Let \(k\) be algebraically closed. Let \(Y\) be the image of the \(2\)-uple embedding of \(\PP^2\) in \(\PP^5\); this is the Veronese surface. A curve is a variety of dimension \(1\). If \(Z \subseteq Y\) is a closed curve, show that there exists a hypersurface \(V \subseteq \PP^5\) with \(V \intersect Y = Z\).
(Set-theoretic and scheme-theoretic intersection)
The equality in this exercise is an equality of algebraic sets. When the plane curve \(C\) has odd degree \(e\), the constructed hypersurface pulls back to the equation \(f^2\) and therefore cuts out the double divisor \(2C\) scheme-theoretically. When \(e\) is even, its pullback is \(f\) and the scheme-theoretic intersection is reduced.
Solution
Let \(\nu:\PP_k^2\to Y\) be the quadratic Veronese embedding, with coordinates \[ [x_0:x_1:x_2]\longmapsto[x_0^2:x_1^2:x_2^2:x_0x_1:x_0x_2:x_1x_2]. \] Write \(S=k[x_0,x_1,x_2]\), \(T=k[y_0,\ldots,y_5]\), and let \(\theta:T\to S\) be this monomial substitution.
1. There is an irreducible homogeneous polynomial \(f\in S\) of degree \(e>0\) with \(\nu^{-1}(Z)=Z(f)\).
The map \(\nu\) is an isomorphism onto \(Y\) by the coordinate calculation in P-AGH212DUPLE. Consequently \(C=\nu^{-1}(Z)\) is a closed irreducible curve in \(\PP^2\). The homogeneous hypersurface criterion gives a positive-degree irreducible homogeneous equation \(f\) for this codimension-one variety. In particular, its full homogeneous ideal is \((f)\).
2. There is a positive-degree homogeneous \(F\in T\) such that\[ \theta(F)= \begin{cases} f,&e\text{ even},\\ f^2,&e\text{ odd}. \end{cases} \]
For every \(m\ge0\), the substitution map \(T_m\to S_{2m}\) is surjective. Indeed, every monomial of degree \(2m\) is a product of \(2m\) variables, with repetition, and pairing these factors expresses it as a product of \(m\) quadratic monomials. All six quadratic monomials occur among the images of the \(y_i\). Since monomials span \(S_{2m}\), the map is surjective.
For even \(e\), apply this with \(m=e/2\) to lift \(f\). For odd \(e\), apply it with \(m=e\) to lift \(f^2\). Both choices have \(m>0\), and the lifted polynomial is nonzero because its image is nonzero.
3. The polynomial \(F\) from step 2 is irreducible.
Suppose \(F=GH\) with nonconstant factors in \(T\). Since \(F\) is homogeneous and \(T\) is a graded domain, both factors are homogeneous: the least and greatest nonzero degrees of a product add, so equality of those degrees in the product forces equality in each factor. The nonzero polynomials \(\theta(G)\) and \(\theta(H)\) then have positive even degrees, twice the respective degrees of \(G\) and \(H\). They are nonzero because their product is the nonzero polynomial in step 2.
The polynomial ring \(S\) is a UFD, and \(f\) is irreducible. If \(e\) is even, the equation \(\theta(G)\theta(H)=f\) contradicts irreducibility of \(f\), since both factors have positive degree. If \(e\) is odd, unique factorization in \(\theta(G)\theta(H)=f^2\) and positivity of the two degrees force \[ \theta(G)=cf,\qquad \theta(H)=c^{-1}f \] for some \(c\in k^\times\). Their degrees would then be the odd number \(e\), contradicting the even degrees just established. Both cases exclude the proposed factorization.
4. The required hypersurface is \(\boxed{V=Z(F)\subseteq\PP_k^5}\), and \(V\cap Y=Z\).
2.Thus P-AGH28HYPERSURFACE makes it a hypersurface in \(\PP^5\).
For every \(a\in\PP^2\), the equation \(F(\nu(a))=0\) is equivalent to \(f(a)=0\), whether the substituted polynomial is \(f\) or \(f^2\). Hence \[ \nu^{-1}(V\cap Y)=Z(\theta(F))=Z(f)=C. \] The bijection \(\nu:\PP^2\to Y\) gives \(V\cap Y=\nu(C)=Z\). Moreover \(V\) does not contain \(Y\), because \(\theta(F)\) is a nonzero homogeneous polynomial and \(I(\PP^2)=(0)\).
5. Q.E.D.
Steps <1>1–<1>3 construct the irreducible hypersurface equation, and step 4 proves the required intersection equality.
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Dependencies
P-AGH212DUPLEThe $d$-uple embedding of $\PP^n$ in $\PP^N$uses