In spherical coordinates, $z=r\cos \theta$ and $x^2+y^2=r^2{\sin }^2\theta$, so the elliptic paraboloid has equationAway from the origin this is $r=\cos \theta /{\sin }^2\theta$, with $0<\theta <\pi /2$. The undivided equation also includes the origin, where angular coordinates are not unique.

In cylindrical coordinates, $x^2+y^2={\rho }^2$, so the elliptic paraboloid is

For the contrapositive of $(b)\Rightarrow (c)$, suppose $\mathrm{char}k=p>0$. ConsiderLet $x_i$ act by multiplication and $D_i$ by formal differentiation. Differentiation preserves the defining ideal, since $D_j(x_i^{p}f)=x_i^p{D}_{j}f$ in characteristic $p$. The relations $[D_i,x_j]={\delta }_{ij}$ hold before quotienting and hence hold on $M$. Its basis consists of the monomials $x^{\alpha }$ with $0\le {\alpha }_i<p$, so ${\dim }_{k}M=p^n>0$. This is the explicit finite-dimensional module for the Weyl algebra in positive characteristic.

Therefore condition (b) fails in positive characteristic and implies characteristic zero. Combining this with the preceding two implications proves $(a)⟺(b)⟺(c)$. The same representation also has nonzero proper kernel, since the infinite-dimensional Weyl algebra cannot inject into a finite-dimensional endomorphism algebra, confirming its failure of simplicity in positive characteristic.

This part proves $(a)\Rightarrow (b)$. If $M\ne 0$ were a finite-dimensional unital module, its action would give a homomorphismThe kernel is a two-sided ideal and is proper because the identity acts as the identity on $M$. A simple ring therefore makes the homomorphism injective. This is impossible: $A_n(k)$ has the infinitely many linearly independent ordered monomials established in part (a), whereas the endomorphism algebra has dimension $({\dim }_{k}M{)}^2$. Thus simplicity excludes every nonzero finite-dimensional module.

There is also a direct characteristic-zero check: the relation $[D_1,x_1]=1$ acts as a matrix identity. Taking its trace gives $0={\dim }_{k}M$ in $k$. In characteristic zero this forces $M=0$. This agrees with the Weyl algebra has no nonzero finite-dimensional modules criterion, but the injectivity argument proves the requested implication without presupposing characteristic zero.

This part proves $(c)\Rightarrow (a)$. Write the Weyl algebra generators as $x_1,\dots ,x_n,D_1,\dots ,D_n$, with $[D_i,x_j]={\delta }_{ij}$ and the other generator commutators zero. The ordered monomials $x^{\alpha }{D}^{\beta }$ form a basis over any field: one may construct the algebra by successively adjoining the commuting derivations $D_i$ to $k[x_1,\dots ,x_n]$, with multiplication $D_{i}f=f{D}_i+\partial f/\partial x_i$, obtaining unique ordered forms. This also makes $A_n(k)$ infinite-dimensional for $n\ge 1$.

Assume $\mathrm{char}k=0$ and let $I$ be a nonzero two-sided ideal. Choose a nonzero element $a=\sum c_{\alpha \beta }{x}^{\alpha }{D}^{\beta }\in I$, and a nonzero term of maximal total degree, say $({\alpha }_0,{\beta }_0)$. The relations giveApply $(\mathrm{ad}D{)}^{{\alpha }_0}(\mathrm{ad}x{)}^{{\beta }_0}$ to $a$. Every lower-degree term vanishes. A term of the same degree survives only if all its exponents dominate $({\alpha }_0,{\beta }_0)$, which forces equality. Thus the resulting element of $I$ isThe factorials are nonzero in characteristic zero. This is a nonzero scalar, so $1\in I$ and $I=A_n(k)$. Therefore $A_n(k)$ is simple in characteristic zero. This scalar extraction by iterated Weyl commutators supplies the full simplicity proof.

