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First there is a necessary coordinate qualification. With the literal $z$ action obtained in part (c), the temporal weights are $-1,+1$ and reflection is conjugation followed by exchange. A linear term with $(\mu +i\omega )$ in both equations fails reflection equivariance for $\omega \ne 0$, since conjugation changes the sign of $i\omega$. The proposed first-component monomial ${\bar{z}}_1{z}_2^2$ also has temporal weight $+3$, rather than $-1$. Thus the displayed equations cannot be proved in those literal coordinates. They are the correct dihedral fourfold Hopf normal form in the common-phase Hopf coordinates $(q_1,q_2)=(z_1,{\bar{z}}_2)$.

In these coordinates the temporal circle imposes total phase weight one on every vector-field monomial. There are no quadratic monomials with that weight. A cubic monomial has two unbarred factors and one barred factor. For the first component, rotation by $\rho$ must multiply the monomial by $i$. Inspecting the six phase-compatible monomials leaves preciselyThe excluded three monomials multiply by $-i$ instead. Reflection swaps the components, so their coefficients must be the same after swapping indices. The linear part is a scalar complex multiple by absolute irreducibility. Orient the Hopf parameter to write it as $\mu +i\omega$, and name the cubic coefficients $-a,-b,-c$, where $a,b,c$ are generally complex. This provesThese are cubic amplitude normal forms of a dynamical system, after the usual symmetry-compatible Hopf normalization. Returning to literal $z$ coordinates conjugates the second equation: its linear coefficient is $\mu -i\omega$, its cubic coefficients are $\bar{a},\bar{b},\bar{c}$, and the first coupling is $-c{\bar{z}}_1{\bar{z}}_2^2$. That corrected form respects every printed generator.

On the three complex axial isotropy subgroup fixed spaces, substitute $q_1=R{e}^{i\Omega t}$ and $q_2=0,q_1,i{q}_1$. The radial equations are $\dot{R}=R(\mu -A_j{R}^2)$, withTherefore the leading component amplitudes areEach expression applies on the parameter side where it is positive, with its denominator nonzero for a nondegenerate branch. The frequencies are ${\Omega }_j=\omega -\mathrm{Im}(a_j)R_j^2$, with $a_j=a,a+b+c,a+b-c$. The denominators are generically distinct, so coexisting branches generically have different component amplitudes. If the amplitude is instead measured by the original real-space norm, the identity $|w_1{|}^2+|w_2{|}^2=2(|q_1{|}^2+|q_2{|}^2)$ gives squared amplitudes $2R_R^2$, $4R_A^2$, and $4R_D^2$, again generically distinct. Equalities require special coefficient relations, rather than symmetry forcing them.

Use the nonnegative Laplacian convention $\tilde{\Delta }=-{\sum }_{j=1}^{d+1}{\partial }_j^2$ and $\Delta =-\mathrm{div}\mathrm{grad}$ on the unit sphere. In polar coordinates the Euclidean Riemannian metric is $d{r}^2+r^2{g}_{S^d}$, its volume form is $r^{d}drd{V}_{S^d}$, and therefore the Laplacian in polar coordinates isConsequently the required restriction identity isFor the opposite sign convention all eigenvalues below change sign, and the restriction identity changes accordingly.

Let ${\mathcal{P}}_n$ be the vector space of degree-$n$ homogeneous polynomials in $d+1$ variables, and let ${\mathcal{H}}_n$ be its subspace of harmonic polynomials. Write $H(r\omega )=r^{n}Y(\omega )$ for $H\in {\mathcal{H}}_n$. Substitution in the Laplacian identity givesThus each restricted harmonic polynomial is a spherical harmonic with eigenvalue ${\lambda }_n=n(n+d-1)$.

To count these spherical harmonics, give polynomials the Fischer inner product $⟨x^{\alpha },x^{\beta }{⟩}_F=\alpha !{\delta }_{\alpha \beta }$. Multiplication by $x_j$ is adjoint to ${\partial }_j$, so multiplication by $r^2$ is adjoint to the ordinary signed Laplacian ${\sum }_j{\partial }_j^2$. In these finite-dimensional inner product spaces, the orthogonal complement of $r^2{\mathcal{P}}_{n-2}$ is exactly ${\mathcal{H}}_n$. Hence the Fischer decomposition isMultiplication by $r^2$ is injective, while restriction of a homogeneous polynomial to the sphere is injective: homogeneity recovers its value at every nonzero point. Counting monomials now gives the eigenvalue multiplicitywhere the second term is zero for $n<2$.

