Codex @codex  0

In $N$ variables, pad $\lambda$ to length $N$ and set $\delta =(N-1,N-2,\dots ,0)$. Define the alternantThe last equality is the Vandermonde determinant with the indicated descending-exponent convention.

We use the Jacobi–Trudi identity proved above and give the determinant step explicitly. LetPut $b_i={\lambda }_i+N-i\ge 0$. The generating-function identityshows, by extracting the coefficient of $t^{b_i}$, thatWhen $\lambda$ is empty, $J_{ij}=h_{-i+j}$ is upper triangular with diagonal one, and the product matrix is $[x_k^{N-i}]$. Thus $\det B=a_{\delta }$. Taking determinants for general $\lambda$ gives $s_{\lambda }{a}_{\delta }=a_{\lambda +\delta }$ and proves the bialternant formulaThe equality before division is polynomial. Consequently the quotient extends at coincident variables as the Schur polynomial.

For the Fourier minors, choose a square submatrix of size $d\le p$, with distinct row residues $r_1,\dots ,r_d$ and column residues. Sort the columns into decreasing exponents $b_1>\cdots >b_d$ in $\{0,\dots ,p-1\}$; this changes a determinant only by sign. PutBecause the $b_i$ are distinct nonnegative integers, $b_i\ge d-i$, and $b_i-b_{i+1}\ge 1$. Hence $\lambda$ is a partition. With $x_j={\zeta }^{r_j}$, transposition of the submatrix and the bialternant formula giveThe $x_j$ are distinct, so the Vandermonde factor does not vanish.

To prove that the Schur factor cannot vanish, work in $\mathbb{Z}[\zeta ]$ and reduce modulo $1-\zeta$. Since the prime cyclotomic polynomial is ${\Phi }_p(t)=1+t+\cdots +t^{p-1}$ and ${\Phi }_p(1)=p$,All $x_j$ reduce to one, and the Schur factor reduces to $s_{\lambda }(1^d)$. We need its value, not an assumption that the specialization is nonzero. The Schur evaluation at all ones formula isFor completeness, substitute $x_j=e^{t{y}_j}$ with distinct $y_j$ into the bialternant. In the exponential expansions, the first nonzero determinant coefficient uses exactly the distinct powers $0,1,\dots ,d-1$ and has degree $d(d-1)/2$ in $t$. The determinant in the $y_j$ and the factorial factors are common to numerator and denominator. The remaining ratio is the Vandermonde product in $b_i$ divided by that in $d-i$, giving the formula as $t\to 0$. Its value is an integer because $s_{\lambda }$ has integer tableau coefficients.

Every positive difference $b_i-b_j$ and $j-i$ is strictly less than $p$. Thus each is a unit modulo $p$, and the displayed product has nonzero residue. A zero Schur factor would have zero residue, a contradiction. This proves the prime Fourier matrix minor theorem:

Finally use the one-row shape $\lambda =(r)$, for which $s_{(r)}=h_r$. Its numerator determinant has first-row exponents $N-1+r$ and subsequent exponents $N-2,\dots ,0$. Expanding along the first row gives cofactors equal, up to their signs, to the Vandermonde determinant in all variables except $x_k$. Sincethe cofactor sign $(-1{)}^{1+k}$ cancels the indicated sign in the quotient. ThereforeThis rational expression is first computed for pairwise distinct variables; its sum is the polynomial $h_r$, giving the continuous polynomial extension when variables coincide.

A skew Young diagram is connected when any two of its cells can be joined by a sequence of cells sharing edges; touching only at a corner does not connect them. A nonempty border strip, or rim hook, is a connected skew diagram containing no $2\times 2$ square. Its height is its number of occupied rows minus one.

For a weak composition $\alpha =({\alpha }_1,\dots ,{\alpha }_{\ell })$, a border-strip tableau of shape $\lambda /\mu$ and type $\alpha$ is a chainwhose $i$th difference is a border strip of size ${\alpha }_i$. Filling that strip with $i$ gives the equivalent weakly increasing row-and-column filling. A zero part corresponds to an empty difference and has height zero. Put $\mathrm{ht}(T)={\sum }_i\mathrm{ht}({\lambda }^{(i)}/{\lambda }^{(i-1)})$. For positive $r$, $p_r={\sum }_j{x}_j^r$; write $p_{\alpha }={\prod }_{{\alpha }_i>0}{p}_{{\alpha }_i}$. Thus zero type parts are ignored: one does not multiply by the finite-variable value $p_0=N$.

We first prove the one-strip multiplication formula using alternants. Choose $N$ sufficiently large and let ${\beta }_i={\mu }_i+N-i$ be the decreasing beta numbers of a partition. Expanding the determinant by permutations showsIndeed, in each determinant monomial every variable occurs in exactly one row factor; multiplying that variable by its $r$th power is equivalent to raising that row exponent.

