Let \(k\) be algebraically closed. A nonempty closed subscheme \(Y\) of \(\PP^n_k\) is called a (strict, global) complete intersection if its saturated homogeneous ideal \(I\) in \(S = k[x_0, \ldots, x_n]\) can be generated by \(r = \codim(Y, \PP^n)\) elements (I, Ex. 2.17).
Let \(Y\) be a nonempty closed subscheme of codimension \(r\) in \(\PP^n\). Then \(Y\) is a complete intersection if and only if there are hypersurfaces, i.e. locally principal subschemes of codimension \(1\), \(H_1, \ldots, H_r\) such that \(Y = H_1 \intersect \cdots \intersect H_r\) as schemes, i.e. \(\mci_Y = \mci_{H_1} + \cdots + \mci_{H_r}\).
If \(Y\) is a complete intersection of dimension \(\geq 1\) in \(\PP^n\), and if \(Y\) is normal, then \(Y\) is projectively normal (Ex. 5.14).
With the same hypotheses as (b), conclude that for all \(l \geq 0\) the natural map \(\Gamma(\PP^n, \OO_{\PP^n}(l)) \to \Gamma(Y, \OO_Y(l))\) is surjective. In particular, taking \(l = 0\), show that \(Y\) is connected.
Now suppose given integers \(d_1, \ldots, d_r \geq 1\) with \(r < n\). Use Bertini’s theorem (8.18) to show that there exist nonsingular hypersurfaces \(H_1, \ldots, H_r\) in \(\PP^n\) with \(\deg H_i = d_i\) such that the scheme \(Y = H_1 \intersect \cdots \intersect H_r\) is irreducible and nonsingular of codimension \(r\) in \(\PP^n\).
If \(Y\) is a nonsingular complete intersection as in (d), show that \[ \omega_Y \cong \OO_Y\left(\sum_i d_i - n - 1\right) . \]
If \(Y\) is a nonsingular hypersurface of degree \(d\) in \(\PP^n\), with \(n\ge2\), use (c) and (e) to show that \(p_g(Y) = \binom{d-1}{n}\). Thus \(p_g(Y) = p_a(Y)\) (I, Ex. 7.2). In particular, if \(Y\) is a nonsingular plane curve of degree \(d\), then \(p_g(Y) = \frac{1}{2}(d-1)(d-2)\).
If \(Y\) is a nonsingular curve in \(\PP^3\) which is a complete intersection of nonsingular surfaces of degrees \(d, e\), then \[ p_g(Y) = \frac{1}{2} de(d + e - 4) + 1 . \] Again the geometric genus is the same as the arithmetic genus (I, Ex. 7.2).
(Positive dimension in the hypersurface genus formula)
Part (f) uses the positive-dimensional hypotheses of (c) and (e). For \(n=1\), a smooth degree-\(d\) hypersurface is a disjoint union of \(d\) reduced points. Its zeroth exterior power of differentials is \(\OO_Y\), so \(h^0(Y,\omega_Y)=d\), whereas its arithmetic genus and \(\binom{d-1}{1}\) both equal \(d-1\). Thus the displayed equality of genera does not extend to that zero-dimensional case.
Hint
For (a), apply the unmixedness theorem to the homogeneous ideal generated by the hypersurface equations. For (b), apply Theorem II.8.23 to the affine cone over \(Y\).
Solution
Write \(\mathfrak m=(x_0,\ldots,x_n)\subset S\). The saturated homogeneous ideal of a closed subscheme is the ideal corresponding to its ideal sheaf, not an arbitrary ideal defining the same Proj. For a nonnegative integer \(a\), use \(\binom{a}{b}=0\) when \(a<b\).
1. If \(J=(F_1,\ldots,F_r)\) is generated by positive-degree homogeneous polynomials and \(\operatorname{Proj}(S/J)\) is nonempty of codimension \(r\), then the \(F_i\) form a regular sequence, \(S/J\) is Cohen–Macaulay of dimension \(n+1-r\), and \(J\) is saturated.
