Here we give another method of calculating the Picard group of a singular curve. Let \(k\) be algebraically closed and let \(X\) be a projective integral curve over \(k\). Let \(\tilde X\) be its normalization, and let \(\pi: \tilde X \to X\) be the projection map (Ex. 3.8). For each point \(P \in X\), let \(\OO_P\) be its local ring, and let \(\tilde\OO_P\) be the integral closure of \(\OO_P\). We use a \(*\) to denote the group of units in a ring.
Show there is an exact sequence \[ 0 \to \bigoplus_{P \in X} \tilde\OO_P^* / \OO_P^* \to \Pic X \mapsvia{\pi^*} \Pic \tilde X \to 0 . \]
Use (a) to give another proof that if \(X\) is a plane cuspidal cubic curve then there is an exact sequence \[ 0 \to \GG_a \to \Pic X \to \ZZ \to 0, \] and if \(X\) is a plane nodal cubic curve there is an exact sequence \[ 0 \to \GG_m \to \Pic X \to \ZZ \to 0 . \]
Hint
For part (a), represent \(\Pic X\) and \(\Pic\tilde X\) as the groups of Cartier divisors modulo principal divisors, and use the exact sequence of sheaves on \(X\) \[ 0\to\pi_*\OO_{\tilde X}^*/\OO_X^* \to\mck^*/\OO_X^* \to\mck^*/\pi_*\OO_{\tilde X}^* \to0. \]
Solution
Put \(K=K(X)=K(\tilde X)\), and write \(\operatorname{CaDiv}(X)\) for the group of Cartier divisors. The Cartier class group is \(\Pic(X)\) because \(X\) is integral [@Har10a, Proposition II.6.15]. In the sequences in part (b), \(\GG_a\) and \(\GG_m\) denote their groups of \(k\)-points, \((k,+)\) and \(k^\times\).
1. The normalization is a nonsingular projective integral curve, and \(\pi\) is an isomorphism outside a finite set \(S\) of closed points.For a closed point \(P\), the ring \(B_P\coloneqq\tilde\OO_P\) is a finite semilocal normal domain over \(A_P\coloneqq\OO_P\), with maximal ideals indexed by the points over \(P\).
The normalization is finite and birational for varieties and preserves projectivity [@Har10a, Exercise II.3.8]. A one-dimensional noetherian normal local domain is a DVR, so \(\tilde X\) is nonsingular [@Har10a, Theorem I.6.2A]. On an affine open \(\Spec A\subseteq X\), the finite module \(\tilde A/A\) is torsion, since both rings have fraction field \(K\). A nonzero element annihilates this module after clearing the denominators of finitely many generators. Thus its support is a proper closed subset of the curve and is finite. Taking a finite affine cover gives \(S\). At a normal point the integral closure changes nothing, so \(\pi\) is an isomorphism on \(X\setminus S\). We may take \(S\) to be the singular locus, since a normal local ring of this curve is regular.
Integral closure commutes with localization, so \((\pi_*\OO_{\tilde X})_P=B_P\). Finiteness over the local ring \(A_P\) makes \(B_P\) semilocal, with precisely the points of \(\pi^{-1}(P)\) as its maximal ideals. Its localizations there are the DVRs \(\OO_{\tilde X,Q}\).
2. For any closed point \(P\) and any integers \(n_Q\), indexed by \(Q\in\pi^{-1}(P)\), there is \(g\in K^*\) with \(v_Q(g)=n_Q\) for every such \(Q\).
Write the maximal ideals of \(B_P\) as \(\mathfrak n_1,\ldots,\mathfrak n_r\), corresponding to \(Q_1,\ldots,Q_r\) over \(P\). For each \(i\), choose a nonzero class in \(\mathfrak n_i/\mathfrak n_i^2\). This vector space is one-dimensional, since localization identifies it with the corresponding quotient in the DVR at \(Q_i\). The Chinese remainder theorem gives \(b_i\in B_P\) with this prescribed class modulo \(\mathfrak n_i^2\) and with \(b_i\equiv1\pmod{\mathfrak n_j}\) for \(j\ne i\). Therefore \(v_{Q_i}(b_i)=1\) and \(v_{Q_j}(b_i)=0\) for \(j\ne i\). The rational function \(g=\prod_i b_i^{n_{Q_i}}\) has the required valuations, including when some exponents are negative.
3. Pullback of Cartier divisors gives a surjection\[ \operatorname{CaDiv}(X)\longrightarrow\operatorname{Div}(\tilde X). \]
Rational local equations on \(X\) remain nonzero in the common function field \(K\) and have unit ratios after pullback. They therefore define a Cartier divisor on \(\tilde X\), identified with a Weil divisor because \(\tilde X\) is nonsingular.
