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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.

For a partition $\nu$ with at most $m$ parts, let $D_{\nu }(V)$ be the Schur module constructed in Question 3, equivalently the image of a Young symmetrizer on $V^{\otimes |\nu |}$. For a weakly decreasing integer tuple $\lambda$, put $k=\max (0,-{\lambda }_m)$ and $\nu =\lambda +k(1,\dots ,1)$. Define the rational Schur module byHere ${\det }^{-k}$ is the one-dimensional representation $g\mapsto \det (g{)}^{-k}$. This is rational, and it is irreducible under the permitted irreducibility assumption; for nonnegative $\lambda$ it is the original polynomial Schur module. Larger shifts give the same module, as will also follow from the character formula below.

Write $s_r(x)={\sum }_{i=1}^m{x}_i^r$ and $p_{\alpha }(x)={\prod }_{r=1}^n{s}_r(x{)}^{{\alpha }_r}$. For a permutation $\sigma$ of the given cycle type, the trace of a permuted tensor power isIn a tensor basis, the trace contracts the matrix entries of $\xi$ around each cycle of $\sigma$; a cycle of length $r$ contributes $\mathrm{tr}({\xi }^r)$. This proof applies to nondiagonalizable endomorphisms as well. The eigenvalues give $\mathrm{tr}({\xi }^r)=s_r(x)$.

The Schur–Weyl duality decomposition, on which $\sigma$ and ${\xi }^{\otimes n}$ act on the two respective factors, gives the same trace asFor a partition, the character extends polynomially to all endomorphisms because the tensor-power action does. This extension is not asserted for determinant-twisted modules at singular matrices.

We now derive the alternant character formula for the general linear group. Put $\delta =(m-1,m-2,\dots ,0)$ andUse the permitted symmetric-group character result, the Frobenius alternant character formula:It concerns characters of $S_n$, rather than assuming the character formula we seek for the general linear group. Substitute the proved trace identity and setThen ${\chi }^{\lambda }(\alpha )={\sum }_{\mu }{c}_{\lambda \mu }{\chi }^{\mu }(\alpha )$ for every conjugacy class. Independence of the irreducible symmetric-group characters forces $c_{\lambda \mu }={\delta }_{\lambda \mu }$.

Each character ${\varphi }_{\mu }$ on diagonal matrices is a symmetric homogeneous polynomial of degree $n$, since conjugation by permutation matrices permutes its arguments. Thus $a_{\delta }{\varphi }_{\mu }$ is alternating of degree $n+\left(\genfrac{}{}{0ex}{}{m}{2}\right)$. Every alternating polynomial of this degree has a unique expansion in alternants $a_{\lambda +\delta }$: a monomial with repeated exponents has zero coefficient, while each strictly decreasing nonnegative exponent vector is uniquely $\lambda +\delta$ for a partition $\lambda$ of $n$ with at most $m$ parts. Its coefficient at $x^{\lambda +\delta }$ is the coefficient of that alternant. The identities for $c_{\lambda \mu }$ therefore give $a_{\delta }{\varphi }_{\mu }=a_{\mu +\delta }$. We have derived the Weyl character formulaFor partitions the quotient is the Schur polynomial, with removable apparent singularities when eigenvalues coincide. For arbitrary dominant integer tuples, multiply the formula for $\nu$ by $(x_1\cdots x_m{)}^{-k}$; this shifts every numerator exponent by $-k$ and yields the same boxed formula. The variables must then be nonzero. The expression also shows independence of the shift used to define the determinant twist. Equality of characters identifies these irreducible modules: the group-algebra image on a direct sum of two irreducibles is finite-dimensional and semisimple, and its span of group operators detects the traces on every simple block.

Finally, the character of a dual representation evaluates the original character at $x_i^{-1}$. Reverse the numerator's columns after making this substitution. The exponents become $-{\lambda }_{m+1-j}+1-j$. Factoring $(x_1\cdots x_m{)}^{-(m-1)}$ converts these into ${\mu }_j+m-j$ for ${\mu }_j=-{\lambda }_{m+1-j}$. The denominator undergoes the identical column reversal and factor, so both signs and factors cancel. Hence ${\varphi }_{\lambda }(x^{-1})={\varphi }_{\mu }(x)$, andThis tuple is again weakly decreasing, so it is precisely the required dominant label.

A Schur module is the value of a Schur functor on a vector space. Over the complex numbers it is either zero or an irreducible polynomial representation of the general linear group. Its character is the Schur polynomial $s_{\lambda }$ in the eigenvalues, and its highest weight is $\lambda$ padded with zeros.