The dimension formula for a homogeneous cone gives \(\dim(S/J)=n+1-r\) [@Har10a, Chapter I, §2]. At the homogeneous maximal ideal this is also the local dimension: every homogeneous irreducible component of the cone contains its vertex. The regular local ring \(S_{\mathfrak m}\) has dimension \(n+1\) and is Cohen–Macaulay. The dimension criterion for regular sequences makes \(F_1,\ldots,F_r\) a regular sequence there. It is also a regular sequence in \(S\). Indeed, each kernel testing regularity is graded, and a nonzero homogeneous element of a graded module cannot become zero upon localization at \(\mathfrak m\): multiplying it by a polynomial with nonzero constant term has a nonzero component in its original degree. The quotient is a Cohen–Macaulay complete intersection, with the stated dimension.
An element of \(J^{\mathrm{sat}}/J\) is annihilated by a power of \(\mathfrak m\). To see this using the saturation definition, choose powers of all the \(x_i\) annihilating it; every monomial of sufficiently large degree is divisible by one of those powers. But \((S/J)_{\mathfrak m}\) has depth \(n+1-r\ge1\). Choose a non-zero-divisor in its maximal ideal; no nonzero element can be killed by a power of that ideal, since it would then be killed by a power of this non-zero-divisor. Thus \((J^{\mathrm{sat}}/J)_{\mathfrak m}=0\). The graded localization observation also applies to this quotient, so it is zero globally. Hence \(J\) is saturated. This excludes an embedded component at the vertex, which would otherwise be invisible on Proj.
2. The equivalence in (a) holds.
If the homogeneous ideal \(I\) is generated by \(r\) elements, it can be generated by \(r\) homogeneous elements. Choose homogeneous lifts of a basis of \(I/\mathfrak m I\); graded Nakayama, or induction on degree, shows that these generate \(I\). Their number is at most \(r\), and the height bound for a generated ideal makes it at least \(\operatorname{height}I=r\). Write these generators as \(F_i\) of degrees \(d_i>0\). Their zero schemes \(H_i=V_+(F_i)\) are hypersurfaces, and sheafifying \(I=(F_1,\ldots,F_r)\) gives precisely the required scheme-theoretic intersection.
Conversely, a nonempty hypersurface in \(\PP^n\) is an effective Cartier divisor. Its invertible ideal is \(\OO(-d)\) for an integer \(d>0\): the Picard group of projective space is generated by \(\OO(1)\), and its inclusion into \(\OO\) is a nonzero homogeneous form of degree \(d\) [@Har10a, Proposition II.6.4, Corollary II.6.16 and Proposition II.5.13]. The degree cannot be negative, and degree zero would give the empty divisor. Thus each given \(H_i\) is defined by a form \(F_i\). Put \(J=(F_1,\ldots,F_r)\). The stated equality of ideal sheaves says \(\operatorname{Proj}(S/J)=Y\) and identifies \(I\) with \(J^{\mathrm{sat}}\) by P-AGH2510SATIDEAL. Step 1 gives \(J^{\mathrm{sat}}=J\), so \(I\) has the required \(r\) generators. For \(r=0\), the only nonempty codimension-zero closed subscheme of the integral scheme \(\PP^n\) is \(\PP^n\) itself, and the empty intersection and zero ideal give the assertion, including \(n=0\).
3. Under the assumptions of (b), the homogeneous coordinate ring \(R=S/I\) is a normal domain.
Step 1 makes \(R\) Cohen–Macaulay of dimension \(\dim Y+1\ge2\), so it satisfies Serre’s condition \((S_2)\). Let \(o\) be the vertex, defined by \(R_+\). The complement of \(o\) is covered by the affine opens \(D(x_i)\), and the standard grading gives \[ R_{x_i}\cong (R_{x_i})_0[x_i,x_i^{-1}]. \] Here \(\operatorname{Spec}(R_{x_i})_0\) is the corresponding affine open of \(Y\) and is normal by assumption. Polynomial extension and localization preserve normality, so each \(R_{x_i}\) is normal. This also holds when a chart has several components: its finitely many normal irreducible components are disjoint, so its coordinate ring is a finite product of normal domains. Apply the polynomial normality lemma used in P-AGH2514PROJNORM, step 4, to each factor.
