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The axisymmetric magnetic mode spans a one-dimensional real representation of axial rotations, so its signed amplitude is a real variable $B$. An equatorial dipole has two real horizontal components. Rotation through an angle $\vartheta$ rotates that pair; encoding the components by one complex amplitude gives $A\mapsto e^{i\vartheta }A$, and rotation through $\pi$ gives $A\mapsto -A$. The velocity mode has the same rotational representation, so write it as $V\mapsto e^{i\vartheta }V$. Magnetic reversal acts asThus the two relevant symmetries are axial rotation and magnetic reversal. Reflection symmetry has not been assumed, so complex coupling coefficients are permitted.

The equations for $B,A$ must be linear in the magnetic variables at fixed velocity. Combining this with rotational equivariance, their general local form iswhere $f$ is real valued and $g,h,j,\ell$ are complex valued. These functions can be understood through their local Taylor expansions; the symmetry argument does not require a convergent infinite series. Reality of $\dot{B}$ explains its conjugate pairing.

For the velocity equation, a monomial $B^p{A}^q(A^{\ast }{)}^r{V}^s(V^{\ast }{)}^t$ must have the same rotational weight as $V$ and be even under magnetic reversal. Its selection rules areA formal sum of all such monomials with complex coefficients is the most general local velocity evolution. At degree one there is $V$; at degree two there is $AB$; at degree three there are $V|V{|}^2,V|A{|}^2,V^{\ast }{A}^2,V{B}^2$. Expanding the magnetic equations similarly gives the equivariant three-mode dynamo normal formThe ${\mu }_i,{\omega }_i,e$ are real, and the remaining coefficients are complex. The minus signs on cubic coefficients are a convention, not an assumption about their signs. Terms such as $B^3$ and $A|A{|}^2$ are forbidden in the magnetic equations by magnetic linearity, even though symmetry alone would allow some of them.

For the normalization of nonzero dynamo coupling coefficients, assume $c_1{c}_3\ne 0$. Make the variable changesThe transformed coefficients are ${\tilde{c}}_1=qv{e}^{i({\chi }_V-{\chi }_A)}{c}_1=1$ and ${\tilde{c}}_3=(q/v)e^{i({\chi }_A-{\chi }_V)}{c}_3$, of modulus one. ThusThe remaining phase is genuine; the two complex coefficients cannot both generally be made positive real. A zero coupling cannot be normalized to one, so the printed normalization presupposes this generic nonzero-coupling case.

Now impose the specified simplification, in the normalized variables, and write $A=r{e}^{i\theta }$, $V=s{e}^{i\varphi }$ with $r,s>0$ and signed real $B$. Put $\psi =\varphi -\theta$. Separating real and imaginary parts givesConstant moduli require the first three right sides to vanish. Since $rs\ne 0$, the first equation also fixes $\cos \psi$, so a continuously evolving phase difference must be constant. Hence the last right side must vanish as well. Both phases then advance at one common frequency, giving a relative equilibrium. These four algebraic conditions and the displayed common-frequency equations govern all the requested nonzero constant-modulus solutions, without needing to solve them.

The pure-velocity branch isThis presumes $d_4\ne 0$; the usual saturated branch has ${\mu }_3,d_4>0$. Near it, set $A=e^{i({\omega }_{3}t+{\varphi }_0)}(u+iv)$, $B=b$, and define $\Delta ={\omega }_2-{\omega }_3$ and $\kappa ={\mu }_1-e{s}_0^2$. Magnetic feedback in the velocity equation is quadratic, so the first magnetic linearization isA branch of the stated relative equilibria has a zero growth rate in this rotating frame. Its determinant therefore gives the magnetic onset on a rotating velocity branch conditionFor ${\mu }_2^2+{\Delta }^2\ne 0$, the zero-mode relation is $u+iv=-c_2{s}_{0}b/({\mu }_2+i\Delta )$, so a nonzero $b$ gives a nonzero $A$ when $c_2\ne 0$. With a transverse eigenvalue crossing and nondegenerate nonlinear saturation this is the generic magnetic bifurcation surface; magnetic reversal pairs its two signs. Degenerate cases require separate treatment. For example, if ${\mu }_2=\Delta =0$ and $c_2\ne 0$, determinant zero is automatic but the stationary $u$ equation forces $b=0$, so that zero mode does not establish the requested branch with both magnetic amplitudes nonzero.