Maximal right ideals of $R/J$ correspond to maximal right ideals of $R$, all of which contain $J$. Their intersection in the quotient is zero. Hence $R/J$ is a right Artinian ring with zero Jacobson radical, and part (a) makes it semisimple.

Part (b) gives the finite radical filtrationEach factor $J^i/J^{i+1}$ is annihilated by $J$, so it is a right module over $R/J$. Every module over a semisimple ring is semisimple: a free module is a direct sum of simple summands of copies of the regular module, and a quotient of a semisimple module is again semisimple, because the images of its simple summands are either zero or simple and span the quotient. A maximal independent family of these simple images spans: any simple image not contained in the sum intersects it trivially and could be added. Each factor is also Artinian, as a subquotient of $R_R$. A semisimple Artinian module has only finitely many simple summands, since an infinite direct sum would give a descending chain by successively deleting summands. Consequently every factor has finite module length.

Concatenating composition series of these finitely many factors makes $R_R$ a module of finite module length. Such a module satisfies both chain conditions, since every strict change consumes a positive amount of length. Therefore $R$ is right Noetherian. This is the Hopkins-Levitzki theorem, obtained from the two preceding results rather than invoked in their proofs.

The descending powers $J\supseteq J^2\supseteq \cdots$ stabilize because they are right ideals. Choose $m\ge 1$ with $J^m=J^{m+1}$ and put $I=J^m$. Then $I=J^{2m}=I^2$.

Suppose $I\ne 0$. Among the right ideals $K$ such that $KI\ne 0$, choose a minimal one; the set is nonempty since $RI=I\ne 0$. The right ideal $KI$ lies in $K$, and $(KI)I=K{I}^2=KI\ne 0$, so minimality forces $KI=K$. Choose $k\in K$ with $kI\ne 0$. Then $(kR)I=kI\ne 0$, and $kR\subseteq K$, so the same minimality gives $K=kR$.

Since $K=KI$ and $I\subseteq J$, its generator satisfies $k={\sum }_{\nu }k{r}_{\nu }{i}_{\nu }=kj$ for an element $j={\sum }_{\nu }{r}_{\nu }{i}_{\nu }\in J$. Thus $k(1-j)=0$. The unit criterion for the Jacobson radical makes $1-j$ invertible, implying $k=0$, contrary to $kI\ne 0$. Therefore $I=0$, andThe minimal-right-ideal proof of radical nilpotence is important here: one cannot apply the finite-generator form of Nakayama lemma directly to $J^m$ before proving it finitely generated. We made a suitable auxiliary right ideal cyclic before using the unit argument.

Use the descending chain condition on the finite intersections of maximal right ideals. It gives a minimal such intersection $K=M_1\cap \cdots \cap M_r$. Intersecting $K$ with any other maximal right ideal cannot decrease it, by minimality. Thus $K$ lies in every maximal right ideal, and $K=J(R)$. When $J(R)=0$, the diagonal mapis injective, and each summand on the right is a simple module.

For completeness, a submodule $N$ of a finite direct sum of simple modules is semisimple. Induct on the number of summands. For $S\oplus T$ with $S$ simple, $N\cap S$ is either $0$ or $S$. In the first case projection into $T$ identifies $N$ with a submodule of $T$, so induction applies. In the second case subtracting the $S$ coordinate shows $N=S\oplus (N\cap T)$, and induction again applies. This also shows finite module length. Apply it to the image of the diagonal map to obtainThis is the semisimplicity of a right Artinian ring with zero radical. It does not assume right Noetherianity in order to prove it.

