= Solution
\b[The assertion is true.] The standard filtration $\mathbb{CP}^0\subset\cdots\subset\mathbb{CP}^n$ gives one cell in each dimension $0,2,\ldots,2n$. 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 $0\leq j\leq n$. 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:
$$
\boxed{H^*(\mathbb{CP}^n;\mathbb Z)\cong\mathbb Z[x]/(x^{n+1}),\qquad |x|=2.}
$$
This proves the <cohomology ring of complex projective space>, rather than only its additive groups. For $n=0$ it is $\mathbb Z$, with $x=0$.
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