Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 114 / 3 / a / Solution

The assertion is true. The standard filtration ${\mathbb{CP}}^0\subset \cdots \subset {\mathbb{CP}}^n$ gives one cell in each dimension $0,2,\dots ,2n$. There are no odd-dimensional cells, so all cellular differentials vanish. The cellular homology theorem and the universal coefficient theorem for cohomology give one copy of $\mathbb{Z}$ in each even degree from zero to $2n$, and zero in every other degree.

Let $x$ be the Poincare dual of a projective hyperplane, with the complex orientation. It has degree two and evaluates to $+1$ on a complex projective line, so it is the positive generator of $H^2$. We use the intersection interpretation of the cup product: the product of duals of oriented submanifolds in transverse position is the dual of their oriented intersection. Distinct transverse complex hyperplanes intersect in ${\mathbb{CP}}^{n-j}$ after $j$ intersections, with positive complex orientation. Thus $x^j$ is the dual of that linear subspace.

Pairing $x^j$ with a transverse linear ${\mathbb{CP}}^j$ gives one positively oriented intersection point. Therefore $x^j$ is a primitive generator of $H^{2j}$, for every $0\le j\le n$. There is no cohomology above dimension $2n$, so $x^{n+1}=0$. These facts show that the surjective graded ring map from $\mathbb{Z}[x]$ has exactly the indicated kernel:This proves the cohomology ring of complex projective space, rather than only its additive groups. For $n=0$ it is $\mathbb{Z}$, with $x=0$.