Characteristic numbers of a product of projective planes

On $M={\mathbb{C}\mathbb{P}}^4$ the ring is $\mathbb{Z}[x]\mathrm{/}(x^5)$ with $⟨x^4,[M]⟩=1$ by [F1], so $x^5=0$ and all higher powers vanish. The total Pontryagin class is $p(TM)=(1+x^2{)}^5=1+5x^2+10x^4+10x^6+5x^8+x^{10}$, which truncates to $1+5x^2+10x^4$; hence $p_1=5x^2$, $p_2=10x^4$, and all other positive classes vanish. Therefore $p_1^2=25x^4$ evaluates to $p_1^2[M]=25⟨x^4,[M]⟩=25$ and $p_2[M]=10⟨x^4,[M]⟩=10$.

On $N={\mathbb{C}\mathbb{P}}^2\times {\mathbb{C}\mathbb{P}}^2$ write $x,y$ for the pullbacks of the generators of the two factors; by [F1] and [F2] the ring is $\mathbb{Z}[x,y]\mathrm{/}(x^3,y^3)$ with $⟨x^2{y}^2,[N]⟩=⟨x^2,[{\mathbb{C}\mathbb{P}}^2]⟩⟨y^2,[{\mathbb{C}\mathbb{P}}^2]⟩=1$, and $p(TN)=(1+x^2{)}^3\times (1+y^2{)}^3$ gives $p_1(TN)=3x^2\otimes 1+1\otimes 3y^2,p_2(TN)=9x^2\otimes y^2\mathrm{.}$ Squaring the first, $p_1^2=9x^4\otimes 1+18x^2\otimes y^2+1\otimes 9y^4$, and the outer terms vanish since $x^3=y^3=0$, while the middle term survives, so $p_1^2[N]=18⟨x^2{y}^2,[N]⟩=18$; similarly $p_2[N]=9⟨x^2{y}^2,[N]⟩=9$.

The resulting matrix with rows $M,N$ and columns $p_1^2,p_2$ is $\left(\begin{array}{cc}25& 10\\ 18& 9\end{array}\right)$, whose determinant is $25\cdot 9-10\cdot 18=225-180=45\mathrm{\ne }0$. This exhibits the nonsingularity of the ordinary degree-eight matrix predicted by the Newton comparison of [F4] and shows that the two classes are separated by their Pontryagin numbers: no nontrivial rational relation between the rows can hold.

Boundary and convention remarks. The monomials $p_1^2$ and $p_2$ are the two partitions of $2$, and monomials of any other total degree evaluate to zero by the conventions of [F3]; in particular the higher Pontryagin classes of the factors vanish by the truncation. For degree zero the point has matrix $(1)$ and the products considered here have positive dimension. The example uses the draft A-page items and the cited suppliers, which carry their own choice declarations.