The six critical Fourier modes are at $\pm {\mathbf{k}}_1,\pm {\mathbf{k}}_2,\pm {\mathbf{k}}_3$, with ${\mathbf{k}}_1+{\mathbf{k}}_2+{\mathbf{k}}_3=0$. Reality of the pattern pairs each positive-wavevector coefficient with the conjugate coefficient at the negative wavevector. The leading critical part of the extended centre manifold parametrization is consequentlyThe displayed six-mode expression is its leading-order projection; nonlinear harmonics on the full centre manifold need not vanish identically.

A translation $\mathbf{d}$ acts by $A_i\mapsto e^{i{\mathbf{k}}_i\cdot \mathbf{d}}{A}_i$. Each monomial in the equation for $A_i$ must therefore carry wavevector ${\mathbf{k}}_i$. Since $-{\mathbf{k}}_{i+1}-{\mathbf{k}}_{i+2}={\mathbf{k}}_i$, the allowed quadratic term is $A_{i+1}^{\ast }{A}_{i+2}^{\ast }$. At cubic order the allowed terms are $A_i|A_i{|}^2,A_i|A_{i+1}{|}^2,A_i|A_{i+2}{|}^2$. Rotation through $60$ degrees acts, with one choice of orientation, asIts square cyclically permutes the amplitudes, equating the coefficients across the three equations. Its cube conjugates every amplitude; equivariance under this half-turn makes all coefficients real. It does not interchange the two cross-coupling coefficients. A reflection would do that and force $b=c$, but that symmetry is expressly absent. Thus the chiral hexagonal amplitude equations areat the retained order, with cyclic indices and real coefficients. The missing opening modulus bar in the last cubic term of the printed formula must be supplied; phase equivariance requires $|A_{i+2}{|}^2$.

Write $A_i=r_i{e}^{i{\varphi }_i}$, with all three amplitudes nonzero, and let $\Phi ={\varphi }_1+{\varphi }_2+{\varphi }_3$. Separating real and imaginary parts givesHence the phase locking of a resonant hexagonal triad obeysFor $\alpha >0$, $\Phi =0$ is phase stable and $\Phi =\pi$ is phase unstable. Two independent translation coordinates remove two phases; since the wavevectors sum to zero, the remaining phase is exactly $\Phi$. Thus a stable fully active phase-locked pattern can be represented by three positive real amplitudes after choosing the spatial origin. The statement is modulo translations, not a restriction on every representative. Boundary states must also be retained: the zero solution is stable for $\mu <0$, and rolls have two zero amplitudes. These are the necessary qualifications to a literal claim that all stable solutions are strictly positive.

In the real nonnegative subspace, a roll isor a cyclic permutation. For hexagons let $s=a+b+c$, which is positive under the assumptions. The equal-amplitude branches solve $\mu +\alpha r-s{r}^2=0$, givingwith only positive roots retained. The upper branch exists for $\mu \ge -{\alpha }^2/(4s)$; the lower branch is positive for $-{\alpha }^2/(4s)\le \mu <0$. They meet at a fold.

For a purported rectangle $(p,p,q)$ with $p,q>0$, subtract its first two equilibrium equations after dividing each by its own amplitude. The quadratic terms agree, and the difference is $(c-b)(p^2-q^2)=0$. Since $(b-a)(c-a)<0$ implies $b\ne c$, it forces $p=q$. A state with exactly two nonzero amplitudes is also impossible because the quadratic term drives the zero third amplitude. Thus there are no nontrivial rectangles.

In fact no other unequal positive equilibria are possible under the straddling cross-couplings. Here is the uniqueness of positive hexagon equilibria with straddling cross-couplings argument. Put $x_i=r_i^2$, $p=r_1{r}_2{r}_3>0$. The equilibrium equations implyConsider $b>a>c$, write $B=b-a>0$, $C=c-a<0$, and relabel cyclically so $x_1$ is maximal. If $x_2\ge x_3$, subtracting equations one and two givesThe left side is positive unless all three are equal, while the right side is nonpositive. If $x_2\le x_3$, subtract equations two and three instead:Now the left side is negative unless all are equal and the right side is nonnegative. Both cases force equality. For $c>a>b$, reverse the cyclic orientation and exchange $b,c$. Thus the only nonzero equilibria in this subspace are the stated rolls and hexagons.

