The quotient $q(w)=\varphi (w)/w$ extends to a holomorphic function at zero with value one. Since $\varphi$ is injective and vanishes at zero, $q$ has no zero in the unit disc. A holomorphic logarithm $\ell$ exists there, with $\ell (0)=0$. Define the odd square-root transform of a normalized univalent function byIt satisfies $h(z{)}^2=\varphi (z^2)$, $h(0)=0$, $h^{\prime }(0)=1$ and $h(-z)=-h(z)$. To check injectivity, $h(z)=h(w)$ gives $\varphi (z^2)=\varphi (w^2)$, hence $z^2=w^2$. If $w=-z$, oddness gives $h(z)=0$, so $z=w=0$. Thus the two arguments always coincide, and $h\in \mathcal{S}$. Choosing the square root by its derivative at zero removes its sign ambiguity.

The Koebe growth theorem states that every normalized univalent function $\varphi \in \mathcal{S}$ satisfiesBoth bounds are sharp. Rotations of the Koebe function, ${\varphi }_{\theta }(z)=z/(1-e^{i\theta }z{)}^2$, attain them along suitable radial directions.

Consider the kernel of a ring homomorphismSince $R$ is a Noetherian ring, this ideal has a finite generating set of an ideal, say $K=(a_1,\dots ,a_t)$. For each $i$, choose $s_i\notin \mathfrak{p}$ with $s_i{a}_i=0$, and putThen $f\notin \mathfrak{p}$ by the prime ideal condition, and $fK=0$. If $K=0$, take the empty product $f=1$.

If $a/f^n$ maps to zero under the ring homomorphism ${\theta }_f$ from a, then $a/1=0$ in $R_{\mathfrak{p}}$ because $f/1$ is a unit there. Hence $a\in K$, so $fa=0$. The vanishing criterion in a module localization now gives $a/f^n=0$ already in $R_f$. ThereforeThis is the finite generation clears localization kernels argument: the finite generating set of an ideal permits a single element outside $\mathfrak{p}$ to annihilate the entire kernel of a ring homomorphism.

Choose $f=1$, so that $R_f=R$ and $f\notin \mathfrak{p}$. For an integral domain, the canonical ring homomorphism $R\to R_{\mathfrak{p}}$ is injective: if $a/1=0$, the vanishing criterion in a module localization gives $sa=0$ for some $s\notin \mathfrak{p}$. Such an $s$ is nonzero, so the absence of zero divisors gives $a=0$. Thus

Since the radical of an ideal $\sqrt{\mathfrak{q}}$ is $(r)$, some positive power of $r$ belongs to $\mathfrak{q}$. Choose the least such exponent $k$. Then $(r^k)\subseteq \mathfrak{q}$ and $r^j\notin \mathfrak{q}$ for $0\le j<k$, with $r^0=1$ excluded because the primary ideal $\mathfrak{q}$ is proper.

Take $x\in \mathfrak{q}$. If $x\notin (r^k)$, choose the largest $j\in \{0,\dots ,k-1\}$ for which $x\in (r^j)$, and write $x=r^{j}u$. Maximality of $j$ implies $u\notin (r)=\sqrt{\mathfrak{q}}$: otherwise $x\in (r^{j+1})$, a contradiction, including when $j=k-1$. The primary ideal condition applied to $r^{j}u\in \mathfrak{q}$ therefore forces $r^j\in \mathfrak{q}$, contrary to the choice of $k$.

Consequently every $x\in \mathfrak{q}$ lies in $(r^k)$, andTogether with i, this classifies the primary ideals for a regular principal prime. The exponent is unique: if $(r^k)=(r^{k+1})$, then $r^k=r^{k+1}c$, and cancellation of the non-zero-divisor $r^k$ would give $1=rc$, contradicting that $(r)$ is proper. No Noetherian hypothesis is needed.

Because $(r)$ is a proper prime ideal, each $(r^k)$ is proper and its radical of an ideal isIndeed, $(r^k)\subseteq (r)$ gives one inclusion of radicals of ideals, while $(rc{)}^k\in (r^k)$ gives the other.

We prove the saturation condition for a primary ideal: if $ab\in (r^k)$ and $a\notin \mathfrak{p}$, then $b\in (r^k)$. For $k=1$ this is the defining property of a prime ideal. For $k>1$, write $ab=r^{k}c$. Since $a\notin (r)$, the prime ideal condition gives $b=r{b}_1$. The element $r$ is a non-zero-divisor, so cancellation givesInduction gives $b_1\in (r^{k-1})$, hence $b\in (r^k)$. Together with the computed radical of an ideal, this proves

In the terminology of a minimal primary decomposition, an associated prime of a primary decomposition of $I$ is a prime ideal occurring as $\sqrt{{\mathfrak{q}}_j}$ in such a decomposition. The first uniqueness theorem for primary decomposition says that this collection is independent of the chosen minimal primary decomposition. Among these prime ideals, an isolated prime of a primary decomposition is minimal under inclusion, equivalently a minimal prime over an ideal $I$; an embedded prime ideal strictly contains another member of the collection. ThusWhen $R$ is a Noetherian ring, these are precisely the associated primes of a module $R/I$. In that case the equivalent definition iswhere $(I:x)$ is a colon ideal. The decomposition terminology applies when a minimal primary decomposition exists; a general ring need not provide one for every ideal.

