Power sums diagonalize the internal product

Put $u_{\rho }:=z_{\rho }{1}_{\rho }$. By [F6], $⟨u_{\rho },{\chi }^{\lambda }⟩={\chi }^{\lambda }(\rho )\in \mathbb{Z}$, and expansion in the orthonormal character basis gives $u_{\rho }={\sum }_{\lambda ⊢n}{\chi }^{\lambda }(\rho ){\chi }^{\lambda }\in R(S_n)$. Thus its characteristic $\mathrm{ch}(u_{\rho })=p_{\rho }$ from [F2] is an admissible integral input for [F3].

The indicators have pointwise product $1_{\rho }{1}_{\sigma }={\delta }_{\rho ,\sigma }{1}_{\rho }$ by [F1]. Hence $u_{\rho }{u}_{\sigma }={\delta }_{\rho ,\sigma }{z}_{\rho }{u}_{\rho }$.

Apply [F3] to the integral virtual characters from step 1.1. Linearity of the characteristic and step 2.1 give $p_{\rho }\ast p_{\sigma }=\mathrm{ch}(u_{\rho }{u}_{\sigma })={\delta }_{\rho ,\sigma }{z}_{\rho }{p}_{\rho }$. This is an identity in ${\mathrm{\Lambda }}^n$, without applying $\ast$ to a rational input.

Extend the integral bilinear product uniquely to ${\mathrm{\Lambda }}_{\mathbb{Q}}^n$ by $(f\mathrm{/}a)\ast (g\mathrm{/}b):=(f\ast g)\mathrm{/}(ab)$ for $f,g\in {\mathrm{\Lambda }}^n$ and positive integers $a,b$. Bilinearity and clearing denominators prove independence of these presentations. The rational power-sum basis supplied by [F2] then makes step 3.1 the asserted diagonalization.

The Schur-basis form is the defining expansion $s_{\lambda }\ast s_{\mu }={\sum }_{\nu }g(\lambda ,\mu ,\nu )s_{\nu }$ quoted in [F3], which is consistent with the definition through $\mathrm{ch}({\chi }^{\lambda })=s_{\lambda }$ and $\mathrm{ch}({\chi }^{\mu })=s_{\mu }$ from [F4]; the structure constants $g(\lambda ,\mu ,\nu )$ are nonnegative integers by [F5]. This proves the stated Schur expansion.