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The off-diagonal auxiliary integrations impose $M^i{}_j=0$ for $i\ne j$, by the Fourier representation of a Dirac delta function. After this constraint, $M=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_N)$ andThe Grassmann Gaussian integral over each off-diagonal pair gives its coefficient. Consequently the ghost determinant isThe prime omits the diagonal directions, which were excluded from the outset. The constant sign depends on the Berezin integral ordering and can be absorbed in normalization. Thus the Vandermonde determinant squared is the eigenvalue measure factor in this matrix diagonalization ghost determinant.

Up to an eigenvalue-independent constant, the remaining integral is $\int {\prod }_{i}d{\lambda }_i{e}^{-S_{\mathrm{eig}}}$, whereThe determinant supplies logarithmic eigenvalue repulsion. An ordering restriction on the eigenvalues changes only a constant factorial; no remaining eigenvalue integral needs to be performed. The off-diagonal bosonic contours and the displayed normalization are understood in the usual Fourier-delta prescription.

Take the Vandermonde determinant $\Delta ={\prod }_{i<j}(x_i-x_j)$. Its square is a symmetric polynomial, hence a polynomial in the elementary symmetric polynomials and belongs to $K$. It is nonzero because the variables are algebraically independent. A transposition of two variables fixes every $e_i$ and therefore fixes $K$, but sends $\Delta$ to $-\Delta \ne \Delta$ in characteristic zero. Thus $\boxed{\alpha =\Delta \notin K,{\alpha }^2\in K}$.

Set $L_i={\lambda }_i+k-i$, for $1\le i\le k$. These are distinct nonnegative integers with $L_1>\cdots >L_k$. Using the falling factorial, the matrix entry can be writtenWhen $k-j>L_i$, the falling factorial vanishes, agreeing with the stipulated convention for a negative factorial argument.

The column polynomials $(X{)}_{k-1},(X{)}_{k-2},\dots ,(X{)}_0$ are monic of successive descending degrees. Subtracting lower-degree combinations of columns turns them into $X^{k-1},X^{k-2},\dots ,1$ without changing the determinant. The Vandermonde determinant therefore givesThe positive sign comes from reversing the usual increasing-degree Vandermonde columns at the same time as writing the differences as $L_i-L_j$.

We now match this denominator with the hook-length formula. The hook lengths in row $i$ are preciselyTo verify the claim, the first hook in the row has length $L_i$. As the column index passes from $j$ to $j+1$, its hook length drops by $1$ plus the number of lower rows ending at column $j$. The skipped lengths are exactly $L_i-L_r$ for those rows: if ${\lambda }_r=j$, then $L_i-L_r={\lambda }_i-j+r-i$. At the right-hand end, lower rows having the same length as row $i$ account for the missing smallest lengths by the same formula. This describes all omitted values, and the remaining count is $L_i-(k-i)={\lambda }_i$.

ConsequentlyCombining the two identities with the hook-length formula proves the determinantal dimension formula

Use the Fourier transform convention $\hat{\phi }(\xi )=\int e^{-ix\cdot \xi }\phi (x)dx$ and $D_j=-i{\partial }_j$, so $\widehat{P(D)u}=P(\xi )\hat{u}$. All distribution pairings use complex bilinearity, and the formal transpose is $P(-D)$, without complex conjugation of the coefficients. If $D_j$ instead denotes ${\partial }_j$, the Fourier multiplier is $P(i\xi )$; the argument below is unchanged after this substitution.

For a nonempty compact convex set $K\subset {\mathbb{R}}^n$, let $H_K(\eta )={\sup }_{x\in K}x\cdot \eta$ be its support function. The Paley–Wiener–Schwartz theorem says that the Fourier transform is a bijection between distributions supported in $K$ and entire functions $F$ on ${\mathbb{C}}^n$ for whichfor some $C>0$ and nonnegative integer $N$. In particular, support in ${\overline{B}}_R$ is equivalent to the bound with $e^{R|\mathrm{Im}\zeta |}$. Convexity is essential in the precise support formulation: the support function of a set equals that of its convex hull.

