unique factorization domain [@Har10a, Proposition II.6.2].

    (A hypersurface complement)

    Let \(k\) be a field, let \(Y\subseteq\PP_k^2\) be an integral curve of degree \(d\ge1\), and put \(X=\PP_k^2\setminus Y\). The divisor restriction sequence is \[ \ZZ\xrightarrow{1\mapsto[Y]}\Cl(\PP_k^2)\longrightarrow\Cl(X)\longrightarrow0. \] Under the degree isomorphism \(\Cl(\PP_k^2)\cong\ZZ\), its first map is multiplication by \(d\). Hence \(\Cl(X)\cong\ZZ/d\ZZ\) [@Har10a, Example II.6.5.1]. In particular, the complement of an integral conic has class group \(\ZZ/2\ZZ\).

    Let \(Z\) be a noetherian integral separated scheme regular in codimension one, and let \(W\subsetneq Z\) be closed. Restriction \(\Cl(Z)\to\Cl(Z\setminus W)\) is surjective, with kernel generated by the prime divisors contained in \(W\). It is an isomorphism if \(W\) has codimension at least two [@Har10a, Proposition II.6.5]. For a noetherian normal domain \(A\), the criterion \(\Cl(\Spec A)=0\) is equivalent to \(A\) being a unique factorization domain [@Har10a, Proposition II.6.2].

    Compute $\Cl(\PP^n)$.
    Compute the class group of the complement of a plane curve of degree $d$.
    Can a class group have torsion?

    Degree-zero divisors on the line and on an elliptic curve

    On \(\PP^1\), any two points satisfy \([a] \sim [b]\) by \(f(z) = (z-a)/(z-b)\). Degree-zero divisors are sums of such differences, so \(\Pic^0(\PP^1) = 0\) and \(\Pic(\PP^1) \cong \ZZ\) by degree.

    On an elliptic curve \(E\), distinct points \(p \neq q\) are never linearly equivalent: \(p - q = \div(f)\) would make \(f : E \to \PP^1\) a degree-one map, hence an isomorphism, contradicting \(g(E) = 1 \neq 0 = g(\PP^1)\). Instead \(\Pic^0(E) \cong E\) by \(p \mapsto [p - O]\), so \(\Pic(E) \cong E \oplus \ZZ\).

    The contrast is the whole content: \(\Pic^0\) measures how far the curve is from being rational, and it vanishes exactly for \(\PP^1\). A curve is rational precisely when it admits two distinct linearly equivalent points, which is the same as admitting a degree-one map to \(\PP^1\).

    The degree argument is the one to reproduce: a principal divisor \(p - q\) gives a rational function with one simple zero and one simple pole, so the induced map to \(\PP^1\) has degree one. For a non-rational example, \(y^2 = x^3 - x\) homogenised to \(y^2 z = x^3 - xz^2\) is the standard witness, and the group law on it is exactly the isomorphism \(E \cong \Pic^0(E)\).

    Show every degree-zero divisor on $\PP^1$ is principal.
    Is every degree-zero divisor on an elliptic curve principal?
    What is $\Pic(E)$ for $E$ elliptic?

    Reduced divisors, multiplicities and pullback

    Reduced divisors, multiplicity at a point, and pullback of divisors

    On a Noetherian integral separated scheme \(X\) regular in codimension one, a prime divisor is a closed integral subscheme of codimension one. A Weil divisor \(D = \sum_i n_i D_i\) with the \(D_i\) distinct prime divisors is effective if all \(n_i \geq 0\), and reduced if every \(n_i \in \{0, 1\}\).

    Integrality of \(X\) gives the function field \(K(X)\). Regularity in codimension one makes the local ring \(\OO_{X,\eta_Y}\) at the generic point of each prime divisor \(Y\) a discrete valuation ring, so every \(f\in K(X)^\times\) has an order \(v_Y(f)\) along \(Y\), which defines \(\operatorname{div}(f)\) (D-5PQ5W).

    (Multiplicity at a point)

    Let \(X\) be a smooth variety and \(D\) an effective divisor with local equation \(f \in \OO_{X,p}\) at a point \(p\). The multiplicity of \(D\) at \(p\) is the largest \(m\) with \(f \in \mfm_p^m\). It is \(0\) when \(p \notin \supp D\), and \(1\) exactly when \(D\) is smooth at \(p\).

    (Pullback)

    Let \(\phi \colon X \to Y\) be a morphism of integral schemes and \(D\) a Cartier divisor on \(Y\) with local equations \(\{(U_i, f_i)\}\), such that \(\phi(X) \not\subseteq \supp D\). The pullback \(\phi^* D\) is the Cartier divisor on \(X\) with local equations \(\{(\phi^{-1} U_i, f_i \circ \phi)\}\). It satisfies \(\OO_X(\phi^* D) \cong \phi^* \OO_Y(D)\), and it respects linear equivalence, so it induces \(\phi^* \colon \Pic Y \to \Pic X\).

    For the plane curve \(D = V(y^2 - x^3)\) in \(\AA^2\), the multiplicity at the origin is \(2\), since \(y^2 - x^3 \in \mfm^2 \setminus \mfm^3\), and the multiplicity is \(1\) at every other point of \(D\). For the double cover \(\phi \colon \AA^1 \to \AA^1\), \(t \mapsto t^2\), the pullback of the divisor \([0]\) is \(2[0]\).

    What is a prime divisor? A reduced divisor?
    For a prime divisor $D$ and $p \in \supp D$, what is the multiplicity of $D$ at $p$?
    What is the pullback of a divisor?
    Which hypotheses on a scheme define Weil divisors and the principal divisor of a rational function?

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