Power sums form a rational but not integral stable basis

For each $r\ge 1$, setting a newly added variable to zero sends the finite rank-$N$ power sum ${\sum }_{i=1}^N{x}_i^r$ to the rank-$(N-1)$ power sum. Hence these are compatible sequences $p_r\in {\mathrm{\Lambda }}^r$ by [F1] and [F2]; the $h_r$ are already the stable homogeneous elements supplied by [F3].

In rank $N$, multiplying the geometric series ${\sum }_{a_i\ge 0}{x}_i^{a_i}{t}^{a_i}$ for $1\le i\le N$ shows from [F2] that $H_N(t):={\sum }_{k\ge 0}{h}_k(x_1,\dots ,x_N)t^k={\prod }_{i=1}^N(1-x_{i}t{)}^{-1}$. Differentiating this finite product gives $H_N^{\mathrm{\prime }}(t)\mathrm{/}{H}_N(t)={\sum }_{i=1}^N{x}_i\mathrm{/}(1-x_{i}t)={\sum }_{r\ge 1}{p}_r(x_1,\dots ,x_N)t^{r-1}$.

Write $H(t):={\sum }_{n\ge 0}{h}_n{t}^n$ in ${\mathrm{\Lambda }}_{\mathbb{Q}}\mathrm{⟦}t\mathrm{⟧}$. The finite identity of step 1.2 holds at every rank, and each coefficient of $H^{\mathrm{\prime }}(t)$, $H(t)$, and ${\sum }_{r\ge 1}{p}_r{t}^{r-1}$ is a stable homogeneous sequence by [F1] and step 1.1. Equality of every rank projection therefore gives $H^{\mathrm{\prime }}(t)=H(t){\sum }_{r\ge 1}{p}_r{t}^{r-1}$. Comparing coefficients of $t^{n-1}$ yields the Newton recurrence $n{h}_n={\sum }_{r=1}^n{p}_r{h}_{n-r}$ for every $n\ge 1$.

Since $h_0=1$, step 2.1 can be solved in either direction as $p_n=n{h}_n-{\sum }_{r=1}^{n-1}{p}_r{h}_{n-r}$ and $h_n=\frac{1}{n}(p_n+{\sum }_{r=1}^{n-1}{p}_r{h}_{n-r})$. Induction shows that each $p_n$ is a polynomial in $h_1,\dots ,h_n$ with leading term $n{h}_n$, and each $h_n$ is a polynomial over $\mathbb{Q}$ in $p_1,\dots ,p_n$ with leading term $p_n\mathrm{/}n$. Because these equations solve the same recurrence for its last unknown and each $n$ is invertible in $\mathbb{Q}$, induction verifies that the two substitutions are inverse through every finite index.

For each finite $m$, the mutually inverse triangular substitutions of step 3.1 identify $\mathbb{Q}[h_1,\dots ,h_m]$ with $\mathbb{Q}[p_1,\dots ,p_m]$. By [F3], the $h_r$ are algebraically independent over $\mathbb{Z}$ and remain so over $\mathbb{Q}$ by clearing denominators; hence the $p_r$ are algebraically independent and generate ${\mathrm{\Lambda }}_{\mathbb{Q}}$ after scalar extension, using the graded direct sum [F1]. A degree-$d$ monomial in variables of weights $\mathrm{deg}{p}_r=r$ is exactly $p_{\lambda }$ for a partition $\lambda ⊢d$ by [F4]; these monomials therefore form a $\mathbb{Q}$-basis of ${\mathrm{\Lambda }}_{\mathbb{Q}}^d$, including $p_{\mathrm{\varnothing }}=1$ at $d=0$.

At $n=1$, step 2.1 gives $h_1=p_1$; at $n=2$ it gives $2h_2=p_1{h}_1+p_2=p_1^2+p_2$, hence $h_2=\frac{1}{2}(p_{(1,1)}+p_{(2)})$. The element $h_2$ belongs to the integral stable ring by [F2] and [F3], while step 4.1 makes $p_{(1,1)}$ and $p_{(2)}$ a $\mathbb{Q}$-basis of degree two. Uniqueness of those rational coordinates and their nonintegral values show that $h_2$ is not in their $\mathbb{Z}$-span, so the power sums do not form an integral basis in degree two.