Elementary and complete families freely generate the stable ring

For each fixed $r$, specializing an added variable to zero sends the finite $e_r$ and $h_r$ to their lower-rank polynomials: terms involving that variable vanish, and the remaining subset or exponent tuples are unchanged. They therefore define homogeneous compatible sequences in ${\mathrm{\Lambda }}^r$ by [F1], [F4], and [F5].

Fix $\lambda ⊢d$ and a rank $N\ge d$. In expanding $e_{{\lambda }^{\mathrm{\prime }}}={\prod }_j{e}_{{\lambda }_j^{\mathrm{\prime }}}$, regard each factor as a column of height ${\lambda }_j^{\mathrm{\prime }}$ and record the distinct variable labels selected by [F4]. For any resulting monomial, relabel variables so its exponents are weakly decreasing and call that exponent partition $\mu$. Among the first $r$ variable labels each column contributes at most $\min (r,{\lambda }_j^{\mathrm{\prime }})$ occurrences. Hence ${\sum }_{i=1}^r{\mu }_i\le {\sum }_j\min (r,{\lambda }_j^{\mathrm{\prime }})={\sum }_{i=1}^r{\lambda }_i,$ so $\lambda ⊵\mu$ by [F3] and [F6]. Since the factors are symmetric, relabeling does not change the coefficient of the orbit sum.

Fix $n\ge 1$. At every finite rank $N\ge 0$, the coefficient of $t^n$ in [F7] gives the recurrence among the rank-$N$ components of $e_i$ and $h_{n-i}$. By [F1], [F4], [F5], and step 1.1, these are the rank-$N$ projections of the corresponding stable products. Since the recurrence holds at every rank, it is the zero sequence in ${\mathrm{\Lambda }}^n$; the constant coefficient is $e_0{h}_0=1$. Thus $E(-t)H(t)=1$ coefficientwise in the stable ring.

Consider the monomial $x_1^{{\lambda }_1}{x}_2^{{\lambda }_2}\cdots$ in rank $N\ge d$. Its first-$r$ exponent sum is ${\sum }_{i=1}^r{\lambda }_i={\sum }_j\min (r,{\lambda }_j^{\mathrm{\prime }})$ for each $r$. In a selection contributing this monomial, each column contributes at most $\min (r,{\lambda }_j^{\mathrm{\prime }})$ to that prefix, so equality of the total forces equality in every column for every $r$. A column of height $h$ must therefore select precisely labels $1,\dots ,h$; this is one selection. Hence the coefficient of the monomial, and therefore of the orbit sum $m_{\lambda }$, in $e_{{\lambda }^{\mathrm{\prime }}}$ is one.

Fix $d>0$ and project to rank $N=d$. By [F10], each stable basis element $m_{\mu }$ projects to the rank-$d$ orbit sum; [F8] identifies its distinct monomial terms, and [F9] says these projections form a $\mathbb{Z}$-basis. Thus the projection identifies stable and finite monomial coefficients. The rank-$d$ expansion of each $e_{{\lambda }^{\mathrm{\prime }}}$ therefore gives the stable transition matrix. Its entries are integers, vanish unless $\lambda ⊵\mu$ by step 1.2, and have diagonal entries one by step 2.2. Ordering the finite dominance poset by a linear extension makes this matrix unitriangular, hence invertible over $\mathbb{Z}$. The degree $d=0$ case is the single basis element $e_{\mathrm{\varnothing }}=m_{\mathrm{\varnothing }}=1$. By [F2], the $m_{\mu }$ are an integral basis; by the conjugation bijection [F3], the products $e_{\kappa }$ are also an integral basis.

Give a variable $E_r$ weight $r$. The degree-$d$ monomials in the polynomial ring $\mathbb{Z}[E_1,E_2,\dots ]$ are exactly $E_{\lambda }$ for $\lambda ⊢d$. Sending $E_r$ to $e_r$ sends these degree-$d$ monomials to the basis $e_{\lambda }$ from step 3.1, so the map is bijective in every degree. Since every polynomial and every element of $\mathrm{\Lambda }$ has finite degree support by [F1], it is an isomorphism of graded rings. Thus the $e_r$ freely generate $\mathrm{\Lambda }$.

Because the $e_r$ freely generate $\mathrm{\Lambda }$ by step 4.1, the assignment $\omega (e_r)=h_r$ extends to a graded ring homomorphism. The coefficient recurrence in step 2.1 and the same identity with $t$ replaced by $-t$ give, for each $n\ge 1$, $h_n-e_1{h}_{n-1}+e_2{h}_{n-2}-\cdots +(-1{)}^n{e}_n=0$ and $e_n-h_1{e}_{n-1}+h_2{e}_{n-2}-\cdots +(-1{)}^n{h}_n=0$. Apply $\omega$ to the first recurrence and induct on $n$, starting from $\omega (h_0)=1=e_0$. If $\omega (h_j)=e_j$ for $j<n$, the resulting equation and the second recurrence have identical terms except for $\omega (h_n)$ and $e_n$, so $\omega (h_n)=e_n$. Hence ${\omega }^2$ fixes each generator $e_n$, and so $\omega$ is an automorphism. It carries the basis $e_{\lambda }$ from step 3.1 to $h_{\lambda }$, proving that the $h_{\lambda }$ form an integral basis and that the $h_r$ are algebraically independent.