Let \(E\) be a divisor on \(\tilde X\). For each \(P\in S\), choose \(g_P\) by step 2 with the coefficients of \(E\) at the points over \(P\). The divisor \(\operatorname{div}(g_P)-E\) has finite support disjoint from \(\pi^{-1}(P)\). Remove its image and the other points of \(S\) to obtain an open neighborhood \(U_P\) of \(P\) on which \[ \operatorname{div}(g_P)|_{\pi^{-1}(U_P)}=E|_{\pi^{-1}(U_P)}. \] On \(U_0=X\setminus S\), the normalization is an isomorphism, so \(E\) gives a Cartier divisor \(D_0\) there. On \(U_P\), use the single rational local equation \(g_P\). Every overlap of these opens lies in \(U_0\), where the corresponding Weil divisors agree. On a nonsingular curve equality of Weil divisors is equality of Cartier divisors, so these local divisors glue. Their pullback is \(E\).
4. The kernel of the map in step 3 is canonically\[ \bigoplus_{P\in X}B_P^*/A_P^*. \]
A Cartier divisor \(D\) with zero pullback is zero on \(U_0\). If \(g_P\) is a rational local equation at \(P\in S\), zero pullback means that \(g_P\) has valuation zero at every point over \(P\). Equivalently, \(g_P\in B_P^*\): an element of \(K\) lies in \(B_P\), respectively \(B_P^*\), precisely when it is regular, respectively a unit, at all maximal ideals of this semilocal ring. Changing a local equation multiplies \(g_P\) by an element of \(A_P^*\). Thus \(D\) determines one class in \(B_P^*/A_P^*\) for each \(P\in S\). If every class is trivial, \(D\) is locally zero everywhere, proving injectivity into this direct sum.
Conversely, choose representatives \(b_P\in B_P^*\) for a tuple of such classes. Shrink an open neighborhood \(U_P\) so that it contains no other singular point and the rational function \(b_P\) has no zero or pole on \(\pi^{-1}(U_P)\). The Cartier divisor with equation \(b_P\) on \(U_P\) and equation \(1\) on \(X\setminus\{P\}\) is defined: on the overlap the curve is nonsingular and \(b_P\) is a regular unit. Its pullback is zero, and its only possibly nonzero local class is the prescribed one at \(P\). Adding these divisors gives the tuple. The construction is independent of representatives by the injectivity just proved. Outside \(S\) the rings \(A_P\) and \(B_P\) agree, and at the generic point both equal \(K\), so their summands vanish.
This is the global kernel of the sheaf inclusion in the source hint: the quotient \(\pi_*\OO_{\tilde X}^*/\OO_X^*\) has stalks \(B_P^*/A_P^*\) and is supported on \(S\). Step 3 establishes the needed surjectivity on global divisors, rather than assuming that global sections preserve a sheaf surjection.
5. Passing to Cartier divisor classes gives the exact sequence in part (a).
Both projective integral curves have only constant global regular functions. Indeed, such a function defines a morphism to \(\PP^1\) whose image is closed by properness and misses \(\infty\). The image is irreducible and hence a point, so the function belongs to \(k\). Consequently \[ \Gamma(X,\OO_X^*)=\Gamma(\tilde X,\OO_{\tilde X}^*)=k^*. \] Principal Cartier divisors on each curve are therefore the image of the same group \(K^*/k^*\), and pullback identifies these two principal-divisor subgroups. In particular, the kernel in step 4 intersects the principal divisors only in zero.
2.Hence\[ 0\longrightarrow\bigoplus_{P\in X}\tilde\OO_P^*/\OO_P^* \longrightarrow\Pic(X)\xrightarrow{\pi^*}\Pic(\tilde X) \longrightarrow0 \]is exact.
6. Let \(X\) be an integral singular plane cubic, with singular point \(P\).Its normalization is \(\PP^1\), and there is a monic quadratic polynomial \(q(t)\) such that, for \(A=\OO_{X,P}\) and its integral closure \(B\),\[ \boxed{B^*/A^*\cong\bigl(k[t]/(q(t))\bigr)^*/k^*.} \]The polynomial \(q\) has a double root in the cuspidal case and two distinct roots in the nodal case.
Choose coordinates with \(P=[0:0:1]\). The cubic equation has the form \(zq_2(x,y)+q_3(x,y)=0\), where the subscripts denote degrees. The quadratic term is nonzero: otherwise the equation is a binary cubic and factors over \(k\), contrary to integrality. Also \(q_2\) and \(q_3\) have no common linear factor, since that would divide the entire equation.