At a roll the active radial eigenvalue is $-2\mu$. The two inactive amplitudes have real-part linearizationIts determinant is ${\mu }^2(a-c)(a-b)/a^2-{\alpha }^2\mu /a<0$. It therefore has a positive eigenvalue: the rolls are unstable for every $\mu >0$, which is stronger than the large-$\mu$ conclusion. In the complex system there is also a neutral translation phase, and the imaginary transverse block has the same determinant.

At a positive hexagon the real-amplitude Jacobian is circulant, with diagonal $d=-2a{r}^2-\alpha r$ and off-diagonals $p=\alpha r-2b{r}^2$, $q=\alpha r-2c{r}^2$. Its uniform eigenvector $(1,1,1)$ has eigenvalueThe other two eigenvectors lie in the zero-sum subspace; using the two nontrivial cube roots of unity givesFor sufficiently large $\mu$, $r\sim \sqrt{\mu /s}$, and the positive term $(b+c-2a)r^2$ dominates the quadratic-coupling term $2\alpha r$. Thus the hexagons are unstable at large $\mu$, while the uniform radial direction on the upper branch is stable. The two translation phases are neutral and the total phase has eigenvalue $-3\alpha r$, so the amplitude instability is genuine.

More precisely, the upper branch is amplitude stable between its fold and the chiral hexagon Hopf thresholdAt this threshold the conjugate pair crosses with nonzero frequency because $b\ne c$, permitting a Hopf bifurcation. The linear calculation locates the crossing; the nonlinear normal form decides the stability and direction of its periodic branch.

The dynamics are bounded in the retained model. For $S=r_1^2+r_2^2+r_3^2$ in the positive phase-locked subspace,Since $r_1{r}_2{r}_3\le (S/3{)}^{3/2}$, the negative quadratic-in-$S$ term dominates at large amplitude. For large positive $\mu$, the origin, rolls and hexagons are all unstable, so generic bounded dynamics cannot end at an asymptotically stable steady pattern. The chiral difference $b-c$ favors successive changes in dominant roll orientation, in the spirit of the Küppers–Lortz instability. Oscillatory competition or periodic switching is the natural behavior near the Hopf transition; linear stability alone does not establish a unique global limit cycle. With $\alpha =0$, coordinate planes are invariant and heteroclinic roll switching is possible. For $\alpha >0$ a missing third mode is regenerated when two others are present, so that exact boundary-plane heteroclinic picture is broken. Finally, sufficiently large $\mu$ is a conclusion of the cubic truncation; far from onset, omitted higher-order terms can change the physical pattern dynamics.

Fix $m$ and take $n\ge m$. On $[0,1]$, every factor $x-r/n$ and every factor $x$ has absolute value at most one. Replacing the factors successively and applying the triangle inequality givesUsing the supremum norm contraction in (a) and the Bernstein falling-factorial identity in (b),For $m=0,1$, the monomials are reproduced exactly. For every fixed $m$, $B_n(x^m)\to x^m$ uniformly on $[0,1]$.

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.

The coordinate ring is $\mathbb{C}[W]=\mathrm{Sym}(W^{\ast })$. The induced action is contragredient on functions:It is a group action by degree-preserving algebra automorphisms, extending the dual action on linear polynomial functions. Its polynomial invariant ring isFor a monic polynomial $f(t)={\prod }_{i=1}^d(t-r_i)$, its polynomial discriminant is $\mathrm{disc}(f)={\prod }_{i<j}(r_i-r_j{)}^2$. This is symmetric in the roots, hence polynomial in the coefficients, and is zero exactly when a root is repeated.

For the alternating group action, let $e_1,\dots ,e_n$ be the elementary symmetric polynomials and put $\Delta ={\prod }_{i<j}(X_i-X_j)$. If $p$ is $A_n$-invariant and $\tau$ is any transposition, decomposeNormality and index two of $A_n$ show that $p_{+}$ is symmetric and $p_{-}$ transforms by the sign character of $S_n$. Any alternating polynomial vanishes when $X_i=X_j$, so every $X_i-X_j$ divides it. These pairwise nonassociate prime factors have product $\Delta$, so $p_{-}=\Delta q$ with $q$ symmetric. The Fundamental theorem of symmetric polynomials yieldsThe sum is direct, since a polynomial that is both symmetric and alternating is zero in characteristic zero. Also ${\Delta }^2=D(e_1,\dots ,e_n)$, where $D$ is the discriminant polynomial ofThus, more precisely than the requested quotient assertion,Surjectivity follows from the direct-sum expression. Divide any putative kernel element by the monic quadratic in $Z$; its remainder is $a(T)+Zb(T)$. The direct sum forces $a(e)=b(e)=0$, and algebraic independence of the $e_i$ gives $a=b=0$. Hence the displayed relation is the entire kernel. The argument also covers $n=2$, where $A_2$ is trivial.