Suppose first that $u$ is a compactly supported distribution with support in $K$. Define $F(\zeta )=⟨u,e^{-ix\cdot \zeta }⟩$, inserting any cutoff function equal to one near $K$ when viewing $u$ on test functions. A fixed cutoff and the finite order of a distribution estimate permit differentiation of the pairing in each complex variable, giving ${\partial }_{\zeta }^{\alpha }F=⟨u,(-ix{)}^{\alpha }{e}^{-ix\cdot \zeta }⟩$. The exponential function's power series converges with every required derivative on a fixed compact set; hence $F$ is an entire function. On real frequencies it agrees with the Fourier transform of a compactly supported distribution.

The exact exponential indicator needs a shrinking cutoff function, rather than one fixed outside $K$. Choose ${\chi }_{\epsilon }=1$ near $K$, with support in $K+B_{2\epsilon }$ and $\Vert {\partial }^{\alpha }{\chi }_{\epsilon }{\Vert }_{\infty }\le C_{\alpha }{\epsilon }^{-|\alpha |}$ for $0<\epsilon \le 1$. Such a cutoff is obtained by convolving the indicator of $K+B_{\epsilon }$ with a mollifier supported in $B_{\epsilon /2}$. On one fixed ball containing all these supports, the finite order of a distribution bound isThe Leibniz rule applied to $\theta ={\chi }_{\epsilon }{e}^{-ix\cdot \zeta }$ therefore givesTaking $\epsilon =(1+|\zeta |{)}^{-1}$ proves the required bound, with $N=2m$ and a fixed additional factor at most $e^2$.

Conversely, suppose $F$ is an entire function with the displayed bound. Its real restriction has polynomial growth, so its inverse Fourier transform defines a tempered distribution byRapid decay of the Schwartz function $\hat{\phi }$ makes this integral absolutely convergent and continuous in the Schwartz space topology.

To determine the support of a distribution, take a test function supported in a half-space $x\cdot \omega \ge H_K(\omega )+\delta$, where $|\omega |=1$ and $\delta >0$. The contour-shift proof of the Paley–Wiener–Schwartz theorem gives, for each $t\ge 0$,Here is the estimate that justifies both the shift and its limiting use. Since $\hat{\phi }(-\xi -it\omega )=\int e^{ix\cdot \xi }{e}^{-tx\cdot \omega }\phi (x)dx$, repeated integration by parts with $1-{\Delta }_x$ yieldsFor fixed $t$, the same estimate is uniform over the imaginary strip from $0$ to $t\omega$. Rotate coordinates so $\omega$ is the first coordinate direction, apply the Cauchy integral theorem on truncated rectangles, and let their real edges tend to infinity. If $2L>N+n$, the boundary integrals vanish and the horizontal integrals converge absolutely; thus the claimed shift is valid. Using $H_K(t\omega )=t{H}_K(\omega )$ now bounds the pairing bywhich tends to zero. A separating hyperplane exists at every point outside the compact convex set $K$. A finite partition of unity on the support of any exterior test function reduces it to such half-spaces, so $\mathrm{supp}u\subseteq K$. Fourier inversion gives uniqueness; the entire transform of this $u$ agrees with $F$ on ${\mathbb{R}}^n$ and hence everywhere, by applying the one-variable identity theorem successively in each coordinate. This completes both directions.

Now apply this theorem to compact support solvability for a constant-coefficient ordinary differential equation. If $P$ is a nonzero constant, its kernel is zero and $u=v/P$ is the unique solution. Otherwise writeThe solution space of $P(-D)\phi =0$ isIndeed these exponentials solve the equation, their initial derivative vectors have a nonzero Vandermonde determinant, and uniqueness for the order-$m$ ordinary differential equation makes them a basis. Pairing $P(D)u=v$ with any such solution proves necessity:

