Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 25I (Algebraic Geometry)

The Hilbert basis theorem makes $k[x_1,\dots ,x_n]$ a Noetherian ring. Reversing the inclusion of ideals shows that its Zariski-closed subsets satisfy the descending-chain condition, so every affine variety is a Noetherian topological space.

Suppose some nonempty closed subset had no finite decomposition into irreducible closed subsets. It would be reducible, so it could be writtenwith $V_1,V_2$ proper closed subsets. At least one of these would again admit no finite irreducible decomposition. Repeating that choice would produce an infinite strictly descending chain of closed subsets, contradicting Noetherianity. ThereforeRemoving any member contained in another leaves exactly the irreducible components.

A point $p$ belongs to $V(I_1+I_2)$ exactly when every polynomial $f_1+f_2$, with $f_j\in I_j$, vanishes at $p$. This implies that every member of each ideal vanishes by taking the other summand to be zero, and the converse follows by addition. HenceThis is the first vanishing loci of ideal sums and intersections identity.

Since $I_1\cap I_2\subseteq I_j$, every point of either $V(I_j)$ lies in $V(I_1\cap I_2)$. Conversely, supposeChoose $f_j\in I_j$ with $f_j(p)\ne 0$. Thenbut $(f_1{f}_2)(p)\ne 0$, so $p\notin V(I_1\cap I_2)$. Therefore

Subtracting the two defining equations givesConsequently the variety isEach member is irreducible. Indeed, after imposing its linear equation, its coordinate ring isThe invertible linear substitutionturns this intowhich is an integral domain. The two curves are distinct and neither contains the other, so they are precisely the irreducible components of two intersecting complex hyperbolas.

The three coordinate axes are the three irreducible components ofTheir common intersection is the origin. Likewise, if ${\ell }_1,{\ell }_2,{\ell }_3$ are distinct linear forms defining the three lines in ${\mathbb{A}}_k^2$, thenand its three irreducible components have only the origin in common. Any isomorphism of algebraic varieties would permute irreducible components and therefore map the origin of $V$ to the origin of $W$.

Every defining polynomial $xy,xz,yz$ has zero differential at the origin, soThe defining polynomial of $W$ is homogeneous of degree three, so its differential also vanishes at the origin, givingAn isomorphism induces an isomorphism of Zariski tangent spaces, which is impossible here. Hence $V$ and $W$ are not isomorphic. This is the tangent-space obstruction between two unions of three lines.