Now let $G$ be finite with no nontrivial linear characters, and write $R=\mathbb{C}[W]$, $S=R^G$. The polynomial ring $R$ is a unique factorization domain. Factor a nonzero invariant $f$ in $R$; invariance permutes the associate classes of its irreducible factors and makes their exponents constant on each orbit. For an orbit $\mathcal{O}$, choose representatives and form its orbit product of polynomial factorsFor every $g$, $g\cdot q_{\mathcal{O}}=\theta (g)q_{\mathcal{O}}$ for a nonzero scalar $\theta (g)$. Applying two group elements proves that $\theta :G\to {\mathbb{C}}^{\times }$ is a homomorphism. The hypothesis forces $\theta =1$, so $q_{\mathcal{O}}\in S$.

This orbit product is prime in $S$. If it divides $ab$ with $a,b\in S$, one factor $p\in \mathcal{O}$ divides $a$ or $b$ in $R$. Invariance of that chosen polynomial makes every factor in the orbit divide it. Their product therefore divides it in $R$, and the quotient is invariant because numerator and denominator are invariant and cancellation is valid in the integral domain $R$. Thus $q_{\mathcal{O}}$ divides $a$ or $b$ in $S$. Every nonzero $f\in S$ is a scalar times a product of these prime orbit products, and the only units of $S$ are nonzero constants, as they are units in $R$. Therefore $\mathbb{C}[W{]}^G$ is a unique factorization domain. This gives a direct proof, without assuming a divisor-class-group theorem.

For a failure in characteristic zero, let the cyclic group of order two act on ${\mathbb{C}}^2$ by $-I$. Its coordinate invariants areEvery invariant monomial has even total degree: its exponents are either both even, or both odd, giving the indicated generators. Reducing powers of $b$ to at most one shows that $ac-b^2$ is the only relation, since $a^i{c}^j$ and $b{a}^i{c}^j$ map to distinct monomials. The three quadratic invariants are irreducible in $S$: each nonconstant invariant has degree at least two, so a product of two nonunits has degree at least four. They are pairwise nonassociate, butgives two different irreducible factorizations. Hence this invariant ring is not a unique factorization domain. The nontrivial sign character is precisely the kind of character excluded in the preceding theorem.

Let $A=\mathbb{C}[x,y,z]$. This polynomial ring is a unique factorization domain of Krull dimension three. For a height-one prime in a unique factorization domain $\mathfrak{p}$, choose a nonzero element of it and an irreducible factor $q$ belonging to $\mathfrak{p}$. In a unique factorization domain, $q$ is prime, so $(q)$ is a nonzero prime ideal contained in $\mathfrak{p}$. Strict containment would give a chain $(0)⊊(q)⊊\mathfrak{p}$ of length two. ThusIf $\mathfrak{p}$ has height three, it is maximal: any strictly larger proper prime would extend a length-three chain and contradict $\dim A=3$. The Weak Hilbert Nullstellensatz over the algebraically closed field $\mathbb{C}$ then givesfor some $a,b,c\in \mathbb{C}$. This proves both requested generator assertions.

For the monomial curve with exponents three, four and five, putAll three map to zero under the given parametrization, so $J\subseteq I=\ker \phi$. Reduce any monomial using $y^2\mapsto xz$, $yz\mapsto x^3$ and $z^2\mapsto x^{2}y$. Each reduction decreases the sum of the exponents of $y,z$, so it terminates. Every polynomial has, modulo $J$, the formIts image is $A_0(t^3)+t^4{A}_1(t^3)+t^5{A}_2(t^3)$. These three terms occupy different exponent classes modulo three, so they cannot cancel. A zero image forces each $A_i=0$. ConsequentlyThe quotient is the domain $\mathbb{C}[t^3,t^4,t^5]$, hence $I$ is a prime ideal. It is finite free over $\mathbb{C}[t^3]$, with basis of a module $1,t^4,t^5$, by the same normal-form argument. Its dimension is therefore one, using that integral extensions preserve Krull dimension. The polynomial-ring height and dimension formula givesFor the generator obstruction, let $\mathfrak{m}=(x,y,z)$. Since $I\subseteq {\mathfrak{m}}^2$, we have $\mathfrak{m}I\subseteq {\mathfrak{m}}^3$. The degree-two initial forms of $g_1,g_2,g_3$ areThey are linearly independent over $\mathbb{C}$ in ${\mathfrak{m}}^2/{\mathfrak{m}}^3$. Thus no nonzero constant linear combination of the $g_i$ belongs to $\mathfrak{m}I$, and their classes are independent in $I/\mathfrak{m}I$. Since the three generators span that quotient,Two global generators would span it by two vectors, which is impossible. This is the initial-form lower bound for ideal generators; it shows why height two does not force two generators, even though height one and height three behave as above.