For sufficiency, choose $R$ with $\mathrm{supp}v\subseteq [-R,R]$ and let $V(\zeta )=⟨v,e^{-ix\zeta }⟩$. The assumed annihilation is precisely $V({\alpha }_j)=0$ for every $j$. Since all roots of a polynomial are simple, $U=V/P$ has removable singularities at every root and is an entire function. Outside one disk, $|P(\zeta )|\ge c(1+|\zeta |{)}^m$. Inside that disk, the extended $U$ is bounded. The polynomial division preservation of exponential type therefore givesThe Paley–Wiener–Schwartz theorem produces a compactly supported distribution $u$ with transform $U$, supported in the same interval. Then $\widehat{P(D)u}=PU=V$, so Fourier inversion gives $P(D)u=v$. ThusThe compactly supported solution is unique: $P\hat{u}=0$ forces the entire transform to vanish off the finitely many roots and hence everywhere. With repeated roots, the corresponding condition would also require derivatives $V^{(k)}(\alpha )=0$, or pairing with $x^k{e}^{-i\alpha x}$, up to one less than each multiplicity.

Work in the prime field ${\mathbb{F}}_p$ and set $r=p-1$. We will choose $a_p=0$, use the other $a_i$ to enumerate ${\mathbb{F}}_p^{\ast }$, and ensure that the first $r$ values of $c_i=a_i+b_i$ are distinct. The zero-sum condition will then force the last value to be the missing field element.

Consider the polynomialIt has total degree of a polynomial $r(r-1)$. Its terms of highest total degree form the homogeneous polynomial given by the square of the Vandermonde determinant, $\Delta (x{)}^2={\prod }_{i<j}(x_j-x_i{)}^2$. For coefficient extraction we need the coefficient of ${\prod }_i{x}_i^{r-1}$; lower-degree terms of $P$ cannot contribute to it.

Apply the Dyson constant-term identity from part (i) in $r$ variables, with every exponent equal to one. Pairing its factors for $i,j$ givesThe constant term is $r!$. Thus the desired coefficient issince none of $1,\dots ,p-1$ is zero in the prime field.

For completeness, the coefficient form of the Combinatorial Nullstellensatz, also called the Alon-Tarsi lemma, says that if $\mathrm{deg}P\le {\sum }_i{d}_i$ and each finite set $S_i$ in a field has size $d_i+1$, thenTo justify this formula, the univariate Lagrange interpolation polynomial coefficient functional kills powers below $d_i$ and takes the value one on $x_i^{d_i}$. Apply the product of these functionals to each monomial of $P$. Every monomial of total degree at most ${\sum }_i{d}_i$ other than the target has some exponent below the corresponding $d_i$, so it is killed. The target survives with coefficient one. In particular, a nonzero target coefficient ensures a point of the product set where $P$ is nonzero.

Use $S_i={\mathbb{F}}_p^{\ast }$ and $d_i=r-1$ for every $i$. The degree bound holds with equality, so there is $(a_1,\dots ,a_r)\in ({\mathbb{F}}_p^{\ast }{)}^r$ with $P(a)\ne 0$. Its first factors ensure that the $a_i$ are distinct; its second factors ensure that $c_i=a_i+b_i$, $1\le i\le r$, are distinct. With $a_p=0$, the $a_i$ enumerate the whole prime field.

Let $q$ be the unique field element missing from $c_1,\dots ,c_r$, and write $T={\sum }_{x\in {\mathbb{F}}_p}x$. Since ${\sum }_i{b}_i=0$,The same sum is $T-q$, hence $q=b_p$. Taking $c_p=b_p=a_p+b_p$ completes the enumeration. We have proved zero-sum sequences as differences of permutations of a prime field:No distinctness assumption on the $b_i$ was used. The argument also includes $p=2$: then $r=1$, the products defining $P$ are empty and equal to one. Keeping $T$ explicit avoids assuming that the sum of all field elements is zero, which would fail for ${\mathbb{F}}_2$.