Choose the coordinate \(x\) so that \(x=0\) is not a tangent direction, and scale the equation so that \(q(t)=q_2(1,t)\) is monic of degree two. Divide \(q_3(1,t)\) by \(q(t)\), obtaining quotient of degree at most one and remainder \(r(t)=a+bt\). Absorb the quotient into \(z\) by a linear coordinate change. The equation becomes \[ zq_2(x,y)+x^2(ax+by)=0,\qquad\gcd(q,r)=1. \] Projection from \(P\) has rational inverse \[ [s:t]\longmapsto[-s q_2(s,t):-t q_2(s,t):s^2(as+bt)]. \] These three cubic forms have no common zero: at a root of \(q_2\) one has \(s\ne0\) and \(as+bt\ne0\). They define a birational morphism \(\PP^1\to X\). Its fibers away from \(P\) contain a single point because projection is its inverse there; the fiber over \(P\) is the finite set of roots of \(q_2\). The morphism is projective, using its closed graph in \(\PP^1\times X\). Its finite fibers therefore imply that it is finite [@Har10a, Exercise III.11.2]. Its source is normal, so the finite birational map is the normalization [@Har10a, Exercise II.3.8]. The points over \(P\) are precisely the roots of \(q\).
On the chart \(z=1\), put \(t=y/x\) in \(K=k(t)\). Then \[ x=-q(t)/r(t),\qquad y=tx,\qquad q(t)+x(a+bt)=0. \] The last equation is monic quadratic in \(t\), so \(C=A[t]=A+At\) is finite over \(A\). Its maximal ideals all lie over \(\mathfrak m=(x,y)A\) and correspond to the roots of \(q\). Indeed, a root \(\alpha\) gives evaluation \(x=y=0\), \(t=\alpha\), while every residue of \(t\) over \(\mathfrak m\) satisfies \(q(t)=0\). The formulas embed \(C\) into the semilocal ring \[ B_0=k[t]_{\{h:\gcd(h,q)=1\}}. \] Each polynomial relatively prime to \(q\) avoids every maximal ideal of \(C\), so is a unit of \(C\). It follows that \(B_0\subseteq C\), and therefore \(C=B_0\). This ring is normal, finite over \(A\), and has fraction field \(K\), so it is \(B\).
Since \(B=A+At\) and \(xt=y\), one has \[ xB=Ax+Ay=\mathfrak m. \] Thus \(I\coloneqq\mathfrak m\) is also an ideal of \(B\). The polynomial \(r\) is a unit in \(B\), so \(I=qB\) and \[ A/I=k,\qquad B/I=k[t]/(q). \] The ideal \(I\) is contained in the Jacobson radicals of both rings. Reduction therefore maps their unit groups onto the unit groups of the quotients, with the same kernel \(1+I\). Taking the quotient gives the claimed unit-group isomorphism, with \(k^*\) embedded as constant polynomials.
The tangent cone at \(P\) is \(q_2\). An ordinary cusp has a repeated tangent and a node has two distinct tangents, giving exactly the stated alternatives for \(q\). There is no other singular point: a line through two singular points would have intersection multiplicity at least four with a cubic, so its restriction polynomial would vanish identically, forcing a line component.
7. For a cuspidal cubic the kernel in part (a) is \((k,+)\), and for a nodal cubic it is \(k^\times\).
In the cuspidal case translate the double root to zero, so that \(k[t]/(q)\cong k[\varepsilon]/(\varepsilon^2)\). Every unit is uniquely \(c(1+u\varepsilon)\) with \(c\in k^*\) and \(u\in k\). Since \[ (1+u\varepsilon)(1+v\varepsilon)=1+(u+v)\varepsilon, \] the quotient by constant units is the additive group \(k\).
In the nodal case write the distinct roots as \(\alpha,\beta\). The Chinese remainder theorem identifies \(k[t]/(q)\) with \(k\times k\) by evaluation at these roots. Its units are \(k^*\times k^*\), and the constants form the diagonal subgroup. The map \((u,v)\mapsto u/v\) identifies the quotient with \(k^*\). These calculations use no characteristic restriction; the distinctness of the node’s tangent directions is the relevant condition.
8. The exact sequences in part (b) follow, with the right-hand map equal to \(\deg\circ\pi^*\).
5.Substituting the two groups from step 7 gives\[ 0\to(k,+)\to\Pic(X)\xrightarrow{\deg\circ\pi^*}\ZZ\to0 \]for the cusp and\[ 0\to k^*\to\Pic(X)\xrightarrow{\deg\circ\pi^*}\ZZ\to0 \]for the node.
9. Q.E.D.
Steps <1>1–<1>5 prove part (a), including the injectivity of the local-unit contribution. Steps <1>6–<1>8 compute that contribution and the quotient for both cubics in part (b).