Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 162 (Intersection Theory)

At the generic point ${\eta }_V$ of $V$, the local ring $R={\mathcal{O}}_{W,{\eta }_V}$ is a one-dimensional Noetherian local domain. Write $r=a/b$ with nonzero $a,b\in R$ and define the order of vanishingThis is independent of the representation. The associated principal $k$-cycle iswhere only finitely many terms are nonzero.

The subgroup ${\mathrm{Rat}}_k(X)\subseteq Z_k(X)$ is generated by these cycles as $W$ ranges over integral $(k+1)$-dimensional subvarieties and $r$ over $R(W{)}^{\times }$. Two $k$-cycles are rationally equivalent when their difference lies in this subgroup, and the Chow group is

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Restriction of cycles from $X$ to $U$ is surjective: every integral subvariety of $U$ is the restriction of its closure in $X$. Its kernel on cycle groups consists exactly of cycles supported on $Z$, hence is the image of $Z_k(Z)\to Z_k(X)$.

This descends to the Chow group level. If a cycle $\alpha$ restricts to zero in $A_k(U)$, express its restriction as a sum of principal divisors on $(k+1)$-dimensional subvarieties of $U$. Taking their closures in $X$ and the same rational functions gives a rationally equivalent cycle whose difference from $\alpha$ is supported on $Z$. Thus $[\alpha ]$ lies in the image of $A_k(Z)$. Since principal divisors restrict to principal divisors, the other composite is zero. This proves the localization sequence for Chow groups

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Homotopy invariance givesor, in cohomological grading, $A^{\ast }({\mathbb{A}}^n)=\mathbb{Z}$ concentrated in degree zero.

Apply the localization sequence successively to $X_{i-1}\subseteq X_i$. Since the Chow group of each stratum is freely generated in the dimensions of its affine components, induction shows that $A_{\ast }(X_i)$ is generated by the closures of the cells appearing up to stage $i$. Hence $A_{\ast }(X)$ is generated by the classes $[V_{i,j}]$ of a cellular decomposition of a scheme.

The standard cells of ${\mathbb{P}}^r$ have one affine space in each dimension $0,\dots ,r$. Consequentlyfor $0\le k\le r$, generated by a linear ${\mathbb{P}}^k$, and zero otherwise. Equivalently, the Chow ring of projective space is $\mathbb{Z}[H]/(H^{r+1})$.

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The exceptional divisor is $E\cong {\mathbb{P}}^{r-1}$, so its Chow groups are generated by the linear subspaces $[{\Gamma }_k]$. The complement $X\setminus E$ is isomorphic through the blow-up map to ${\mathbb{P}}^r\setminus \{p\}$. The localization sequence for $\{p\}\subseteq {\mathbb{P}}^r$ shows that the latter's Chow groups are generated by the restrictions of linear spaces $H_j$ through $p$ for $1\le j\le r$; its zero-dimensional Chow group vanishes.

Under the complement isomorphism, $H_j\setminus \{p\}$ corresponds to the open part of the strict transform ${\tilde{H}}_j$. A second localization sequence, now for $E\subseteq X$, shows that $A_{\ast }(X)$ is generated by the lifts $[{\tilde{H}}_j]$ together with the images $[{\Gamma }_k]$ from $A_{\ast }(E)$, exactly as claimed.

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Let $\zeta =c_1({\mathcal{O}}_{\mathbb{P}(E)}(1))$. The projective bundle formula for Chow groups is the mapFlat pullback raises dimension by $e$, and intersection with ${\zeta }^i$ lowers it by $i$, so every summand has dimension $k$.

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Compactify the total space of $E$ as $\overline{E}=\mathbb{P}(E\oplus {\mathcal{O}}_X)$; its hyperplane at infinity is $\mathbb{P}(E)$ and its open complement is $E$. Under the projective bundle formula for $\overline{E}$, the pushforward from that hyperplane spans the summands containing positive powers of $\zeta$. The localization quotient therefore retains the zeroth summandwhose restriction to $E$ is precisely the vector-bundle pullback ${\pi }^{\ast }:A_k(X)\to A_{k+e+1}(E)$. It survives injectively. In fact this proves the stronger homotopy invariance of Chow groups: ${\pi }^{\ast }$ is an isomorphism.

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From the tautological exact sequence and the Whitney formula in the Chern class formalism,Thus $c_i(\xi )$ is a sum of terms ${\zeta }^j{p}^{\ast }{c}_{i-j}(E)$ with $0\le j\le i$. The projective-bundle pushforward satisfiesThe projection formula now makes every term vanish when $i<e$. For $i=e$, only the term ${\zeta }^e{p}^{\ast }\alpha$ survives, giving

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Write $S(\zeta )={\sum }_{j\ge 0}{\zeta }^j$. By the projective bundle formula, it is enough to verify the identity on a basis element $\beta ={\zeta }^q\cap p^{\ast }\alpha$, $0\le q\le e$. Expandand use the pushforward identities obtained from part iii. Multiplication by $c(p^{\ast }E)S(\zeta )$ is the inverse triangular operation, becauseIt recovers ${\zeta }^q{p}^{\ast }\alpha$ in dimension $k$. By linearity, every $\beta \in A_k(\mathbb{P}(E))$ satisfies

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The Euler sequence on ${\mathbb{P}}^m$ isThe Whitney product formula for the Total Chern class therefore givesin $A^{\ast }({\mathbb{P}}^m)=\mathbb{Z}[H]/(H^{m+1})$.

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If $P_X(t)=\chi (X,{\mathcal{O}}_X(t))$ is the Hilbert polynomial of a pure $k$-dimensional closed subscheme, thenEquivalently, if $H=c_1(\mathcal{O}(1))$, the degree of a projective scheme isA generic complementary linear subspace meets $X$ in a nonempty zero-dimensional scheme whose length is this number. It is therefore a positive integer.

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Let $N=N_{X/{\mathbb{P}}^m}$ be the normal bundle, of rank $m-n$. Since $TX$ is trivial, its exact sequence and the Euler-sequence calculation giveIf $m<2n$, put $i=m-n+1$, so $1\le i\le n$. Since $i$ exceeds the rank of $N$, $c_i(N)=0$, yet the formula givesIntersecting with $H^{n-i}$ and integrating over $X$ givescontradicting positivity of the degree. Hence $m\ge 2n$.

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Now $m=2n$ and $N$ has rank $n$. Write $d=\mathrm{deg}X$, so the pushforward of the fundamental class isThe self-intersection formula and the calculation of $c(N)$ giveOn the other hand, pulling back $d{H}^n$ makes the left side $d{H}^n\cap [X]$. Integrating over $X$ givesSince $d>0$, cancellation yields

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