A generic linear projection over an infinite field is not enough for the stated arbitrary field $k$. Instead use Noether normalization by weighted substitutions to make the defining polynomial monic in one coordinate.

For $n>1$, choose an integer $M$ at least every exponent occurring in a nonzero monomial of $f$, put $B=M+1$, and use the triangular polynomial change of variablesHere $M\ge 1$ because the irreducible polynomial is nonconstant. This is an automorphism, with inverse $y_i=x_i-x_n^{B^i}$, $y_n=x_n$. A monomial $x_1^{{\alpha }_1}\cdots x_n^{{\alpha }_n}$ contributes a highest $y_n$ powerThese weights are all distinct: the exponents are base-$B$ digits bounded by $M$. All other terms in the expansion have smaller $y_n$ degree than that monomial's weight. Therefore the highest-degree $y_n$ term in the transformed $f$ comes from exactly one original monomial and has a nonzero coefficient in $k$, independent of the other $y_i$. No assumption about the cardinality or characteristic of $k$ enters this argument.

Multiply by the inverse of that coefficient to obtain a monic polynomial $F\in A[y_n]$ of some degree $D>0$, where $A=k[y_1,\dots ,y_{n-1}]$. The same construction for $n=1$ simply rescales $f$ to be monic over $A=k$.

The monic polynomial quotient is finite free because division by $F$ gives a unique representative of degree less than $D$. Thusas $A$-modules. Uniqueness follows because any nonzero multiple of a monic $F$ has $y_n$ degree at least $D$. In particular the map $A\to A[y_n]/(F)$ is injective.

The induced morphismis finite, since its coordinate algebra is a finite module, and flat, since tensoring with a finite free module is a finite direct sum of copies of the original module and preserves exactness. This is a finite flat Noether normalization of an affine hypersurface, with the explicit projection coordinates $y_i=x_i-x_n^{B^i}$. The construction actually applies to any nonconstant polynomial, not only an irreducible one.

The general modular form, the nonuniform Frankl-Wilson theorem, can be stated as follows. Let $p$ be a prime number, and let $L\subseteq {\mathbb{F}}_p$ have $s$ elements. Suppose $\mathcal{F}\subseteq \mathcal{P}([n])$ has $|A|\mathrm{mod}p\notin L$ for every member, while $|A\cap B|\mathrm{mod}p\in L$ for any distinct members. ThenThe uniform Frankl-Wilson theorem sharpens this to $|\mathcal{F}|\le \left(\genfrac{}{}{0ex}{}{n}{s}\right)$ when all members have size $k$ and $0\le s\le \min \{k,n-k\}$. Both forms require a prime modulus and exclusion of the self-intersection residue.

For the general proof, work over the finite field ${\mathbb{F}}_p$. Associate to each member $A$ the intersection polynomialAt the characteristic vectors of sets $1_B$, this is zero if $B\ne A$, and nonzero if $B=A$. Thus these polynomials, regarded as functions on the Boolean hypercube, are linearly independent: evaluating any relation at $1_A$ isolates its coefficient. Apply multilinear reduction on the Boolean cube, replacing every positive power of a variable by that variable. Values on the Boolean hypercube remain unchanged, and the resulting multilinear polynomials have degree at most $s$. Their ambient space has a monomial basis consisting of $x_S={\prod }_{i\in S}{x}_i$ for $|S|\le s$, with dimension ${\sum }_{j=0}^{\min (s,n)}\left(\genfrac{}{}{0ex}{}{n}{j}\right)$. This proves the general bound, including $s=0$.

