Let \(Y \subseteq \PP^n\) be a nonempty algebraic set, and let \(\theta: \AA^{n+1} \sm \ts{(0,\ldots,0)} \to \PP^n\) send the point with affine coordinates \((a_0,\ldots,a_n)\) to the point with homogeneous coordinates \(\tv{a_0 : \cdots : a_n}\). The affine cone over \(Y\) is \[ C(Y) = \inverseof{\theta}(Y) \union \ts{(0,\ldots,0)} . \]
Show that \(C(Y)\) is an algebraic set in \(\AA^{n+1}\) whose ideal equals \(I(Y)\), regarded as an ordinary ideal of \(k[x_0,\ldots,x_n]\).
Show that \(C(Y)\) is irreducible if and only if \(Y\) is.
Show that \(\dim C(Y) = \dim Y + 1\).
The projective closure \(\overline{C(Y)}\) in \(\PP^{n+1}\) is called the projective cone over \(Y\).
Solution
Part 1. The set \(C(Y)\) is algebraic because it is of the form \(C(Y) = V(I(Y))\). That the ideal is again \(I(Y)\): a polynomial \(f\) vanishing on \(C(Y)\) vanishes at every \(\vector{a} \neq \vector{0}\) of the cone, and reading \(\vector{a}\) as homogeneous coordinates on \(\PP^n\) shows \(f\) vanishes on \(Y\). Conversely, if \(f \in I(Y)\) is homogeneous then \(f(\lambda \vector{a}) = \lambda^{\deg f} f(\vector{a}) = 0\), so \(f\) vanishes on \(C(Y) \sm \ts{\vector{0}}\). Every homogeneous polynomial of positive degree also vanishes at \(\vector{0}\), so \(f \in I(C(Y))\).
Part 2. An algebraic set is irreducible exactly when its ideal is prime, and by part 1 the two ideals coincide: \(I(Y) = I(C(Y))\).
Part 3. Using the previous exercise, \[ \dim C(Y) = \dim S(C(Y)) = \dim S(Y) = \dim Y + 1 . \]
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P-AGH26HOMDIMThe homogeneous coordinate ring satisfies $\dim S(Y) = \dim Y + 1$uses