Let \(k\) be algebraically closed.
The intersection of two varieties need not be a variety. Let \(Q_1, Q_2 \subseteq \PP_k^3\) be the quadric surfaces given by \(x^2 - yw = 0\) and \(xy - zw = 0\). Show that \(Q_1 \intersect Q_2\) is the union of a twisted cubic curve and a line.
Even when the intersection of two varieties is a variety, the ideal of the intersection need not be the sum of the ideals. Let \(C \subseteq \PP_k^2\) be the conic given by \(x^2 - yz = 0\), and let \(L\) be the line given by \(y = 0\). Show that \(C \intersect L\) is a single point \(P\), but that \(I(C) + I(L) \neq I(P)\).
Solution
The intersections in the statement are intersections of algebraic sets. For part (a), use coordinates \([x:y:z:w]\) on \(\PP^3\); part (b) uses \([x:y:z]\) on \(\PP^2\).
1. The two components in (a) are\[ \boxed{T=\{[s^2t:st^2:t^3:s^3]:[s:t]\in\PP^1\},\qquad R=Z(x,w).} \]
The parametrization of \(T\) consists of the four degree-three monomials in \(s,t\), with the pure cube \(s^3\) placed last. It is therefore the twisted cubic, and in particular a closed irreducible curve. The set \(R\) is the projectivization of the two-dimensional coordinate subspace with \(x=w=0\), hence is a line by P-AGH211LINEAR.
Substitution of the parametrization of \(T\) gives \(x^2=yw\) and \(xy=zw\). Both equations also vanish on \(R\). Thus \(T\cup R\subseteq Q_1\cap Q_2\).
Conversely, on the chart \(w\ne0\), normalize \(w=1\). The equations give \(y=x^2\) and \(z=xy=x^3\), so the point is \([x:x^2:x^3:1]\), the point of \(T\) with \([s:t]=[1:x]\). On \(w=0\), the first equation forces \(x=0\), so the point belongs to \(R\). These cases exhaust the projective intersection and prove \(Q_1\cap Q_2=T\cup R\).
2. The two quadrics are varieties, but their intersection is reducible.
The polynomial \(x^2-yw\) is primitive as a polynomial in \(y\) over the UFD \(k[x,z,w]\), because \(x^2\) and \(w\) are relatively prime. It is linear and irreducible over the fraction field of that UFD, so Gauss’s lemma makes it irreducible in \(k[x,y,z,w]\). The same argument applies to \(xy-zw\), viewed as a primitive linear polynomial in \(z\) over \(k[x,y,w]\). Both homogeneous equations therefore define irreducible projective hypersurfaces by P-AGH28HYPERSURFACE.
The point \([1:1:1:1]\) lies in \(T\) but not \(R\). The point \([0:1:0:0]\) lies in \(R\) but not \(T\): on \(T\), the equation \(w=s^3=0\) forces \(s=0\), giving only \([0:0:1:0]\). Thus neither of the two nonempty closed subsets \(T,R\) contains the other. Their union from step 1 is reducible, and those subsets are its two irreducible components by P-AGH25NOETHERIAN. This verifies the stated phenomenon in (a).
3. In (b),\[ \boxed{C\cap L=\{P\},\quad P=[0:0:1],\qquad I(C)+I(L)=(x^2,y)\subsetneq(x,y)=I(P).} \]
On \(L\), the conic equation becomes \(x^2=0\), hence \(x=0\). The remaining projective coordinate \(z\) must be nonzero, giving exactly \(P=[0:0:1]\).
The polynomial \(x^2-yz\) is primitive and linear in \(y\) over \(k[x,z]\), so the Gauss-lemma argument of step 2 makes it irreducible. The homogeneous vanishing ideals are therefore \[ I(C)=(x^2-yz),\qquad I(L)=(y),\qquad I(P)=(x,y) \] by P-AGH24CORRESPONDENCE and the coordinate-subspace calculation in P-AGH211LINEAR. Consequently \[ I(C)+I(L)=(x^2-yz,y)=(x^2,y). \] This does not contain \(x\). Indeed, reduction modulo \(y\) would otherwise give \(x\in(x^2)\) in \(k[x,z]\), which would force \(1=xh\) for some polynomial \(h\), an impossibility after setting \(x=0\). Since \(x\in I(P)\), the containment is strict. The radical of the sum is \((x,y)\); before taking the radical, its quotient has the nonzero square-zero element given by the class of \(x\). This proves all of (b).
4. Q.E.D.
Steps <1>1–<1>2 give the reducible intersection in (a), and step 3 gives the point intersection with unequal ideals in (b).