Straightforward. The forward map pads a partition of length \(\ell \leq N\) with \(N - \ell \) trailing zeros, and the inverse map strips trailing zeros. The padded tuple is weakly decreasing because the original partition is weakly decreasing and its entries are positive (hence \(\geq 0\)). The composition in both directions is the identity: padding then stripping recovers the original partition, and stripping then padding recovers the original \(N\)-partition (since an \(N\)-partition is antitone, all zeros are trailing).

The bound \(\lambda _i \leq \sum _j \lambda _j\) holds because each summand is nonnegative. The second claim uses the bound to show all parts must be zero when the size is.

The length is the cardinality of \(\{ i : \lambda _i \neq 0\} \), which is a subset of \([N]\).

Antitone is preserved under addition since each coordinate sum preserves the inequality. The size formula follows from rearranging a sum.

Sum the componentwise inequalities.

The Young diagram is the disjoint union of rows, where row \(i\) has \(\lambda _i\) cells. Thus \(|Y(\lambda )| = \sum _i \lambda _i = |\lambda |\).

By definition, row \(i\) consists of cells \((i, j)\) with \(0 \leq j {\lt} \lambda _i\), which has exactly \(\lambda _i\) elements.

Forward: if \(\mu _i \leq \nu _i\) for all \(i\), then \(j {\lt} \mu _i\) implies \(j {\lt} \nu _i\), so \((i,j) \in Y(\mu )\) implies \((i,j) \in Y(\nu )\). Backward: if \(Y(\mu ) \subseteq Y(\nu )\) and \(\mu _i {\gt} \nu _i\) for some \(i\), then \((i, \nu _i) \in Y(\mu ) \setminus Y(\nu )\), a contradiction.

The skew diagram is a subset of the Young diagram, and each row contributes \(\max (\lambda _i - \mu _i, 0)\) cells. The identity with sizes uses telescoping.

Direct computation of sums.

Sorting depends only on the multiset of values, which is invariant under permutation of the indices.

The sum of entries is preserved by sorting.

A weakly decreasing list is already sorted.

An \(N\)-partition is antitone by definition, so the result follows from Lemma 6.48.

Immediate from the definitions of monomial and product of powers.

By Lemma 6.50, \(x^a = \mathrm{monomial}(a, 1)\), and the coefficient extraction from a single monomial is immediate.

The identity \(x^{a+b} = x^a \cdot x^b\) follows from \(x_i^{a_i + b_i} = x_i^{a_i} \cdot x_i^{b_i}\) and distributivity of products.

By Lemma 6.50, this reduces to the standard result that the total degree of a monomial equals the sum of its exponents.

Forward: if the sorts are equal as \(N\)-partitions, then the sorted lists agree pointwise, hence are equal as lists, hence as multisets — and the sorted list of a multiset equals the multiset itself. Backward: if the multisets of values are equal, sorting them gives the same list.

\(\operatorname {sort}(a)\) is an \(N\)-partition, so by Lemma 6.49 it is its own sort.

\(\operatorname {sort}(\mu ) = \mu \) by Lemma 6.49, and each \(\mu _i \leq |\mu | {\lt} |\mu | + 1\) by Lemma 6.32.

For any permutation \(\sigma \in S_N\), we have \(\sigma \cdot m_\mu = \sum _{a : \operatorname {sort}(a) = \mu } x^{a \circ \sigma ^{-1}}\). Composition with \(\sigma ^{-1}\) is a bijection on \(\{ a \in \mathbb {N}^N : \operatorname {sort}(a) = \mu \} \) by Lemma 6.46, so the sum is unchanged.

The tuple \(\mu \) itself is in the sort preimage of \(\mu \) (by Lemma 6.57), and no other element of the sort preimage contributes to the coefficient of \(x^\mu \).

Every \(a\) in the sort preimage of \(\nu \) satisfies \(\operatorname {sort}(a) = \nu \). If \(a = \mu \) (as a tuple), then \(\operatorname {sort}(\mu ) = \mu \neq \nu \) (by Lemma 6.49), a contradiction. So no element of the sort preimage of \(\nu \) equals \(\mu \), hence the coefficient is \(0\).

Each monomial \(x^a\) in the sum defining \(m_\mu \) has degree \(\sum _i a_i = |\operatorname {sort}(a)| = |\mu |\) by Lemma 6.47.

This combines parts (a), (b), and (c).

We have \(e_n = \sum _{S \subseteq \{ 1,\ldots ,N\} ,\, |S|=n} \prod _{i \in S} x_i\). Each product \(\prod _{i \in S} x_i = x^{\chi _S}\) where \(\chi _S\) is the indicator function of \(S\). Every indicator function of an \(n\)-element subset sorts to \((1, \ldots , 1, 0, \ldots , 0)\) (with \(n\) ones). Conversely, every \(0\)-\(1\) tuple in the sort preimage of \((1, \ldots , 1, 0, \ldots , 0)\) is the indicator function of some \(n\)-element subset. This gives a bijection between \(n\)-element subsets and the sort preimage, proving the result.