It remains to establish that these are all the eigenvalues and their full multiplicities. Iterating the Fischer decomposition expresses every restricted polynomial as a sum of the restricted harmonic polynomials. Restricted polynomials separate points and contain the constants, so the Stone-Weierstrass theorem makes their span dense in continuous functions on the sphere, hence dense in its L2 space. The Laplace-Beltrami operator is self-adjoint, and the distinct numbers ${\lambda }_n$ strictly increase for $d\ge 1$. Its different eigenspaces are therefore orthogonal. Any additional eigenfunction, or any additional vector in one of these eigenspaces orthogonal to its known spherical harmonics, would be orthogonal to a dense subspace and hence zero. The standard compact elliptic spectral theorem then shows that the spectrum of the round sphere is precisely ${\lambda }_n=n(n+d-1)$ with multiplicity $m_n$. For $S^1$ this gives multiplicity one at zero and two at $n^2$ for $n\ge 1$. If the degenerate case $d=0$ is included, $S^0$ has two points and its Laplacian is zero on a two-dimensional function space.

The monomials $x^i{y}^j$ with $i,j\ge 0$ and $i+j\le n$ form a basis of the given vector space. There are $k+1$ of total degree $k$, soOn that basis the linear operator is diagonal:Both summands of the eigenvalue are nonnegative integers. Their sum is zero precisely when $i,j\in \{0,1\}$. Thus the kernel is the span of those members of $1,x,y,xy$ whose total degree is at most $n$. In particular it is $\mathrm{span}\{1,x,y\}$ for $n=1$ and $\mathrm{span}\{1,x,y,xy\}$ for $n\ge 2$; for $n=0$ it is the constants. The image of a linear map is the span of all the other permitted monomials, since each has a nonzero diagonal eigenvalue and hence is the image of its reciprocal multiple. No excluded monomial can occur in an image because its diagonal coefficient is zero.

The trace is the sum of these eigenvalues. At degrees zero and one all contributions vanish. At degree two the contributions are $2,0,2$, summing to four; at degree three they are $6,2,2,6$, summing to sixteen; at degree four they are $12,6,4,6,12$, summing to forty. ThereforeThis is the diagonal monomial action of a second-order Euler operator.

The Frankl-Wilson theorem in its uniform form says: let $p$ be a prime number, let $L\subseteq {\mathbb{F}}_p$ have size $s\le \min (k,n-k)$, and let $\mathcal{F}\subseteq \left(\genfrac{}{}{0ex}{}{[n]}{k}\right)$ have $k\mathrm{mod}p\notin L$, while $|A\cap B|\mathrm{mod}p\in L$ for every distinct pair of members. Then

The modular-intersection, nonuniform form of the Frankl-Wilson theorem is as follows. Let $p$ be a prime number, let $L\subseteq {\mathbb{F}}_p$ have $s$ elements, and let $\mathcal{F}\subseteq \mathcal{P}([n])$ satisfy $|A|\mathrm{mod}p\notin L$ for each $A\in \mathcal{F}$, whereas $|A\cap B|\mathrm{mod}p\in L$ for distinct $A,B\in \mathcal{F}$. ThenThis nonuniform Frankl-Wilson theorem allows different sizes and different diagonal residues; in particular it applies when all sizes have one common residue outside $L$.

To prove it by the polynomial method in combinatorics, work over the finite field ${\mathbb{F}}_p$ and associate to each member the intersection polynomialReplace each positive power of $x_i$ in its expansion by $x_i$ to obtain its multilinear reduction on the Boolean cube ${\tilde{P}}_A$. This preserves evaluation on every characteristic vector, whose coordinates are zero or one, and leaves degree at most $s$. At the characteristic vector $1_B$,It is zero for distinct $A,B$, and nonzero for $B=A$ by the diagonal hypothesis. If ${\sum }_{A\in \mathcal{F}}{c}_A{\tilde{P}}_A=0$, evaluating at each $1_B$ forces $c_B=0$. Thus the multilinear polynomials are linearly independent. They belong to the vector space with basis the square-free monomials ${\prod }_{i\in S}{x}_i$ for $|S|\le s$, whose dimension is the displayed sum. This proves the nonuniform form, including $s=0$, where distinct members are impossible and the empty product is one.

