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Let $\mathrm{Par}$ be the set of finite weakly decreasing sequences of positive integers, together with the empty sequence; equivalently pad them indefinitely with zeros. The size is $|\lambda |={\sum }_i{\lambda }_i$, the length counts positive parts, and $\mathrm{Par}(n)$ consists of size-$n$ integer partitions. Their Young diagrams have cells $(i,j)$ with $1\le j\le {\lambda }_i$.

Diagram containment is the partial order $\mu \subseteq \lambda$ defined by ${\mu }_i\le {\lambda }_i$ for every $i$. Componentwise minimum and maximum give its meet and join, so this is Young's lattice. Its covers add one cell. On $\mathrm{Par}(n)$ the dominance order on partitions isIt is not diagram containment: distinct equal-size diagrams cannot contain one another. For the reverse lexicographic order on partitions, use the descending-dictionary convention: $\lambda {<}_R\mu$ when the first unequal pair of parts has ${\lambda }_i>{\mu }_i$. Thus for size four its increasing list is $(4),(3,1),(2,2),(2,1,1),(1,1,1,1)$. This is a total order opposite to increasing dictionary order on integer partitions, not the last-differing-exponent order on polynomial monomials.

Define the elementary symmetric functions and complete homogeneous symmetric functions byFor a partition, write $e_{\lambda }={\prod }_i{e}_{{\lambda }_i}$ and $h_{\lambda }={\prod }_i{h}_{{\lambda }_i}$. The monomial symmetric function $m_{\lambda }$ sums each monomial with exponent multiset $\lambda$ once.

Here is a proof of the Fundamental theorem of symmetric polynomials sufficient for the stable algebra. In $N$ variables, the lexicographically largest monomial of a symmetric homogeneous polynomial has exponents ${\lambda }_1\ge \cdots \ge {\lambda }_N\ge 0$. Otherwise exchanging an inverted pair of variables produces a larger monomial with the same coefficient. The polynomialhas that monomial as its largest one, with coefficient one. Subtract a suitable multiple and continue; there are only finitely many monomials of the fixed degree, so the procedure terminates and proves generation. Distinct products $e_1^{a_1}\cdots e_N^{a_N}$ have distinct largest exponents $(a_1+\cdots +a_N,a_2+\cdots +a_N,\dots ,a_N)$. The largest such monomial in any nonzero polynomial relation cannot cancel, proving algebraic independence. Stabilizing in each degree, where only $e_r$ with $r$ no greater than that degree can occur, gives

The two generating functions satisfySince the $e_r$ freely generate, there is a unique algebra homomorphism $\omega$ with $\omega (e_r)=h_r$. Apply it to the last identity:It follows that $\omega (h_r)=e_r$ and ${\omega }^2(e_r)=e_r$. Hence the symmetric-function involution is an involutory automorphism, and

For the forgotten coefficients, it is useful to derive the required duality explicitly. Put $p_r={\sum }_i{x}_i^r$. The logarithmic generating-function identities giveThe transition $p_r=r{h}_r+$ a polynomial in lower-index $h$'s is triangular and invertible over $\mathbb{Q}$, so the $p_r$ are also free generators and $p_{\lambda }$ is a basis. Define the Hall inner product of symmetric functions byExpand the same formal kernel in two ways:The last expansion is a reproducing kernel for this inner product. Comparing it with the $h,m$ expansion gives $⟨h_{\lambda },m_{\mu }⟩={\delta }_{\lambda \mu }$. One can see this without assuming any Schur orthogonality: pair either kernel with $h_{\nu }(y)$; the reproducing property gives $h_{\nu }(x)$, and uniqueness of its $h$ expansion proves the duality. The signs in $\omega (p_{\lambda })$ show that $\omega$ is an isometry; together with ${\omega }^2=1$ this makes it self-adjoint.

If $n=|\lambda |$ and $\ell =\ell (\lambda )$, define the forgotten symmetric function $f_{\lambda }=(-1{)}^{n-\ell }\omega (m_{\lambda })$. Its monomial coefficient is consequentlyIt remains to compute this coefficient rather than merely invoke duality. Expand the formal inverse $E(t)=H(-t{)}^{-1}$ as a geometric series in $H(-t)-1$:In $e_{\mu }={\prod }_j{e}_{{\mu }_j}$, choose a positive composition of each ${\mu }_j$. A term is $h_{\lambda }$ precisely when the concatenated list is a distinct rearrangement $\alpha$ of the parts of $\lambda$. Its boundaries at the end of each chosen composition are exactly the prescribed prefix sums of $\mu$. Conversely such a rearrangement uniquely determines the compositions by cutting at those prefix sums. Every one of its signs is $(-1{)}^{\sum {\mu }_j-\ell }=(-1{)}^{n-\ell }$, canceled by the prefactor in $a_{\lambda \mu }$. ThusIdentical parts are not artificially distinguished. For example $f_{(2,1)}=m_{(2,1)}+2m_{(3)}$: only the order $(2,1)$ refines $(2,1)$, while both orders refine the single block $(3)$.