Let \(k\) be algebraically closed. A variety \(Y\) of dimension \(r\) in \(\PP^n\) is a strict complete intersection if \(I(Y)\) can be generated by \(n-r\) elements. It is a set-theoretic complete intersection if \(Y\) can be written as the intersection of \(n-r\) hypersurfaces.
Let \(Y \subseteq \PP^n\) be a variety with \(Y = Z(\mfa)\) for a homogeneous ideal \(\mfa\), and suppose \(\mfa\) can be generated by \(q\) elements. Show that \(\dim Y \geq n - q\).
Show that a strict complete intersection is a set-theoretic complete intersection.
The converse of (b) is false. Let \(Y\) be the twisted cubic curve in \(\PP^3\). Show that \(I(Y)\) cannot be generated by two elements, and find hypersurfaces \(H_1, H_2\) of degrees \(2\) and \(3\) with \(Y = H_1 \intersect H_2\).
The source records as open the question whether every closed irreducible curve in \(\PP^3\) is a set-theoretic intersection of two surfaces [@Har10a, Exercise I.2.17(d)].
Solution
Use \(S=k[x_0,\ldots,x_n]\) and \(\mathfrak m=S_+=(x_0,\ldots,x_n)\). All varieties in this exercise are nonempty and irreducible, and the intersections are set-theoretic.
1. Under the hypotheses of (a), \(\dim Y\ge n-q\).
The projective Nullstellensatz gives \(I(Y)=\sqrt{\mfa}\), since \(Y\) is nonempty, by P-AGH23ZIPROPS, part (d). The ideal \(I(Y)\) is prime by P-AGH24CORRESPONDENCE, so it is the unique minimal prime over \(\mfa\). Krull’s height theorem for an ideal generated by \(q\) elements gives \(\operatorname{height}I(Y)\le q\) in the noetherian polynomial ring [@Har10a, Chapter I, §1]. The polynomial-ring dimension formula and the cone dimension formula give \[ \dim Y=\dim(S/I(Y))-1 =(n+1-\operatorname{height}I(Y))-1\ge n-q \] [@Har10a, Chapter I, §1 and Exercise I.2.6]. This applies even when the given \(q\) generators were not homogeneous.
2. Every strict complete intersection is a set-theoretic complete intersection.
Put \(c=n-r\) and suppose \(I=I(Y)\) can be generated by \(c\) elements. There is a homogeneous generating set with exactly \(c\) elements. Indeed, a homogeneous basis of \(I/\mathfrak m I\) has at most \(c\) elements, since any generating set of \(I\) spans this quotient. Lift that basis homogeneously to \(g_1,\ldots,g_a\). These lifts generate \(I\) by induction on degree: after subtracting a linear combination of same-degree basis lifts, a homogeneous element lies in \(\mathfrak m I\), where it is a sum of positive-degree multiples of lower-degree elements of \(I\). The latter elements are generated by the lifts by induction. Thus \(a\le c\), while the height bound from step 1, or Krull’s height theorem applied to \(I\), gives \(a\ge\operatorname{height}I=c\). Consequently \(a=c\).
For each \(g_i\), choose an irreducible homogeneous factor \(f_i\in I\). Such a factor exists because \(I\) is prime; factors of a homogeneous polynomial in a graded domain are homogeneous by the least-and-greatest-degree argument in P-AGH28HYPERSURFACE. Then \[ I=(g_1,\ldots,g_c)\subseteq(f_1,\ldots,f_c)\subseteq I, \] so these ideals are equal. Each \(Z(f_i)\) is a hypersurface containing \(Y\), and their intersection is \(Z(I)=Y\). For \(c=0\), the ideal is zero and the empty intersection is \(\PP^n\). This proves (b) with hypersurfaces that are themselves irreducible varieties.
3. The ideal of the twisted cubic cannot be generated by two elements.
Use the coordinates of P-AGH216INTERSECT and write \[ Y=\{[s^2t:st^2:t^3:s^3]:[s:t]\in\PP^1\}\subseteq\PP^3_{[x:y:z:w]}. \] Its ideal has no nonzero constant or linear component. For the linear component, substitution of \(ax+by+cz+dw\) gives \[ as^2t+bst^2+ct^3+ds^3, \] which vanishes identically only if all four coefficients are zero, since these are distinct monomials and \(k\) is infinite.
On the other hand, the three quadrics \[ F=x^2-yw,\qquad B=xy-zw,\qquad C=y^2-xz \] belong to \(I(Y)\) by direct substitution and are linearly independent, as witnessed by their respective monomials \(x^2,xy,y^2\). The degree-two part of \(\mathfrak m I(Y)\) is zero because \(I(Y)\) has no components below degree two. Thus their classes remain independent in \(I(Y)/\mathfrak m I(Y)\). Any ideal generating set maps to a spanning set of that vector space over \(S/\mathfrak m=k\), so it must contain at least three elements. In particular no two elements, homogeneous or otherwise, generate \(I(Y)\).
4. The hypersurfaces requested in (c) are\[ \boxed{H_1=Z(x^2-yw),\qquad H_2=Z(y^3+wz^2-2xyz).} \]
3.Their degrees are two and three, respectively.The polynomial \(F\) is irreducible by the primitive-linear-polynomial argument in P-AGH216INTERSECT.The polynomial \(G\), regarded as a polynomial in \(w\) over \(k[x,y,z]\), is primitive: its coefficient \(z^2\) and constant term \(y^3-2xyz\) are relatively prime, since \(z\) does not divide the latter.It is linear over the fraction field, so Gauss’s lemma makes it irreducible.Thus both zero sets are hypersurfaces in the required sense.
The polynomial identities \[ (xy-zw)^2=wG+y^2F,\qquad (y^2-xz)^2=yG+z^2F \] hold by expansion, including in characteristic two. At a common zero of \(F\) and \(G\), they force \(xy-zw=0\) and \(y^2-xz=0\). On \(w\ne0\), set \(w=1\); the first two quadratic equations give \(y=x^2\) and \(z=x^3\), hence a point of \(Y\). On \(w=0\), \(F=0\) forces \(x=0\), and \(y^2-xz=0\) then forces \(y=0\). The only remaining projective point is \([0:0:1:0]\), also in \(Y\). Therefore \(H_1\cap H_2\subseteq Y\). The opposite inclusion follows from their vanishing on the parametrization, so equality holds. Combined with step 3, this proves that the twisted cubic is set-theoretically, but not strictly, a complete intersection.
5. Q.E.D.
Step 1 proves (a), step 2 proves (b), and steps <1>3–<1>4 prove both requests in (c).