The Frobenius characteristic is an isometric graded ring isomorphism

Membership $\mathrm{ch}(R(S_n))\subseteq {\mathrm{\Lambda }}^n$: for $\mu ⊢n$, [F8] gives ${\phi }^{\mu }={\sum }_{\lambda }{K}_{\lambda \mu }{\chi }^{\lambda }\in R(S_n)$ and [F5] gives $\mathrm{ch}({\phi }^{\mu })=h_{\mu }\in {\mathrm{\Lambda }}^n$. Since $(K_{\lambda \mu })$ is invertible over $\mathbb{Z}$ [F9], each ${\chi }^{\lambda }={\sum }_{\mu }(K^{-1}{)}_{\mu \lambda }{\phi }^{\mu }$ and, by linearity of $\mathrm{ch}$ [F3], $\mathrm{ch}({\chi }^{\lambda })={\sum }_{\mu }(K^{-1}{)}_{\mu \lambda }{h}_{\mu }\in {\mathrm{\Lambda }}^n$. As the ${\chi }^{\lambda }$ span $R(S_n)$ over $\mathbb{Z}$ [F10], $\mathrm{ch}(R(S_n))\subseteq {\mathrm{\Lambda }}^n$ for every $n$, hence $\mathrm{ch}(R_S)\subseteq \mathrm{\Lambda }$.

Injectivity: if $f={\sum }_n{f}_n\in R_S$ satisfies $\mathrm{ch}(f)=0$, then each degree component $\mathrm{ch}(f_n)$ vanishes by degree preservation [F3]; the isometry formula [F4] then gives $⟨f_n,f_n{⟩}_{S_n}=⟨\mathrm{ch}(f_n),\mathrm{ch}(f_n){⟩}_H=0$, and vanishing of ${\sum }_{\rho ⊢n}\mathrm{\mid }{f}_n(\rho ){\mathrm{\mid }}^2\mathrm{/}{z}_{\rho }$ forces $f_n=0$; hence $f=0$ and $\mathrm{ch}$ is injective on $R_S$.

Multiplicativity and unit: for homogeneous $f\in R(S_m)$, $g\in R(S_n)$ one has $\mathrm{ch}(f\circ g)=\mathrm{ch}(f)\mathrm{ch}(g)$ [F6]; for general $f={\sum }_m{f}_m$, $g={\sum }_n{g}_n$ bilinearity of $\circ$ [F2] and linearity of $\mathrm{ch}$ [F3] give $\mathrm{ch}(f\circ g)={\sum }_{m,n}\mathrm{ch}(f_m)\mathrm{ch}(g_n)=\mathrm{ch}(f)\mathrm{ch}(g)$. For the trivial character $e$ of $S_0$ one has $\mathrm{ch}(e)=e(\mathrm{\varnothing })p_{\mathrm{\varnothing }}\mathrm{/}{z}_{\mathrm{\varnothing }}=1$ since $e(\mathrm{\varnothing })=1$ and $z_{\mathrm{\varnothing }}=p_{\mathrm{\varnothing }}=1$.

Surjectivity onto $\mathrm{\Lambda }$: step 1.1 shows $\mathrm{ch}(R_S)\subseteq \mathrm{\Lambda }$, and $h_{\mu }=\mathrm{ch}({\phi }^{\mu })\in \mathrm{ch}(R_S)$ for every partition $\mu$ [F5]; since $\{h_{\mu }:\mu ⊢d\}$ is a $\mathbb{Z}$-basis of ${\mathrm{\Lambda }}^d$ for every $d$ [F7], the image contains a $\mathbb{Z}$-basis of $\mathrm{\Lambda }$ and therefore equals $\mathrm{\Lambda }$.

By steps 1.1, 2.1 and 1.2 the map $\mathrm{ch}:R_S\to \mathrm{\Lambda }$ is a bijective degree-preserving $\mathbb{Z}$-linear map, hence a $\mathbb{Z}$-module isomorphism; by step 1.3 it is multiplicative and sends the unit to $1$. Transporting the ring axioms of $\mathrm{\Lambda }$ along the bijection: for $f,g,h\in R_S$, associativity and commutativity of $\circ$ follow from $\mathrm{ch}((f\circ g)\circ h)=\mathrm{ch}(f)\mathrm{ch}(g)\mathrm{ch}(h)=\mathrm{ch}(f\circ (g\circ h))$ and $\mathrm{ch}(f\circ g)=\mathrm{ch}(g\circ f)$ together with injectivity of $\mathrm{ch}$, distributivity is the bilinearity of $\circ$ [F2], and $e\circ f=f$ because $\mathrm{ch}(e\circ f)=1\cdot \mathrm{ch}(f)$; so $R_S$ is a commutative graded $\mathbb{Z}$-algebra and $\mathrm{ch}$ is a graded ring isomorphism. The isometry clause is [F4].

Both $R_S$ and $\mathrm{\Lambda }$ are free $\mathbb{Z}$-modules, graded with finitely generated homogeneous components; a $\mathbb{Z}$-module isomorphism between them remains an isomorphism after tensoring with $\mathbb{Q}$ or $\mathbb{C}$, with inverse ${\mathrm{ch}}^{-1}\otimes \mathrm{i}\mathrm{d}$. Hence $\mathrm{ch}\otimes \mathbb{Q}:R_{S,\mathbb{Q}}\to {\mathrm{\Lambda }}_{\mathbb{Q}}$ and $\mathrm{ch}\otimes \mathbb{C}:R_{S,\mathbb{C}}\to {\mathrm{\Lambda }}_{\mathbb{C}}$ are isomorphisms.