For completeness, obtain the uniform sharpening without dividing by factorials in a finite field. For $s\ge 1$ put $a=k\mathrm{mod}p$ and $g(x)={\sum }_i{x}_i-a$. The functions $x_{S}g$ with $|S|<s$ are linearly independent. Indeed, a relation gives $gh=0$ with $\mathrm{deg}h<s$, so $h$ is supported only on weights in $H=\{j\in [0,n]:j\equiv a(\mathrm{mod}p)\}$. With boundary weights $-1,n+1$, this set has a gap of at least $s+1$: either consecutive allowed weights differ by $p\ge s+1$, or its terminal gap does, since $n\ge k+s\ge a+s$. The alternating sum of $h$ over any interval of $s$ free coordinates vanishes by its degree. Across that gap only one endpoint level can contribute, forcing $h$ to vanish there. Delete that level and repeat across the enlarged gap until $H$ is empty. Hence $h=0$, and independence of the square-free monomials gives the claim. Adjoin these functions to the $f_A$. Evaluation at each family vector eliminates the coefficients of $f_A$, since $g$ vanishes there; the claim eliminates all remaining coefficients. Counting dimensions yieldsThe case $s=0$ permits at most one member directly. The gap argument is an instance of the modular layer vanishing lemma.

For the final application, enumerate the distinct sets as $A_1,\dots ,A_m$ and let $v_i=1_{A_i}\in {\mathbb{R}}^n$ be their characteristic vectors of sets. If $m\le 1$, the desired bound is immediate for $n\ge 1$. Otherwise $|A_i|\ge k$, because $A_i$ intersects another member in $k$ points. At most one member can have size $k$: two distinct $k$-sets cannot have intersection size $k$. The Gram matrix of these vectors isFor real coefficients $c_i$,All terms are nonnegative, with $k>0$. If the expression vanishes, every coefficient corresponding to a set of size greater than $k$ is zero. At most one coefficient remains, and the first term forces it to be zero as well. Thus the characteristic vectors of sets are linearly independent in ${\mathbb{R}}^n$, proving the constant-intersection family boundThis proof explicitly covers the possible member of size exactly $k$; assuming all diagonal corrections were strictly positive would miss that case. Positivity of $k$ is essential: when $k=0$, the empty set and all $n$ singleton sets give $n+1$ members.

Order the index set of the open cover $\mathcal{U}=(U_a)$. The Čech cochain groups areand their differential is the alternating sum of restrictions:Terms obtained by deleting two indices cancel in pairs, so ${\delta }^2=0$. The Čech cohomology is ${\check{H}}^p=\ker \delta /\mathrm{im}\delta$, with zero incoming differential in degree zero. Its zeroth group is the group of global sections by sheaf gluing.

The acyclic cover theorem applies if every nonempty finite intersection is acyclic for $\mathcal{F}$. For a quasi-coherent sheaf on a variety it is sufficient that every such intersection is affine; on a separated variety, any affine open cover has this property. On ${\mathbb{P}}^n$, take the $n+1$ standard charts $U_i=D_{+}(X_i)$. Every intersection is a principal open in an affine chart, hence affine. The Čech complex has no terms above degree $n$, provingThis is the cohomological dimension bound from an affine cover and requires only quasi-coherence, not finite generation.

For the remaining projective calculations assume $n\ge 1$. There is a necessary zero-dimensional exception to the negative-twist assertion: ${\mathbb{P}}^0$ is a point, every twist is trivial there, and $H^0({\mathbb{P}}^0,\mathcal{O}(m))\cong k$ even for $m<0$.

Put $S=k[X_0,\dots ,X_n]$ with its usual grading. The twisting sheaf on projective space is the sheaf associated with the shifted graded module $S(m)$; on an intersection $U_I={\bigcap }_{i\in I}{U}_i$ its sections areOn $U_i$ a generator is $e_i=X_i^m$, with transition $e_j=(X_j/X_i{)}^m{e}_i$. These are regular units on overlaps and satisfy the cocycle relation, so they glue an invertible sheaf for every integer $m$, including negative $m$.

A global section is a compatible family of these homogeneous fractions, hence a single element of degree $m$ in ${\bigcap }_i{S}_{X_i}\subseteq \mathrm{Frac}S$. For $n\ge 1$, $X_0$ and $X_1$ are relatively prime in the unique factorization domain $S$, so $S_{X_0}\cap S_{X_1}=S$; therefore the full intersection is $S$. It follows thatFor $m\ge 0$ its dimension is $\left(\genfrac{}{}{0ex}{}{n+m}{n}\right)$. The proof uses all-chart compatibility; regularity on one chart alone would permit poles on its complement.