Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 21 / 2

Any isomorphism would permute irreducible components and preserve their intersections. In each configuration, the origin is the unique point lying on all three components, so an isomorphism must take origin to origin.

The coordinate ring of the spatial configuration isThe displayed ideal is indeed the intersection of the three coordinate-axis ideals. Its generators have no linear parts, so at the origin the residue classes of $x,y,z$ form a basis of $\mathfrak{m}/{\mathfrak{m}}^2$. Hence its algebraic cotangent space and Zariski tangent space have dimension three.

For the plane configuration, the coordinate ring is $B=k[s,t]/(st(s-t))$. The three linear factors are distinct in every characteristic, so this principal ideal is radical. It likewise has no linear part, but now $\mathfrak{n}/{\mathfrak{n}}^2$ has basis $s,t$ and dimension two. An isomorphism of local rings induces an isomorphism of these cotangent spaces, which is impossible:Thus the varieties are not isomorphic. This is embedding dimension distinguishes three-branch curves: matching incidence of branches does not determine their local algebra.

Parametrize the line as $\ell (t)=x+t(x^{\prime }-x)$ and set $F(t)=f(\ell (t))$, a polynomial of degree at most three. The Jacobian criterion at a singular point of a hypersurface gives vanishing of its equation and all first derivatives. ConsequentlyThus $t^2$ and $(t-1{)}^2$ both divide $F$. They are coprime in every characteristic, so their degree-four product divides a polynomial of degree at most three. It follows that $F=0$, provingThis is the secant line through singular points of a cubic hypersurface argument. If the given defining polynomial has repeated factors, use its square-free part for the reduced variety's Jacobian criterion: vanishing of that polynomial and its gradient at a singular point implies vanishing of the original polynomial's gradient there as well, by the product rule. The same restriction argument therefore still applies.