Integral cohomology ring of complex projective space

Identification: by [F1] and [F2] the projective space of ${\mathbb{C}}^{N+1}$ is the space of complex lines with the tautological bundle $\gamma$, and by [F3] the projective bundle of the trivial rank-$(N+1)$ bundle over a point is the same space with the same tautological line; hence $x=e({\gamma }_{\mathbb{R}})$ is the page's class for ${\mathbb{C}\mathbb{P}}^N$.

The sphere bundle: by [F4] the pair $(\mathrm{\ell },v)$ with $v$ a unit vector in the line $\mathrm{\ell }$ determines $v$ and is determined by it, so $S({\gamma }_{\mathbb{R}})\cong S^{2N+1}$; by [F6] and [F7] its integral cohomology is $\mathbb{Z}$ in degrees $0$ and $2N+1$ and vanishes in all other positive degrees.

Apply [F7] to the free homology groups in [F8]. It gives $H^{2k}({\mathbb{C}\mathbb{P}}^N;\mathbb{Z})\cong \mathbb{Z}$ for $0\le k\le N$ and zero cohomology in every other degree, in particular in degrees $2N+1$ and $2N+2$. The Gysin sequence [F5] of the rank-two oriented bundle ${\gamma }_{\mathbb{R}}$ reads $\cdots \to H^{i-2}({\mathbb{C}\mathbb{P}}^N)\stackrel{⌣x}{\to }{H}^i({\mathbb{C}\mathbb{P}}^N)\stackrel{p^{\ast }}{\to }{H}^i(S^{2N+1})\to \cdots$. For $2\le i\le 2N$, step 1.2 makes the middle sphere group zero, as well as the sphere group immediately preceding $H^{i-2}$; exactness therefore makes $⌣x$ an isomorphism. Starting from $H^0=\mathbb{Z}$, these isomorphisms show that $x^k$ generates $H^{2k}$ for every $0\le k\le N$. The independently established vanishing $H^{2N+2}=0$ gives $x^{N+1}=0$ without any circular appeal to the Gysin sequence.

Inclusion compatibility: the inclusion ${\mathbb{C}}^{m+1}\subseteq {\mathbb{C}}^{N+1}$ carries the tautological line of ${\mathbb{C}\mathbb{P}}^m$ to the restriction of the tautological line of ${\mathbb{C}\mathbb{P}}^N$ by [F2], so naturality of the Euler class [F3] gives $x_m={\iota }^{\ast }{x}_N$.

Ring structure: by step 2.1 the group $H^{2k}$ is the infinite cyclic group generated by $x^k$ for $0\le k\le N$, all other groups vanish, and $x^{N+1}=0$. The product satisfies $x^a\cdot x^b=x^{a+b}$, so every class is represented uniquely by a polynomial of degree at most $N$ and the graded ring is $\mathbb{Z}[x]\mathrm{/}(x^{N+1})$. With step 2.2 this is the assertion.

Boundary cases. For $N=0$ the projective space is a point, $x=0$ and the ring is $\mathbb{Z}[x]\mathrm{/}(x)=\mathbb{Z}$, as required; the sphere $S^1$ has the stated cohomology by step 1.2. For $N=1$ the computation gives $H^{\ast }(S^2)=\mathbb{Z}[x]\mathrm{/}(x^2)$, the standard result. The coefficient ring $\mathbb{Z}$ is nonzero and the Gysin sequence is used only in degrees $\le 2N+2$, all of which lie in the range controlled by step 1.2. AC enters only through [A1].