Assume the coefficients are sufficiently differentiable in the region considered, with no material discontinuity or caustic at the front. Let $t=T(\mathbf{x})$ describe arrival time, $\mathbf{p}=\nabla T$ the slowness vector, and $\mathbf{n}$ the unit elastic wave propagation normal. For any field, its jump is the value immediately behind the front minus that immediately ahead. If the first nonzero jump occurs in the kth time derivative, $k\ge 2$, writeA smooth background may be added without changing these leading jumps. Since $F_j^{\prime }=F_{j-1}$, differentiation expresses spatial jumps in terms of $p_j$ and time-derivative jumps. The restriction $k\ge 2$ lets displacement field and its first time derivative remain continuous and avoids an imposed impulsive source at the front.

Insert this local expansion in $\rho {\partial }_t^2{u}_i={\partial }_j(c_{ijpq}{\partial }_p{u}_q)$. The coefficient of $F_{k-2}$ isFor isotropic elasticity the slowness acoustic matrix is $\mu |\mathbf{p}{|}^{2}I+(\lambda +\mu )\mathbf{p}{\mathbf{p}}^T$. Its longitudinal eigenvalue is $(\lambda +2\mu )|\mathbf{p}{|}^2$. The P branch therefore gives the eikonal equation and polarizationThe next coefficient, $F_{k-1}$, isThe matrix on the left is symmetric and annihilates $\mathbf{n}$. Project onto $n_i$ to obtain its solvability condition. Put $a=a^{[k]}$. The two projected terms can be evaluated without dropping gradients of mass density or moduli:These identities use $|\mathbf{n}{|}^2=1$, hence $n_q{\partial }_p{n}_q=0$, and $(\lambda +2\mu )/\alpha =\rho \alpha$. Their curvature terms cancel, leavingMultiplying by $a$ proves the requested P-wavefront discontinuity transport law,The wave amplitude here is real for a real jump; a complex wave amplitude uses the corresponding modulus-squared law. An elastic ray follows $d\mathbf{x}/ds=\mathbf{n}$, with arclength $s$ and $dt/ds=1/\alpha$. If $J$ is the infinitesimal cross-sectional area of a neighboring ray tube, $d\log J/ds=\nabla \cdot \mathbf{n}$. Thus $\rho \alpha a^{2}J$ is constant along that tube, the ray-tube conservation for a P-wave jump. This defines the geometrical spreading and shows why the wave amplitude changes with material properties even before attenuation is considered. At a caustic the simple ray-tube wave amplitude is singular and a uniform local description is needed.

Let the positive undisturbed surface tension be $\bar{\sigma }={\sigma }_0-A{\Gamma }_0>0$. An insoluble surfactant perturbation $g$ produces ${\sigma }_1=-Ag$. The tangential Marangoni stress changes the surface condition to $\mu u_z(h)={\sigma }_x$. Solving the lubrication theory momentum equation now givesAt leading order, conservation of insoluble surfactant on a moving interface reduces to ${\Gamma }_t+(\Gamma u_s{)}_x=0$. Corrections from the changing circumference are smaller by $h_0/a$. Likewise the normal-pressure contribution $-{\sigma }_1/a$ gives a coupling smaller by $h_0/a$ than the retained tangential Marangoni stress, so it is omitted at the same order.

For thickness and concentration amplitudes $H,g$ proportional to $e^{\beta t+ikx}$, put $D=a^{-2}-k^2$ and use $p_1=-\bar{\sigma }DH$. The two conservation equations becomeSet $s=k^2{a}^2$, $\lambda =h_0^3\bar{\sigma }/(3\mu a^4)$ and $\alpha =h_{0}A{\Gamma }_0/(\mu a^2)$. The uncoupled capillary growth and elastic relaxation rates are $L=\lambda s(1-s)$ and $S=\alpha s$. The off-diagonal product is $3LS/4$, so the determinant is $(\beta -L)(\beta +S)+3LS/4$. Thus the surfactant stabilization of an annular liquid lining relation isThis proves the dispersion relation before its lettered deductions. The physical assumption $\bar{\sigma }>0$ is needed for the claimed capillary instability band.

