Products of complex projective spaces have an invertible Pontryagin-number matrix

Newton power-sum classes. The recurrence defines $s_m$ uniquely in $\mathbb{Z}[p_1,\dots ,p_m]$: the coefficient of $p_m$ is $(-1{)}^{m-1}m$ and the remaining terms involve only $p_1,\dots ,p_{m-1}$. Substitution of cohomology classes is therefore well defined, natural under pullback, and stable under adjoining trivial summands, since it is a finite polynomial expression in the Pontryagin classes. For a formal total class $P(t)=1+{\sum }_{a\ge 1}{p}_a{t}^a$ with $P(0)=1$ the recurrence is equivalent to the coefficient identity $t{P}^{\mathrm{\prime }}(t)P(t{)}^{-1}={\sum }_{m\ge 1}(-1{)}^{m-1}{s}_m(p)t^m$, because $t{P}^{\mathrm{\prime }}=\left({\sum }_{a\ge 1}a{p}_a{t}^a\right)$ and $P^{-1}$ is the unique inverse series. Newton's identities [F4] identify these polynomials with the power sums of the Chern roots when the $p_a$ are elementary symmetric functions, so no choice of roots or splitting space is needed. The product rule shows that the logarithmic derivative of $P_E{Q}_F$ is the sum of the logarithmic derivatives of $P_E$ and $P_F$; comparing coefficients and using the rational Whitney multiplicativity [F4] gives $s_m(E\oplus F)=s_m(E)+s_m(F)$ over $\mathbb{Q}$.

Values on projective factors. For ${\mathbb{C}\mathbb{P}}^{2j}$ the total class is $P(t)=(1+x^{2}t{)}^{2j+1}$ by [F1], a finite polynomial; differentiating, $t{P}^{\mathrm{\prime }}\mathrm{/}P=(2j+1)x^{2}t(1+x^{2}t{)}^{-1}=(2j+1){\sum }_{m\ge 1}(-1{)}^{m-1}{x}^{2m}{t}^m$, so $s_m(T{\mathbb{C}\mathbb{P}}^{2j})=(2j+1)x^{2m}$ in the truncated ring. On $P_J$, the product decomposition and external-product multiplicativity of [F5], the integral class identity of [F2], the additivity of step 1.1 and naturality give $s_m(T{P}_J)={\sum }_{b=1}^r(2j_b+1)x_b^{2m}$, where $x_b$ is the pullback of the generator of the $b$-th factor.

Refinement vanishing. Expanding $s_I={\prod }_a{s}_{i_a}$ by step 2.1, $s_I(T{P}_J)={\sum }_{\phi }{\prod }_{a=1}^t(2j_{\phi (a)}+1)x_{\phi (a)}^{2i_a},$ the sum over all assignments $\phi$ of the labeled parts of $I$ to the $r$ factors. A term depends on $\phi$ only through the block sums $n_b={\sum }_{\phi (a)=b}{i}_a$ and the multiplicities $k_b=\mathrm{\#}\{a:\phi (a)=b\}$, and equals ${\prod }_b(2j_b+1{)}^{k_b}{x}_b^{2n_b}$; since ${\sum }_b{n}_b=k={\sum }_b{j}_b$, any unequal block sums force $n_b>j_b$ for some $b$, making that term zero in the truncated ring [F1]. If all $n_b=j_b$, the degree-matched cross-product pairing [F3] evaluates the term on $[P_J]$ as ${\prod }_b(2j_b+1{)}^{k_b}$, because every factor has $⟨x_b^{2j_b},[{\mathbb{C}\mathbb{P}}^{2j_b}]⟩=1$ [F1]. Thus $B_{I,J}\mathrm{\ne }0$ requires the block sums of $I$ to be exactly the parts $j_b$ of $J$, which is the stated refinement relation; otherwise $B_{I,J}=0$.

Diagonal and triangularity. Take $I=J$, so both have $r$ parts. A contributing assignment must assign parts to factors with block sums $j_b$, and since there are $r$ parts and $r$ factors each block is nonempty; hence each factor receives exactly one part, necessarily its own $j_b$. Each of the ${\prod }_d{m}_d(J)!$ permutations of equal parts gives such an assignment and contributes the factor ${\prod }_b(2j_b+1)$ of step 2.1, so $B_{J,J}=({\prod }_d{m}_d(J)!){\prod }_b(2j_b+1)\mathrm{\ne }0$. Since proper refinement is a strict partial order, choose a total order of the partitions of $k$ extending it, with every proper refinement earlier; by step 3.1 the matrix $B$ in that order is upper triangular with the nonzero displayed diagonal, hence invertible over $\mathbb{Q}$.

Newton substitution and invertibility of $A$. The recurrence shows $s_m=(-1{)}^{m-1}m{p}_m+$ (a polynomial in $p_1,\dots ,p_{m-1}$), so recursively $p_m$ is a rational polynomial in $s_1,\dots ,s_m$ with $p_m\equiv \pm s_m\mathrm{/}m$ modulo lower weight, and the substitution is weight-preserving and triangular with nonzero diagonal in the same order. On weight $k$ the monomials in the $p$-classes correspond exactly to the partitions of $k$, so this is a finite invertible rational row transformation $C$ expressing the $s$-classes in the $p$-classes; explicitly $B_{I,J}={\sum }_K{C}_{I,K}{A}_{K,J}$ and $C$ is invertible, hence $A=C^{-1}B$ is invertible over $\mathbb{Q}$ as well.

Degree-eight and degree-zero checks. For ${\mathbb{C}\mathbb{P}}^4$, $p(T{\mathbb{C}\mathbb{P}}^4)=(1+x^2{)}^5$ gives $p_1=5x^2$, $p_2=10x^4$, so $p_1^2[{\mathbb{C}\mathbb{P}}^4]=25$ and $p_2[{\mathbb{C}\mathbb{P}}^4]=10$. For $N={\mathbb{C}\mathbb{P}}^2\times {\mathbb{C}\mathbb{P}}^2$, the class identity of [F2] and the multiplicativity of the external product [F5] give $p_1=3x^2\otimes 1+1\otimes 3y^2$ and $p_2=9x^2\otimes y^2$, hence $p_1^2=9x^4\otimes 1+18x^2\otimes y^2+1\otimes 9y^4$ with only the middle term surviving the evaluations ($x^3=y^3=0$, and $x^4,y^4$ exceed the top degrees), so $p_1^2[N]=18$ and $p_2[N]=9$; the displayed two-by-two matrices and the determinant $-45$ of $A$ follow, and the $s$-matrix row is $s_2=p_1^2-2p_2$ with values $25-20=5$ and $18-18=0$. For $k=0$ the empty partition gives the point, $A$ and $B$ are the one-by-one matrix $(1)$, and the conventions of [F3] cover the empty product. AC is used only through the cited suppliers, which carry their own declarations.