The vertex has codimension \(\dim R\ge2\). All local rings of \(R\) at primes of height at most one therefore lie in this normal complement and are regular. Thus \(R\) satisfies \((R_1)\), and Serre’s normality criterion makes \(R\) normal, including at the vertex.
In particular \(R_{R_+}\) is a domain. The map \(R\to R_{R_+}\) is injective: if \(a\ne0\), multiplying its lowest nonzero homogeneous component by the nonzero constant term of any element outside \(R_+\) cannot give zero. Therefore \(R\) itself is a domain. As a normal domain it is integrally closed, proving projective normality. No connectedness or integrality of \(Y\) was assumed in this argument.
4. Part (c) holds, and under its hypotheses \(H^0(Y,\OO_Y)=k\) and \(H^0(Y,\OO_Y(l))=0\) for \(l<0\).
The projectively normal coordinate ring in step 3 has \[ R_l\cong H^0(Y,\OO_Y(l))\qquad(l\ge0) \] by P-AGH2514PROJNORM, part (d). The quotient \(S_l\to R_l\) is surjective, proving the requested restriction-map surjectivity. At \(l=0\) it identifies \(H^0(Y,\OO_Y)\) with \(k\). A decomposition into two nonempty open-and-closed subschemes would give a nontrivial idempotent global function, impossible in a field. Thus \(Y\) is connected; it is also integral, since it is the Proj of the domain \(R\).
For the extra negative-degree assertion, let \(l=-q<0\) and suppose \(s\ne0\) is a section of \(\OO_Y(-q)\). Choose two distinct closed points of the positive-dimensional integral projective variety \(Y\) and a linear form \(h\) vanishing at one but not the other. The product \(h^q s\) is a nonzero global section of \(\OO_Y\): its generic value is nonzero because \(Y\) is integral. It vanishes at the first point, whereas every global function is constant. This is a contradiction, so no such \(s\) exists.
5. The hypersurfaces required in (d) exist.
Construct \(Y_j=H_1\cap\cdots\cap H_j\) inductively, starting from \(Y_0=\PP^n\). Suppose \(Y_{j-1}\) is smooth and integral of dimension \(n-j+1\ge2\). The complete degree-\(d_j\) system gives the Veronese embedding of \(\PP^n\) and an embedding of \(Y_{j-1}\) by restriction. By Bertini's hyperplane theorem, applied to these two embedded smooth varieties, a dense open subset of the same space of degree-\(d_j\) forms gives hypersurfaces smooth in \(\PP^n\) and meeting \(Y_{j-1}\) smoothly in codimension one [@Har10a, Theorem II.8.18]. Intersect the two dense open subsets and choose a point in their intersection; the parameter space is irreducible and \(k\) is algebraically closed. This yields a smooth hypersurface \(H_j\) and a nonempty smooth \(Y_j\) of codimension \(j\) in \(\PP^n\). Nonemptiness follows because a hypersurface of positive degree meets every positive-dimensional projective variety not contained in it [@Har10a, Chapter I, §7].
Step 2 makes \(Y_j\) a global complete intersection. Its dimension \(n-j\) is positive, and smoothness makes it normal. Steps <1>3–<1>4 therefore make it connected and integral. This completes the induction through \(j=r\) and proves all the requirements, including smoothness of each ambient \(H_j\). For \(r=0\), take \(Y=\PP^n\).