For $B_0>0$, change to logarithmic magnitude and its correctly transformed probability density function:Using ${\partial }_B=B^{-1}{\partial }_x$, the radial equation becomesThis is diffusion with constant drift. Set $y=x-3Dt$ and $g(y,t)=f(y+3Dt,t)$; then $g_t=D{g}_{yy}$. The initial Dirac delta transforms to $f(x,0)=\delta (x)$, so the heat-kernel solution givesDividing by $B$ yields the lognormal magnetic-field amplification lawExpanding the square and using $B=B_0{e}^x$ gives the equivalent requested formThe logarithm is a Gaussian random variable with mean $3Dt$ and variance $2Dt$, which proves normalization and spreading directly. The exponential moments of that Gaussian distribution give $⟨B^q⟩=B_0^q{e}^{Dq(q+3)t}$, independently checking the energy growth. The limit at $t\downarrow 0$ is the initial Dirac delta distribution, not a pointwise finite function.

Write $ϵ=l/d_i\ll 1$ and $\delta =v_A\tau /l\ll 1$. For magnetic amplitudes of order $B_0$, ion acceleration is of order $v_A^2/l$. Thus its change over the fast time isThe ions are consequently fixed to leading order on this timescale. Their first nonzero response is driven by the momentum equation, with negligible self-advection for an initially resting ion flow; one need not assert that the dimensional Lorentz acceleration literally vanishes. In fast-time, Alfvén-velocity dimensionless variables, the ion equation reads ${\partial }_{\hat{t}}\hat{\mathbf{u}}=O(\delta )$, giving ${\partial }_{\hat{t}}\hat{\mathbf{u}}=0$ at leading order.

Meanwhile the electron-current velocity is $j/(en)\sim v_A{d}_i/l=v_A/ϵ$, larger than a bulk flow of order $v_A$. The ion-advection term in induction is smaller than Hall induction by $O(ϵ)$ even if that slow flow is retained. The leading magnetic equation is thereforewith stationary ions at leading order. This is the fast-time electron-MHD limit of Hall magnetohydrodynamics and is precisely electron magnetohydrodynamics.

The already obtained fast-wave eigenvector has $u/b=|q/\omega |\sim 1/(k{d}_i)\ll 1$, confirming that ion motion cannot affect its leading frequency. The closed magnetic equation therefore suffices to obtain that branch without using momentum to close it. Its timescale is ${\tau }_w\sim 1/(v_A{d}_i|k_{\parallel }|k)$; compatibility with ${\tau }_w\ll l/v_A$ requires $|k_{\parallel }|d_i\gg 1$ when $l\sim k^{-1}$. Arbitrarily small $k_{\parallel }$ can violate this fast-time assumption even though $k{d}_i\gg 1$. The slow branch does not obey the stated ordering and is not retained by the fixed-ion reduction.

When $k{d}_i\gg 1$, expand the square root in the exact frequencies. The fast branch haswhile the slow branch has ${\Omega }_{\mathrm{slow}}=|q|[(k{d}_i{)}^{-1}+O((k{d}_i{)}^{-3})]$. HenceThe fast dispersive wave is the mode sought in the question's subsequent fast-time limit. In a strongly oblique wave, $k\simeq k_{\perp }$, so its frequency scales as $|\omega |\simeq |k_{\parallel }|v_A{k}_{\perp }{d}_i$. The slow branch instead tends to $(|k_{\parallel }|/k){\Omega }_i$, where the ion cyclotron frequency is ${\Omega }_i=v_A/d_i=e{B}_0/(m_{i}c)$.

There is a terminology qualification: in the pressure-free incompressible Hall equations used here, the fast branch is conventionally a whistler wave. The question calls the requested small-scale waves kinetic Alfvén waves, but a general kinetic-Alfvén dispersion includes pressure/kinetic ordering absent from these equations. The boxed dispersion is the result for the printed model; no extra pressure response or kinetic term has been silently added.

