Do isomorphic varieties have isomorphic affine cones?
Solution
1. For the standard embedding\[ \PP^1\subseteq\PP^1, \]the affine cone is\[ \widehat{\PP^1}=\AA^2. \]
The homogeneous ideal of \(\PP^1\) in its own homogeneous coordinate ring \[ \CC[s,t] \] is zero. By definition, the affine cone is the zero locus of the same homogeneous ideal in \(\AA^2\), hence all of \(\AA^2\).
2. Under the quadratic Veronese embedding\[ \nu_2:\PP^1\longrightarrow\PP^2, \qquad [s:t]\longmapsto[s^2:st:t^2], \]the image is the conic\[ C=V(XZ-Y^2). \]
Every image point satisfies \[ XZ-Y^2=s^2t^2-(st)^2=0. \] Conversely, \(\nu_2\) is the quadratic Veronese closed immersion, so it is an isomorphism of \(\PP^1\) onto its image. Since the image is an irreducible projective curve contained in the irreducible conic \(V(XZ-Y^2)\), the two coincide.
3. The affine cone over the Veronese image is\[ \widehat C = V(XZ-Y^2) \subseteq \AA^3. \]
The conic \(C\subseteq\PP^2\) is defined by the homogeneous ideal \[ (XZ-Y^2). \] The affine cone is the zero locus of that same homogeneous ideal in \(\AA^3\).
4. The cone \(\widehat C\) is singular at the origin.
Its coordinate ring is \[ A=\CC[X,Y,Z]/(XZ-Y^2). \] At the origin let \[ \mfm=(X,Y,Z)\normal A. \] The defining relation has degree \(2\), so it gives no linear relation in \[ \mfm/\mfm^2. \] Hence \[ \dim_\CC\mfm/\mfm^2=3. \] On the other hand, \(A\) is a hypersurface domain of dimension \[ 3-1=2. \] Thus the local ring \(A_\mfm\) has embedding dimension \(3\) and Krull dimension \(2\), so it is not regular. Therefore the vertex is singular.
5. The two affine cones are not isomorphic.
The affine plane \(\AA^2\) is smooth, hence all of its local rings are regular. By step 4, \(\widehat C\) has a nonregular local ring at its vertex. An isomorphism of varieties induces isomorphisms of local rings and therefore preserves regularity. Consequently \[ \AA^2\not\cong\widehat C. \]
But both projective varieties are abstractly \(\PP^1\): one is the standard copy of \(\PP^1\) and the other is its Veronese image. Hence isomorphic projective varieties can have non-isomorphic affine cones.
6. Q.E.D.
Steps <1>1–<1>3 construct the affine cones of two embeddings of \(\PP^1\), and steps <1>4–<1>5 show that they are not isomorphic. The answer is \(\boxed{\text{no}}\): the affine cone depends on the projective embedding.