6. For the smooth complete intersection in (e),\[ \boxed{\omega_Y\cong\OO_Y\left(\sum_{i=1}^r d_i-n-1\right).} \]
The defining equations give a surjection \[ \bigoplus_{i=1}^r\OO_Y(-d_i)\longrightarrow\mathcal I_Y/\mathcal I_Y^2. \] The conormal sheaf is locally free of rank \(r\), since \(Y\) and \(\PP^n\) are smooth [@Har10a, Theorem II.8.17]. The source has the same rank, so the surjection is an isomorphism: over a local ring its square matrix is invertible after reduction modulo the maximal ideal, hence has unit determinant. Take determinants in the exact conormal sequence \[ 0\to\mathcal I_Y/\mathcal I_Y^2\to\Omega_{\PP^n/k}|_Y\to\Omega_{Y/k}\to0. \] Using \(\omega_{\PP^n}=\OO_{\PP^n}(-n-1)\) and the determinant identity in D-MODCONORM gives \[ \OO_Y(-n-1)\cong\OO_Y\left(-\sum_i d_i\right)\otimes\omega_Y, \] which rearranges to the stated formula.
7. For the hypersurface in (f),\[ \boxed{p_g(Y)=p_a(Y)=\binom{d-1}{n}.} \]
By steps <1>3–<1>4 this smooth positive-dimensional hypersurface is integral and projectively normal. Step 6 gives \(\omega_Y=\OO_Y(a)\) with \(a=d-n-1\). If \(a<0\), step 4 gives \(p_g(Y)=0\), in agreement with the binomial convention. If \(a\ge0\), then \(a<d\). The ideal generated by its degree-\(d\) equation has no degree-\(a\) component, so step 4 gives \[ p_g(Y)=\dim_k(S/(F))_a=\dim_kS_a =\binom{a+n}{n}=\binom{d-1}{n}. \]
The homogeneous quotient sequence has Hilbert polynomial \[ P_Y(q)=\binom{q+n}{n}-\binom{q+n-d}{n}, \] where both binomial expressions here are polynomials in \(q\). At zero, the second polynomial is \[ \binom{n-d}{n}=(-1)^n\binom{d-1}{n}, \] as follows by reversing the signs of its \(n\) factors. Consequently \[ p_a(Y)=(-1)^{n-1}(P_Y(0)-1)=\binom{d-1}{n} \] [@Har10a, Chapter I, §7]. For \(n=2\) this is \((d-1)(d-2)/2\), the plane-curve formula.
8. For the curve in (g),\[ \boxed{p_g(Y)=p_a(Y)=1+\frac{de(d+e-4)}2.} \]
1.In this two-element case, its graded resolution is\[ 0\to S(-d-e)\xrightarrow{h\mapsto(-Gh,Fh)} S(-d)\oplus S(-e)\xrightarrow{(a,b)\mapsto Fa+Gb} S\to R\to0. \]For exactness at the middle term, \(Fa+Gb=0\) implies \(b\in(F)\) because \(G\) is a non-zero-divisor modulo \(F\); substituting \(b=Fh\) then gives \(a=-Gh\).The first map is injective since \(S\) is a domain.
Put \(a=d+e-4\). Steps 4 and 6 identify \(H^0(Y,\omega_Y)\) with \(R_a\) when \(a\ge0\), and make both zero when \(a<0\). Taking the degree-\(a\) part of this resolution gives \[ \begin{aligned} p_g(Y) &=\binom{d+e-1}{3}-\binom{e-1}{3}-\binom{d-1}{3}\\ &=1+\frac{de(d+e-4)}2. \end{aligned} \] The last term \(S_{a-d-e}=S_{-4}\) vanishes. The displayed binomials agree with \(b(b-1)(b-2)/6\) also at \(b=0,1,2\), so direct expansion proves the final equality for every \(d,e\ge1\), including the line and conic cases.
The same resolution gives \[ P_Y(q)=\binom{q+3}{3}-\binom{q-d+3}{3} -\binom{q-e+3}{3}+\binom{q-d-e+3}{3} =deq-\frac{de(d+e-4)}2. \] For a curve, \(p_a(Y)=1-P_Y(0)\). This is the same value as the computed geometric genus, proving (g).
9. Q.E.D.
Steps <1>1–<1>2 prove (a), step 3 proves (b), step 4 proves (c), and steps <1>5–<1>8 prove (d)–(g), respectively.