Solving the polarization quadratic gives all four signed roots:A sign-independent description is the pair of positive frequency magnitudesFor wavelengths long compared with the ion skin depth, $k{d}_i\ll 1$, both magnitudes approach $|k_{\parallel }|v_A$. The Hall correction is then small and the usual Alfvén wave dispersion ${\omega }^2=k_{\parallel }^2{v}_A^2$ is recovered. In an incompressible system the two transverse polarizations include the shear and pseudo-Alfvén wave sectors; away from this limit Hall coupling selects circular helicity combinations. A small parallel wavenumber alone does not remove the relative Hall splitting: the expansion parameter is the full $k{d}_i$.

Linearize about zero bulk flow and the uniform guide field. Put $\mathbf{b}=\delta \mathbf{B}/\sqrt{4\pi \rho }$, and take plane waves proportional to $e^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$. Let $k_{\parallel }=\mathbf{k}\cdot \hat{\mathbf{z}}$, $k=|\mathbf{k}|$, $q=k_{\parallel }{v}_A$, andBoth $\mathbf{u}$ and $\mathbf{b}$ are perpendicular to $\mathbf{k}$. Projecting momentum onto that plane eliminates total pressure. The remaining magnetic-tension term givesIn the induction equation, the Hall contribution is $-c(B_0{\partial }_z)(\nabla \times \delta \mathbf{B})/(4\pi en)$; its two derivative factors of $i$ in the Fourier mode produce a positive coefficient multiplying $\mathbf{k}\times \mathbf{b}$. ThusEliminating $\mathbf{u}$ givesOn the transverse plane, $(i\mathbf{k}\times {)}^2=k^2$, so applying that operator again proves the incompressible Hall-MHD wave dispersion:Equivalently, in a helicity decomposition of a transverse Fourier mode with $i\mathbf{k}\times {\mathbf{b}}_h=hk{\mathbf{b}}_h$, $h=\pm 1$, each polarization obeys ${\omega }^2-hqk{d}_i\omega -q^2=0$. For $k_{\parallel }=0$ the projected linear restoring terms vanish and these modes have zero frequency; the propagating calculation assumes $k_{\parallel }\ne 0$.

True as a major quantitative test of the hot Big Bang, with a significant lithium qualification. Big Bang nucleosynthesis predicts several abundances from one main baryon-density parameter using measured nuclear and weak rates, rather than fitting a separate primordial composition for every species. Comparisons involve distinct observing environments and an independent cosmic microwave background determination at a much later epoch. Their overall agreement is powerful; it is not a claim that every light-element measurement is mutually consistent without stellar or observational corrections. The cosmological lithium problem must be included in that assessment.

Use the relative potential $\Psi ={\Phi }_{\ast }-\Phi$ and integrate the stated, untruncated Maxwell-Boltzmann velocity distribution over ${\mathbb{R}}^3$. The normalization cancels its three Gaussian integrals:The gravitational potential sign reverses under the relative-potential convention, so the spherical Poisson equation for Newtonian gravity isAlso ${\Psi }^{\prime }=-Gm(r)/r^2$, with $m^{\prime }=4\pi r^2\rho$. Differentiating the density relation yieldsAn ideal gas with constant temperature has $P=\rho k_{B}T/m_0$. Its hydrostatic equilibrium equation is precisely the same whenTogether with the same enclosed-mass equation, central density and boundary conditions, this proves the isothermal stellar and gas density equivalence. In dimensionless terms, $y=-[\Psi -\Psi (0)]/{\sigma }^2$ and $x=r\sqrt{4\pi G\rho (0)}/\sigma$ obey $x^{-2}(x^2{y}^{\prime }{)}^{\prime }=e^{-y}$ with $y(0)=y^{\prime }(0)=0$ in both systems.