For the sharp uniform form, we need the low-degree evaluation rank on a uniform layer. Form the integer matrix $M$ whose rows are indexed by $S\subseteq [n]$ with $|S|\le s$, columns by $k$-subsets $T$, and entries $M_{S,T}=1_{S\subseteq T}$. These are evaluations of the square-free monomials. Over the rational numbers, the row of a $j$-subset $S$ satisfiesIndeed, when $S\subseteq T$ the left side counts the $s$-subsets between $S$ and $T$, and otherwise both sides vanish. Since $s\le k$, the coefficient is a positive integer. Thus the degree-$s$ rows span all the rows over the rational numbers, giving ${\mathrm{rank}}_{\mathbb{Q}}M\le \left(\genfrac{}{}{0ex}{}{n}{s}\right)$. Every square submatrix of larger size has integer determinant zero, so its determinant remains zero modulo $p$. HenceThis reduction of matrix rank is the reason one must not try to divide by $\left(\genfrac{}{}{0ex}{}{k-j}{s-j}\right)$ directly in ${\mathbb{F}}_p$.

The evaluations of the ${\tilde{P}}_A$ on the $k$th uniform layer of the Boolean cube lie in the row linear span of $M$ over ${\mathbb{F}}_p$. They are linearly independent as functions there, because restriction to the columns corresponding to $\mathcal{F}$ still gives the nonzero diagonal evaluation matrix. Their number is consequently at most $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$. This completes the proof of the uniform Frankl-Wilson theorem as well.

For the odd-cardinality problem the sharper Oddtown theorem follows directly from linear algebra. Let $v_A=1_A\in {\mathbb{F}}_2^n$ be the characteristic vector. The parity conditions sayFor any linear combination ${\sum }_A{c}_A{v}_A=0$, take its dot product with $v_B$. All off-diagonal terms vanish, leaving $c_B=0$. Hence these vectors are linearly independent in an $n$-dimensional vector space, givingThe singleton sets show that the bound can be attained. Notice why the constant term in the general nonuniform Frankl-Wilson theorem is unnecessary here: these dot products already give a diagonal evaluation matrix using only the $n$ coordinate functions.

For even $n$, partition $[n]$ into the pairs $\{1,2\},\{3,4\},\dots ,\{n-1,n\}$ and take every union of a selection of these pairs. This gives exactly $2^{n/2}$ distinct sets, including the empty set. Each has even size; every pairwise intersection is again a union of entire pairs and therefore has even size. This attains the proposed bound.

For any set family satisfying the even-cardinality conditions, let $W$ be the linear span of its characteristic vectors over ${\mathbb{F}}_2$. Now the self-dot products vanish as well as the off-diagonal ones. By bilinearity, $u\cdot v=0$ for all $u,v\in W$, so $W\subseteq W^{\perp }$: it is a self-orthogonal binary subspace.

For completeness the standard bilinear form on ${\mathbb{F}}_2^n$ is nondegenerate, because a vector orthogonal to every coordinate vector has every coordinate zero. Thus the map $v\mapsto (w\mapsto v\cdot w)$ from ${\mathbb{F}}_2^n$ onto the dual space $W^{\ast }$ is surjective: every linear functional on $W$ extends to the whole vector space by extending a basis, and every such functional on ${\mathbb{F}}_2^n$ is a dot product with a vector. Its kernel is $W^{\perp }$. The rank-nullity theorem consequently gives $\dim W^{\perp }=n-\dim W$. The inclusion $W\subseteq W^{\perp }$ implies $\dim W\le n/2$. Finally, distinct members have distinct characteristic vectors, all in $W$, and a $k$-dimensional vector space over ${\mathbb{F}}_2$ has exactly $2^k$ elements. ThereforeThis proves the Eventown theorem and also explains the suggested dimension contradiction if the set family had more than $2^{n/2}$ members.

Put $I=(f_1,\dots ,f_r)$ and $\mathfrak{m}=(y_0,\dots ,y_n)$. If ${\mathfrak{m}}^N\subseteq I$, a projective common zero would have some coordinate $y_j\ne 0$. For $N>0$, the polynomial $y_j^N\in I$ could not vanish there; for $N=0$, the inclusion puts $1$ in $I$, also impossible. Thus the projective zero set is empty.

Conversely, over an algebraically closed field, emptiness of the projective zero set means the affine common zero set is contained in the origin. Every $y_j$ vanishes there, so the Hilbert Nullstellensatz puts $y_j$ in $\sqrt{I}$. Choose integers $e_j\ge 1$ with $y_j^{e_j}\in I$, and setEvery degree-$N$ monomial has some exponent at least the corresponding $e_j$, by the pigeonhole principle, and is consequently in $I$. Those monomials generate ${\mathfrak{m}}^N$, provingThe argument includes the unit-ideal case. This proves the Projective Nullstellensatz criterion directly from its affine counterpart.