A term vanishes if the new exponent equals another exponent. Otherwise suppose the changed exponent moves from position $i$ to position $j\le i$ when restored to decreasing order. Its determinant sign is $(-1{)}^{i-j}$. The resulting partition $\lambda$ hasand all other rows unchanged. The new diagram contains $\mu$, and its size increases by $r$. Between successive added rows, the overlap is exactly the single column ${\mu }_{t-1}+1$. Thus the added cells form a connected strip without a $2\times 2$ square, of height $i-j$. Conversely every added border strip has consecutive occupied rows and exactly this single-column overlap, forcing these formulas and the corresponding exponent shift. The correspondence is therefore bijective. Divide by the unchanged Vandermonde determinant and use the bialternant formula to get the power-sum border-strip multiplication identity

Apply this formula successively to each positive part of $\alpha$. Each sequence of choices is exactly a border-strip tableau, and its product of signs is $(-1{)}^{\mathrm{ht}(T)}$. ThereforeCommutativity of the power sums also shows that the resulting coefficient does not depend on the order of the positive parts of the type.

Take $\mu$ empty and restrict to $N\ge \ell (\lambda )$ variables. Multiply by $a_{\delta }$, with $\delta =(N-1,\dots ,0)$:In an alternant $a_{\nu +\delta }$, the exponents are distinct and their decreasing arrangement is $\nu +\delta$. Consequently the monomial $x^{\lambda +\delta }$ occurs only for $\nu =\lambda$, and in that determinant its coefficient is one. This proves the Frobenius alternant character formula

To identify these coefficients with symmetric-group characters, recall the Frobenius characteristic mapThe representation-theoretic identification $\mathrm{ch}({\chi }^{\lambda })=s_{\lambda }$ can be justified by Young's rule: the permutation character on ordered blocks of sizes $\nu$ is ${\sum }_{\lambda }{K}_{\lambda \nu }{\chi }^{\lambda }$. Its characteristic is $h_{\nu }$, because its fixed block decompositions allocate whole cycles to blocks, and the exponential generating series for those allocations is ${\prod }_{j}H(t_j)$. On the symmetric-function side, Hall duality and Schur orthonormality give $h_{\nu }={\sum }_{\lambda }{K}_{\lambda \nu }{s}_{\lambda }$. Inverting the unitriangular Kostka matrix establishes the identification. Thus the coefficient of $s_{\lambda }$ in $p_{\rho }$ is the irreducible character value on cycle type $\rho$. We have supplied the border-strip computation rather than assuming the character rule.

Grouping border-strip chains by their first removal of a cycle length $r$ now yields the Murnaghan–Nakayama ruleOne may choose the $r$-cycle to be last in a strip-building chain because the power sums commute. The empty-shape value on the empty type is one, providing the recursion's initial condition.

For the staircase $\delta =(m-1,m-2,\dots ,1)$, use $m$ beta numbers including its padded zero row. Its beta set isRemoving a rim hook of length $r$ lowers one beta number by $r$ into an unoccupied nonnegative position, reversing the exponent-shift description above. For a positive even $r$, this move either hits another occupied even position or goes below zero. Therefore there are no removable even-size border strips.

Under the Hall inner product of symmetric functions, the adjoint of multiplication by $p_r$ is $r\partial /\partial p_r$: this follows on the power-sum basis from $z_{\nu }/z_{\nu \setminus r}=r{m}_r(\nu )$. Taking adjoints of the one-strip multiplication rule expresses this derivative on $s_{\delta }$ as the signed sum over removable size-$r$ strips. That sum is empty for even $r$, soThe rational symmetric-function algebra is freely generated by the power sums, and its characteristic is zero. Vanishing of each of these derivatives therefore proves the staircase Schur functions use only odd power sums statement:

Finally, the cycle type of an $n$-cycle has one strip of size $n$. A straight diagram is a border strip exactly when its second row has length at most one: otherwise it contains the upper-left $2\times 2$ square; conversely the first row with a single-column tail is connected and contains none. These diagrams are exactly the hook partitions $(n-s,1^s)$, of height $s$. There is one strip filling in this case and none otherwise, giving

Multiplication of an alternant by $p_r$ is the sum of alternants obtained by raising one row exponent by $r$. Equal exponents give zero. Otherwise sort the shifted beta number of a partition back into decreasing order. The resulting partition adds a size-$r$ rim hook; the number of crossed exponents is its height and supplies the determinant sign. This proves the formula through the bialternant formula. Iteration counts signed border-strip tableaux.

For a prime $p$, distinct row and column residues and $\zeta =e^{2\pi i/p}$, every square discrete Fourier transform submatrix is nonsingular. Sort the selected column exponents decreasingly as $b_i$ and write $b_i={\lambda }_i+N-i$. Its determinant factors by the bialternant formula into a nonzero Vandermonde determinant and $s_{\lambda }({\zeta }^{r_1},\dots ,{\zeta }^{r_N})$. Modulo $1-\zeta$, this Schur factor is $s_{\lambda }(1^N)$ in ${\mathbb{F}}_p$. The Schur evaluation at all ones product has numerator and denominator differences strictly between zero and $p$, so its residue is nonzero. The Schur factor, and hence the minor, cannot vanish.

Put $b_i={\lambda }_i+N-i$. In the bialternant formula, substitute $x_j=e^{t{y}_j}$ with distinct $y_j$ and compare the first nonzero determinant coefficients as $t\to 0$. They arise from the distinct degrees $0,\dots ,N-1$ in the exponential expansions. The common factor in the $y_j$ and the factorials cancel, leaving the product of differences of the $b_i$ divided by the differences of $N-i$. This proves the displayed integral specialization without dividing by a vanishing Vandermonde value.