This calculation uses the distribution as printed for all velocities. Imposing an additional cutoff at positive relative energy changes the integral to ${\rho }_1[e^W\mathrm{erf}\sqrt{W}-2\sqrt{W}/\sqrt{\pi }]$, where $W=\Psi /{\sigma }^2\ge 0$, and no longer gives an exactly isothermal gas profile. An untruncated self-gravitating isothermal sphere also needs an outer modification for finite total mass. The function $\mathrm{erf}$ here is the error function. Finally, the half-line Gaussian integral printed in the note lacks a factor: the correct value is ${\int }_0^{\infty }{e}^{-\alpha x^2}dx=\frac{1}{2}\sqrt{\pi /\alpha }$.

The Gaussian integral over the full vertical velocity line is $\sqrt{2\pi }{\sigma }_z$. Thus the mass density of the self-gravitating isothermal slab isNeglecting radial derivatives makes the Poisson equation for Newtonian gravity ${\Phi }_{zz}=4\pi G\rho (z)$. Substituting $\Phi ={\sigma }_z^2\varphi$ and $z=z_0\zeta$, with $z_0^2={\sigma }_z^2/(8\pi G{\rho }_0)$, givesMultiply by ${\varphi }^{\prime }$ and integrate. The central value and symmetry under $z\mapsto -z$ fix the constant of integration:On the positive-$\zeta$ side, Newtonian gravity makes ${\varphi }^{\prime }\ge 0$. Put $y=e^{\varphi /2}$; then $4(y^{\prime }{)}^2=y^2-1$ with $y(0)=1$. Integration gives $2\mathrm{arcosh}y=\zeta$, and even reflection supplies the other side. ThereforeDirect differentiation verifies both the equation and central conditions. Uniqueness of the smooth initial value problem rules out other branches meeting those conditions. The surface density is $4{\rho }_0{z}_0$, and the conventional slab scale height is $2z_0$; the factor of two is only a definition of the length scale.

Let $d=k-1$ and let $e_m$ denote the elementary symmetric polynomial of degree $m$ in the $d$ interior knots $t_{i+1},\dots ,t_{i+d}$, with $e_0=1$. Expand the two sides of the Marsden identity as polynomials in $x$:andComparing the coefficient of $x^{d-m}$ and cancelling the common sign gives the monomial B-spline coefficientsEquivalently,In particular $a_i^{(0)}=1$, giving partition of unity on the interior interval. For $k\ge 2$, $a_i^{(1)}=(t_{i+1}+\cdots +t_{i+k-1})/(k-1)$, and the highest-degree coefficient is the product of those interior knots. The formulas also cover $k=1,m=0$ through empty products.

For fixed $t$, put $g_t(u)=(u-t{)}_{+}^{k-1}$, a truncated power function. Begin with distinct increasing knots. The two interpolation polynomials ${\ell }_i(\cdot ,t)$ and ${\ell }_{i+1}(\cdot ,t)$ agree at their $k-1$ shared knots. Their difference has degree at most $k-1$, soFor $k=1$ the empty product is one and the same assertion holds for constant interpolants. The leading coefficient of a degree-at-most-$k-1$ interpolant on $k$ nodes is its order-$k-1$ divided difference, by the Newton interpolation polynomial. ThereforeThis proves the Lee interpolation identityThe identity holds for every real $x,t$, with the usual convention for the order-zero truncated power. Admissible repeated knots are handled by the corresponding confluent divided differences and consistent one-sided limits at breakpoints; taking these limits preserves the Lee interpolation identity.

Sum over $i$ to telescope:If $t_k<t<t_{n+1}$, the first $k$ knot values of $g_t$ are zero, so ${\ell }_1=0$. At the last $k$ knots $t_{n+1},\dots ,t_{n+k}$, the function $g_t$ agrees with the polynomial $(u-t{)}^{k-1}$, so uniqueness of polynomial interpolation gives ${\ell }_{n+1}(x,t)=(x-t{)}^{k-1}$. Thus the Marsden identity is