We have \(h_n = \sum _{\substack{[a, , \in , \mathbb , {, N, }, ^, N, , \\ , , a, _, 1, , +, , \cdots , +, , a, _, N, , =, , n]}} x^a\). We partition the sum by \(\operatorname {sort}(a)\): the tuples \(a\) with \(\sum a_i = n\) are partitioned into groups indexed by \(N\)-partitions \(\mu \) of size \(n\), and each group contributes \(m_\mu \).

We have \(p_n = \sum _i x_i^n\). Each \(x_i^n\) equals \(x^{e_i \cdot n}\) where \(e_i\) is the \(i\)-th standard basis vector scaled by \(n\). All such tuples \((0, \ldots , 0, n, 0, \ldots , 0)\) sort to \((n, 0, \ldots , 0)\). When \(n {\gt} 0\), these are the only elements of the sort preimage of \((n, 0, \ldots , 0)\), because if \(\operatorname {sort}(a) = (n, 0, \ldots , 0)\) then the multiset of values of \(a\) contains exactly one \(n\) and \(N-1\) zeros, so \(a\) must be a standard basis vector times \(n\).

Direct from the definition \(p_0 = \sum _{i=1}^N x_i^0 = N\).

The zero partition has size \(0\), so each entry \(a_j\) of any \(a\) in \(\operatorname {sortPreimage}(0)\) satisfies \(a_j {\lt} 0 + 1 = 1\), forcing \(a_j = 0\). Thus the sort preimage is \(\{ (0, \ldots , 0)\} \) and \(x^0 = 1\).

The permutation action on \(f\) sends \(f(x_1, \ldots , x_N)\) to \(f(x_{\sigma (1)}, \ldots , x_{\sigma (N)})\). The coefficient of \(x^a = x_1^{a_1} \cdots x_N^{a_N}\) in \(\sigma \cdot f\) equals the coefficient of \(x^{a \circ \sigma } = x_1^{a_{\sigma (1)}} \cdots x_N^{a_{\sigma (N)}}\) in \(f\), which is exactly the stated identity.

Since \(f\) is symmetric, \(\sigma \cdot f = f\). By Proposition 6.72, the coefficient of \(x^{\sigma \cdot d}\) in \(\sigma \cdot f\) equals the coefficient of \(x^d\) in \(f\).

Equal sorts imply equal multisets of values (by Lemma 6.54), which means \(d_1\) and \(d_2\) differ by a permutation \(\sigma \). The result then follows from Lemma 6.73.

By Lemma 6.73, \([x^{\sigma \cdot d}] f = [x^d] f \neq 0\).

Unfold the definition and apply Lemma 6.46.

This combines parts (a) through (d).

Let \(f \in \mathcal{S}\). Write \(f = \sum _{d \in \mathrm{supp}(f)} [x^d] f \cdot x^d\). Partition the support by \(\operatorname {sort}(d)\): for each \(N\)-partition \(\mu \), group together all \(d\) with \(\operatorname {sort}(d) = \mu \). Since \(f\) is symmetric, all monomials in the same group have the same coefficient (by Lemma 6.74), which can be factored out. Since \(f\) is symmetric, the support is closed under permutation, so the full sort preimage of each \(\mu \) is present. The remaining sum of monomials in each group equals \(m_\mu \), showing \(f\) is in the span.

We verify that both sides have the same coefficient for each exponent \(d\). On the right-hand side, extract the coefficient of \(x^d\): using Lemmas 6.60 and 6.61, only the term with \(\lambda = \operatorname {sort}(d)\) survives, contributing \([x^{\lambda }] f\). By Lemma 6.74, \([x^{\lambda }] f = [x^d] f\) since \(\operatorname {sort}(d) = \operatorname {sort}(\lambda )\).

Extract the coefficient of \(x^\nu \) for each \(N\)-partition \(\nu \) of size \(n\); only the \(\mu = \nu \) term survives (by Lemmas 6.60 and 6.61), isolating each coefficient.

Let \(f \in \mathcal{S}_n\). Since \(f\) is symmetric, it lies in the span of all \(m_\mu \) (by Theorem 6.79). Since \(f\) is homogeneous of degree \(n\), applying the degree-\(n\) homogeneous component extracts exactly the \(m_\mu \) with \(|\mu | = n\) (by Lemma 6.62, \(m_\mu \) is homogeneous of degree \(|\mu |\), so the homogeneous component is \(m_\mu \) when \(|\mu | = n\) and \(0\) otherwise).

Combines linear independence (Theorem 6.82) and spanning (Theorem 6.83).