The conjugates of $\omega$ are $\omega ,ldots,{\omega }^{p-1}$. Fromwe obtain, at every nontrivial $p$th root $\alpha$,The product of the roots is one and ${\prod }_{a=1}^{p-1}(1-{\omega }^a)={\Phi }_p(1)=p$. The derivative formula for a polynomial discriminant therefore yields the discriminant of a prime cyclotomic power basis:

Since $p\equiv 1(\mathrm{mod}4)$, the sign is positive. The Vandermonde determinantbelongs to $L$ and satisfies ${\Delta }^2=p^{p-2}$. Hence

HereThe L1 norm and uniform norm satisfyso the identity from the $d_{\infty }$ metric to the $d_1$ metric is one-Lipschitz and therefore continuous. The two metrics do not induce the same topology on all of $C[0,1]$. Forone has $d_1(f_n,0)=1/(2n)\to 0$ but $d_{\infty }(f_n,0)=1$. Thus $d_1$ convergence need not imply uniform convergence.

A map $F:(X,d_X)\to (Y,d_Y)$ is Lipschitz continuous if there is $L<\infty$ such thatfor all $x,x^{\prime }\in X$. Evaluation at $a\in [0,1]$ is one-Lipschitz because

Choose distinct $a_0,\dots ,a_n\in [0,1]$ and defineThe Vandermonde determinant is nonzero, so $T$ is a bijection. It is one-Lipschitz for the two uniform metrics. If ${\ell }_i$ are the associated Lagrange cardinal polynomials, thenand henceThus $T^{-1}$ is also Lipschitz.

Let ${\hat{\mathcal{P}}}_n$ be the polynomials whose values lie in $[-1,1]$. Its image under $T$ lies in the bounded cube $[-1,1{]}^{n+1}$. It is closed: if $T(p_j)\to v$, then the Lipschitz inverse gives uniform convergence $p_j\to p=T^{-1}(v)$, and passing to the limit pointwise preserves $|p(x)|\le 1$. By the Heine-Borel theorem, $T({\hat{\mathcal{P}}}_n)$ is compact, and the continuous inverse carries compactness back to ${\hat{\mathcal{P}}}_n$. Therefore

Use Shamir's secret sharing over the finite field ${\mathbb{F}}_p$. The Chair independently chooses $a_1,\dots ,a_{r-1}$ from the uniform distribution on a finite set ${\mathbb{F}}_p$ and forms the random polynomialChoose distinct nonzero $x_1,\dots ,x_s\in {\mathbb{F}}_p$, with $s$ equal to the number of discs to be issued, and inscribe $(x_i,f(x_i))$ on disc $i$.

Any $r$ discs give $r$ distinct values of a polynomial of degree at most $r-1$. Their Vandermonde determinant is nonzero, so polynomial interpolation recovers $f$ uniquely and hence recovers the secret as $N=f(0)$.

Conversely, fix any $r-1$ observed discs and any candidate secret $n\in {\mathbb{F}}_p$. Interpolating through those $r-1$ points and $(0,n)$ gives exactly one polynomial of degree at most $r-1$. Thus every candidate $n$ is compatible with the observed discs in exactly the same number of ways. Because the coefficients were chosen uniformly, the conditional distribution of the secret is unchanged after seeing any $r-1$ discs. This proves the Perfect secrecy of Shamir's threshold scheme.

Let $f(x)=a{\prod }_{i=1}^n(x-{\alpha }_i)$. Its polynomial discriminant isThe squared Vandermonde determinant is a symmetric polynomial in the roots. By the Fundamental theorem of symmetric polynomials, it is a polynomial in their elementary symmetric functions, which are the coefficients of $f$ by Vieta formulas; hence $\Delta (f)\in K$. Equivalently, the resultant formula puts it directly in $K$.

For $f=x^4+rx+s$, a resultant calculation with $f^{\prime }=4x^3+r$ gives

Let the distinct values of the faithful character $\chi$ be $a_1,\dots ,a_r$. If an irreducible character $\psi$ occurred in none of $1,\chi ,\dots ,{\chi }^{r-1}$, thenfor $0\le n<r$. The Vandermonde matrix is invertible, so every $S_j=0$. But faithfulness implies that $\chi (g)=\chi (1)$ only for $g=1$: in a unitary realization, equality of the trace with the dimension forces every eigenvalue to be one. For that fiber $S_j=\psi (1)>0$, a contradiction.

