Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 3 / 20F (Algebraic Geometry)

For $X\subset {\mathbb{A}}^n$ with vanishing ideal $I(X)$, the Zariski tangent space at $P$ isIt is the kernel of the Jacobian matrix of generators for the ideal, or intrinsically the dual of ${\mathfrak{m}}_P/{\mathfrak{m}}_P^2$. A point is a singular point of an algebraic variety when this tangent dimension exceeds the local dimension of the variety; otherwise it is nonsingular. For an irreducible affine variety the local dimension at a closed point equals its dimension.

Write $F=xyz+yzw+zwx+wxy$. A projective hypersurface singularity must have $F=0$ and all four first partial derivatives zero. If all coordinates were nonzero, writeOn $F=0$, differentiation gives $F_x=-xyzw/x^2\ne 0$, a contradiction. Thus some coordinate is zero; by symmetry take $w=0$. The surface equation gives $xyz=0$, so a second coordinate is zero, say $z=0$. Then $F_z=xy$, so a third coordinate is zero. The remaining four projective possibilities indeed have every first derivative zero. Therefore all singular points areOver the usual characteristic-zero ground field, each is an ordinary double point: near $[1:0:0:0]$, the affine equation has nondegenerate quadratic part $yz+zw+wy$, and symmetry treats the others.

Use the two charts of the blowup of the affine plane at the origin. In the chart $y=xv$, the total transform isso the proper transform has equation $1-x{v}^3=0$. It misses the exceptional divisor $x=0$ and is smooth, since its derivative with respect to $x$ is $-v^3\ne 0$ there.

In the chart $x=yu$, the total transform is $y^2(u^2-y)$, so the proper transform isIts equation $u^2-y=0$ has partial derivative $-1$ with respect to $y$, so it is smooth even at its unique intersection $(u,y)=(0,0)$ with the exceptional divisor. The two chart descriptions agree on their overlap. Thus one blowup makes the cusp nonsingular, although it is tangent to the exceptional divisor there.