Given a curve \(Y\) of degree \(d\) in \(\PP^2\), show that there is a nonempty open subset \(U\) of \(\dualof{(\PP^2)}\) in its Zariski topology such that each \(L \in U\) meets \(Y\) in exactly \(d\) points.
Hint: Show that the set of lines in \(\dualof{(\PP^2)}\) which are either tangent to \(Y\) or pass through a singular point of \(Y\) is contained in a proper closed subset.
This result shows that one could have defined the degree of \(Y\) to be the number \(d\) such that almost all lines in \(\PP^2\) meet \(Y\) in \(d\) points, where “almost all” refers to a nonempty open subset of the dual projective space.
Solution
For a point \(P\in\PP^2\), write \[ \dualof{P}=\{L\dualof{\in(\PP^2)}:P\in L\}. \] This is a projective line in the dual plane.
If \(d=1\), then \(Y\) is a line. Removing from \(\dualof{(\PP^2)}\) the single point representing \(Y\) leaves a nonempty open set, and every remaining line meets \(Y\) in exactly one point. Thus assume below that \(d>1\).
1. The singular locus \(\Sing Y\) is finite.
The singular locus is a proper closed subset of the irreducible curve \(Y\). Every proper closed subset of a noetherian irreducible curve has dimension zero and therefore consists of finitely many closed points. Hence \[ \Sing Y=\{P_1,\ldots,P_m\} \] for some \(m\ge0\).
2. The set of lines passing through a singular point of \(Y\) is a proper closed subset of \(\dualof{(\PP^2)}\).
By step 1 it is \[ B_{\mathrm{sing}}=\dualof{P_1}\cup\cdots\cup \dualof{P_m}. \] Each \(\dualof{P_i}\) is a projective line in the dual projective plane, hence closed and proper. Their finite union is closed and has dimension at most one, so it is proper.
3. The set of tangent lines to nonsingular points of \(Y\) is contained in a proper closed subset of \(\dualof{(\PP^2)}\).
By Exercise I.7.3, the tangent-line assignment \[ \gamma:\Reg Y\dualof{\longrightarrow(\PP^2)} \] is a morphism, and the closure of its image is the dual algebraic set \(\dualof{Y}\). Since \(\Reg Y\) is a curve, the dimension of the closure of its image is at most one. Thus \[ \dualof{Y}\dualof{\subsetneq(\PP^2)}, \] which has dimension two. Every tangent line at a nonsingular point lies in \(\dualof{Y}\) by construction.
4. There is a nonempty open subset of the dual plane consisting of lines which meet no singular point and are tangent nowhere along \(\Reg Y\).
Put \[ B=\dualof{Y}\cup B_{\mathrm{sing}}. \] By steps <1>2–<1>3, \(B\) is a finite union of closed subsets of dimension at most one. Hence it is a proper closed subset of the irreducible two-dimensional projective plane \(\dualof{(\PP^2)}\). Therefore \[ U=\dualof{(\PP^2)}\setminus B \] is nonempty and Zariski open. By its definition, a line \(L\in U\) passes through no singular point of \(Y\) and is not tangent to \(Y\) at any nonsingular point.
5. Every line \(L\in U\) meets \(Y\) in exactly \(d\) distinct points.
4. Moreover \(L\) is not the tangent line at \(P\), so Exercise I.5.4 together with the tangent characterization of I.7.3 gives\[ (L.Y)_P=1. \]
The line \(L\) is not a component of \(Y\): if \(Y\) itself is a line, then the corresponding point of the dual plane belongs to \(\dualof{Y}\) and is excluded from \(U\); otherwise irreducibility already prevents a line component. Exercise I.5.4(c) therefore gives the total intersection number \[ \sum_{P\in L\cap Y}(L.Y)_P=d. \] Every summand equals one, so the number of distinct intersection points is exactly \(d\). Thus \[ \boxed{\#(L\cap Y)=d\qquad(L\in U)}. \]
6. Q.E.D.
Step 4 constructs the required nonempty open subset \(U\), and step 5 proves the asserted cardinality for every line represented by a point of \(U\).