If a monic polynomial $f$ has roots ${\alpha }_1,\dots ,{\alpha }_n$ in a splitting field, its polynomial discriminant isIt is a symmetric polynomial in the roots, so it belongs to the base field; it is nonzero exactly when $f$ is a separable polynomial.

Putthe Vandermonde determinant. A root permutation $\sigma$ sends $D$ to $\mathrm{sgn}(\sigma )D$. Hence the Galois group of a polynomial is contained in the alternating group $A_n$ exactly when every Galois automorphism fixes $D$, which by the fixed-field property is equivalent to $D\in K$. This implies that $\mathrm{disc}(f)=D^2$ is a square in $K$. Conversely, if $\mathrm{disc}(f)=d^2$ for $d\in K$, then $D^2=d^2$, so $D=\pm d\in K$. The Galois group therefore fixes $D$ and consists of even permutations. This proves the discriminant criterion for an alternating Galois group.

Now letThe rational root theorem shows that none of $\pm 1,\pm 2$ is a root, so this cubic is irreducible over $\mathbb{Q}$. The discriminant of a depressed cubic isThis is not a square in $\mathbb{Q}$, so the Galois group of an irreducible cubic is $S_3$ over $\mathbb{Q}$.

Over $K=\mathbb{Q}(\sqrt{-19})$, the discriminant is the squareThe cubic remains irreducible: a root in the degree-two extension $K/\mathbb{Q}$ would generate over $\mathbb{Q}$ both a degree-three field, by irreducibility, and a subfield of a degree-two field, contradicting the tower law for field extensions. Its Galois group over $K$ is consequently $A_3\cong C_3$.

Conditioned on any $r-1$ shares in Shamir's secret sharing, each candidate secret in ${\mathbb{F}}_p$ remains equally likely. Appending $(0,N)$ to those shares determines exactly one degree-at-most-$(r-1)$ polynomial for every candidate $N$, by the nonzero Vandermonde determinant.

For three distinct times, the columns $1,t,t^2$ are linearly independent by the Vandermonde determinant. Thus a quadratic Poisson regression for the log mean can reproduce any three positive time-group means. A grouped goodness-of-fit test of that mean form has zero residual degrees of freedom. Replicated observations still allow checks of the Poisson distribution and overdispersion within each time group; saturation of the mean does not establish the response distribution.

A symmetric polynomial is unchanged by permutations of its variables. The Fundamental theorem of symmetric polynomials expresses it as a polynomial in the elementary symmetric polynomials. The square of the Vandermonde determinant is an important example.

The left side is an alternating polynomial of total degree one more than $\Delta (x)$. Dividing by the Vandermonde determinant gives a symmetric homogeneous polynomial of degree one, necessarily $a{\sum }_i{x}_i+bt$. At $t=0$, $a=1$; differentiation in $t$ and the Euler theorem for homogeneous functions give $b=\left(\genfrac{}{}{0ex}{}{m}{2}\right)$. Thus the identity holds as a polynomial identity, even at repeated coordinates. At shifted partition coordinates with $t=-1$, it proves the removable-corner recurrence in the hook-length formula.

For a nonnegative weight $w$ and monic polynomials $p_j$ of degree $j$, this determinant is ${\prod }_i\sqrt{w(x_i)}{\prod }_{i<j}(x_j-x_i)$. Its square is the product of the weights times the squared Vandermonde determinant. For normalized weighted polynomials it differs by the product of their norm factors.

Every length-$p$ sequence in the prime field ${\mathbb{F}}_p$ with sum zero, including sequences with repetitions, is the coordinatewise difference of two enumerations of that field. Apply the Combinatorial Nullstellensatz on $({\mathbb{F}}_p^{\ast }{)}^{p-1}$ to the product of the Vandermonde determinant in variables $x_i$ and its translate by the first $p-1$ sequence terms. The Dyson constant-term identity gives a nonzero top coefficient, and the zero-sum condition supplies the final missing field element. The argument includes $p=2$.