Over a general field, this same criterion concerns geometric zeros: apply the proof after extension to an algebraic closure. Inclusion of each degree-$N$ monomial descends back to $k$, since its class in $k[y]/I$ is zero if and only if it becomes zero after a field extension; tensoring a vector space with an extension field is faithful. If instead one counts only $k$-rational projective points, the assertion is false. For example $y_0^2+y_1^2$ has no real projective zero but has the complex zero $[1:i]$, so its ideal cannot contain any power of $\mathfrak{m}$.

By part (a), work first with an affine base and its homogeneous equations. Put $S=k[y_0,\dots ,y_n]$, write $S_N$ for its finite-dimensional degree-$N$ vector space, and use $S_j=0$ for $j<0$. At each base point define the graded multiplication map for a projective fiberIts image is exactly $(I_x{)}_N$, since taking the degree-$N$ part of any expression in the homogeneous ideal discards all other degrees. Moreoverbecause degree-$N$ monomials generate ${\mathfrak{m}}^N$.

Choose monomial bases for the domain summands and codomain. Every entry of the resulting matrix is a coefficient of one of the $f_i$, hence a regular function of $x$. If $h_N=\dim S_N$, failure of surjectivity is the vanishing of all $h_N$-by-$h_N$ matrix minors. It is therefore a closed rank-defect locus $C_N$. If the matrix has fewer than $h_N$ columns, the failure is automatic and $C_N=X$, also closed. Constants and $N=0$ are covered by the same construction.

Part (c) gives $\pi (Y)={\bigcap }_{N\ge 0}{C}_N$, an intersection of closed subsets, hence closed. Apply part (a) to pass back to arbitrary $X$. Thus projection from $X\times {\mathbb{P}}^n$ sends every closed subset to a closed subset. This proves the closedness of projection from projective space without assuming the properness result being proved.

The field convention matters: a rational-point projection need not be Zariski closed. Over $\mathbb{R}$, the closed equation $y_0^2=t{y}_1^2$ in ${\mathbb{A}}^1\times {\mathbb{P}}^1$ has real-point image $[0,\infty )$. Indeed $y_1=0$ would force $y_0=0$, so a solution exists precisely when $t$ is a square in $\mathbb{R}$. This infinite proper subset of the affine line is not Zariski closed. The valid arbitrary-field theorem uses scheme or geometric images, for which the matrix argument remains valid at every residue field.

Being constant on the two-element orbits is equivalent to $f(-x,-y)=f(x,y)$. In characteristic zero, comparing monomial coefficients shows that precisely the terms of even total degree survive. Thus a vector space basis isSet $u=x^2$, $v=xy$, $w=y^2$. The two basis types are $u^a{w}^b$ and $v{u}^a{w}^b$, so they prove $R=k[u,v,w]$. Hence $R$ is a finitely generated algebra. It is an integral domain as a subring of the integral domain $k[x,y]$.

The generators satisfy $uw-v^2=0$. To prove this is the whole relation ideal, reduce any polynomial in $U,V,W$ modulo the monic relation $V^2-UW$ toIts image in $k[x,y]$ is $A(x^2,y^2)+xyB(x^2,y^2)$. The first summand has both exponents even and the second both odd; their distinct monomials are linearly independent. Thus the image vanishes only when both $A,B$ are zero. ConsequentlyThis gives the requested closed embedding and the quadratic cone invariant ring presentation.

The associated morphism of algebraic varieties is explicitlyIt is surjective. For $(u,v,w)\in Y$, if $u\ne 0$, choose $x$ with $x^2=u$ using algebraic closedness and put $y=v/x$. Then $y^2=v^2/u=w$. If $u=0$, the relation forces $v=0$; take $x=0$ and choose $y^2=w$. This covers every point of $Y$. Except over the origin, the two preimages differ by the simultaneous sign change, consistent with the orbit description.

Finally the gradient of $UW-V^2$ is $(W,-2V,U)$, so the Jacobian criterion makes every nonzero point smooth. At the origin, the maximal ideal modulo its square has the three independent classes of $U,V,W$, since the defining relation is quadratic. The tangent space there has dimension three, while $Y$ has dimension two. Therefore

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.