Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 162 / 3

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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