Let \(Y \subseteq \AA^n\) be an affine variety, and identify \(\AA^n\) with the open set \(U_0 \subseteq \PP^n\) by the homeomorphism \(\phi_0\). The closure \(\bar{Y}\) of \(Y\) in \(\PP^n\) is called the projective closure of \(Y\).
Write \(S \da k[x_0,\ldots,x_n]\), and let \[ \alpha: S^h \to k[y_1,\ldots,y_n], \quad f \mapsto f(1, y_1, \ldots, y_n), \qquad \beta: k[y_1,\ldots,y_n] \to S^h, \quad g \mapsto x_0^{\deg g} \, g\qty{\frac{x_1}{x_0}, \ldots, \frac{x_n}{x_0}} . \]
Show that \(I(\bar{Y})\) is the ideal generated by \(\beta(I(Y))\).
Let \(Y \subseteq \AA^3\) be the twisted cubic \(Y = \ts{(t,t^2,t^3) \st t \in k}\). Its projective closure \(\bar{Y} \subseteq \PP^3\) is called the twisted cubic curve. Find generators for \(I(Y)\) and for \(I(\bar{Y})\), and use this example to show that if \(f_1,\ldots,f_r\) generate \(I(Y)\), then \(\beta(f_1),\ldots,\beta(f_r)\) need not generate \(I(\bar{Y})\).
Solution
Part 1. We want \(I(\bar{Y}) = \gens{\beta(I(Y))}\). Write \(H_i = Z(x_i) \subseteq \PP^n\) and embed \(\AA^n \injects \PP^n\) along \(U_0 \da \PP^n \sm H_0 = \ts{x_0 = 1}\), so that \((x_1,\ldots,x_n) \mapsto \tv{1 : x_1 : \cdots : x_n}\).
The inclusion \(I(\bar{Y}) \subseteq \gens{\beta(I(Y))}\). Let \(g \in I(\bar{Y})\), which we may take homogeneous, so \(g(a_0,\ldots,a_n) = 0\) whenever \(\tv{a_0 : \cdots : a_n} \in \bar{Y}\). In particular \(g\) vanishes on \(\bar{Y} \intersect U_0 = Y\), where \(x_0 \neq 0\), so writing a point as \(\tv{1 : b_1 : \cdots : b_n}\) with \(b_i = a_i/a_0\) we get \(g(1, b_1, \ldots, b_n) = 0\), and \((b_1,\ldots,b_n)\) is a point of \(Y\). So set \(f(y_1,\ldots,y_n) \da g(1, y_1, \ldots, y_n)\), which vanishes on \(Y\) and hence lies in \(I(Y)\). Then \[ \beta(f)(x_0,\ldots,x_n) = x_0^{\deg f} f\qty{\frac{x_1}{x_0}, \ldots, \frac{x_n}{x_0}} = x_0^{\deg g} \, g\qty{1, \frac{x_1}{x_0}, \ldots, \frac{x_n}{x_0}} = g(x_0, x_1, \ldots, x_n) . \] Note that this argument places \(g\) in the ideal generated by \(\beta(I(Y))\); it need not be literally in the image of \(\beta\).
The inclusion \(\gens{\beta(I(Y))} \subseteq I(\bar{Y})\). Let \(f = \beta(g)\) with \(g \in I(Y)\), so \[ f(x_0,\ldots,x_n) = x_0^{\deg g} \, g\qty{\frac{x_1}{x_0}, \ldots, \frac{x_n}{x_0}} . \] For a point \((a_1,\ldots,a_n) \in Y\) we get \(f(1, a_1, \ldots, a_n) = g(a_1,\ldots,a_n) = 0\), so \(f\) vanishes on \(Y\). Since \(f\) is homogeneous, its zero locus is closed in \(\PP^n\), and a closed set containing \(Y\) contains \(\bar{Y}\). Hence \(f \in I(\bar{Y})\).
Part 2. The curve \(Y\) has codimension \(2\), so we expect two equations, and indeed \[ I(Y) = \gens{ f_1 = x^2 - y,\ f_2 = x^3 - z } . \] Homogenize with respect to a new variable \(w\), working in \(k[w,x,y,z]\): \[ \beta(f_1) = w^2\qty{ \qty{\frac{x}{w}}^2 - \frac{y}{w} } = x^2 - yw, \qquad \beta(f_2) = w^3\qty{ \qty{\frac{x}{w}}^3 - \frac{z}{w} } = x^3 - w^2 z . \] So \(J \da \gens{\beta(f_1), \beta(f_2)} = \gens{ wy - x^2,\ x^3 - w^2 z }\).
On the other hand, \(\bar{Y}\) is the image of \(\tv{s : t} \mapsto \tv{s^3 : s^2 t : s t^2 : t^3}\), and homogenizing a Gröbner basis for \(I(Y)\) gives \[ I(\bar{Y}) = \gens{ wy - x^2,\ wz - xy,\ xz - y^2 } \subseteq k[w,x,y,z] . \] Inspection shows \(xz - y^2 \notin J\): every element of \(J\) has \(x\)-adic valuation either \(0\) or at least \(2\), while \(xz - y^2\) does not. So the homogenizations of a generating set of \(I(Y)\) do not generate \(I(\bar{Y})\).
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P-AGHTWCUBICThe twisted cubic is an affine variety of dimension oneuses