A minimal primary decomposition is a finite expressionin proper primary ideals such that the prime ideals ${\mathfrak{p}}_j=\sqrt{{\mathfrak{q}}_j}$ are pairwise distinct and no primary ideal can be omitted. More explicitly, for every $j$,Thus minimality means both irredundancy and distinct radicals of ideals; it does not assert that the primary ideals themselves are uniquely determined. For $I=R$, one may use the empty intersection convention.

For a time-dependent type-two generating function for a canonical transformation, the differential identitygives $K=H+{\partial }_t{F}_2$, with the partial derivative taken at fixed $(q,P)$. Here $F_2=\cos q\sin P$ has no explicit time dependence, so the Hamiltonian correction isAs always, adding a function of time alone to the new Hamiltonian leaves Hamilton's equations unchanged. Choosing a time-independent type-two generating function for a canonical transformation fixes that irrelevant freedom and gives the displayed $\Lambda =0$.

On the interior of the principal inverse-cosine branch, $0<q<\pi$, the given canonical transformation hasIntegrating $Q={\partial }_P{F}_2$ suggests $F_2=\cos q\sin P+C(q)$. Its other derivative is ${\partial }_q{F}_2=-\sin q\sin P+C^{\prime }(q)$; comparison with $p$ requires $C^{\prime }(q)=0$. Hence a suitable type-two generating function for a canonical transformation isBoth derivative equations reproduce the specified transformation exactly. Its mixed derivative is $F_{2,qP}=-\sin q\cos P$, so it defines a local canonical transformation where $0<q<\pi$ and $\cos P\ne 0$, equivalently where $|Q/\cos P|<1$. At $q=0,\pi$ or $\cos P=0$, the stated formulas fail to give a regular coordinate chart. These restrictions explain the square-root branch and the local meaning of canonicity. This is a trigonometric type-two canonical transformation.

A type-two generating function for a canonical transformation uses the mixed variables $(q,P)$ and definesFor a local invertible canonical transformation, require $F_{2,qP}\ne 0$ so that $p=F_{2,q}(q,P)$ can be solved for $P$. The differential identity $d{F}_2=pdq+QdP$ gives $d(F_2-PQ)=pdq-PdQ$, expressing preservation of the canonical symplectic form.

Work on $I>0$, with a local choice of the periodic angle $\theta$. The inverse formulas areIn particular, ${\theta }_q=p/(2I)$, ${\theta }_p=-q/(2I)$, $I_q=\alpha q$ and $I_p=p/\alpha$. Their Poisson bracket isThe other fundamental Poisson brackets vanish by antisymmetry. Thus this is a canonical transformation for every $\alpha >0$, on local charts away from the origin. At $I=0$ the angle is undefined.

In the new coordinates the harmonic oscillator Hamiltonian isFor $I$ to be a conserved quantity throughout phase space, Hamilton's equations require $\dot{I}=-{\partial }_{\theta }{H}_0=0$ identically. The two coefficients must coincide. For the usual $m>0$ and $\omega >0$, positivity of $\alpha$ selectsThese are harmonic oscillator action-angle variables; the action variable is $I=H_0/\omega$.

The canonical momentum of the harmonic oscillator is $p=\partial L_0/\partial \dot{q}=m\dot{q}$. Substituting $\dot{q}=p/m$ into the Legendre transform in mechanics gives

Since $H$ is a conserved quantity, the Hamiltonian energy balance gives $H_t=0$. Since $f$ is a conserved quantity, $f_t+\{f,H\}=0$. Take an explicit partial derivative with respect to time, holding the phase space coordinates fixed:The last Poisson bracket vanishes because $H_t=0$. ThusThis is the time derivative of a conserved quantity for an autonomous Hamiltonian. The conservation hypotheses are understood to hold for every trajectory, as identities on phase space.

Set $D={\partial }_t+\{\cdot ,H\}$. The hypotheses mean $Df=Dg=0$ as identities on phase space. By the Jacobi identity for the Poisson bracket,Also, the product rule applied to the defining coordinate formula givesAdding these identities provesConsequentlyThis proves the Poisson bracket of time-dependent conserved quantities theorem, including explicit time dependence of $f$ and $g$.

The Poisson bracket is antisymmetric, so $\{H,H\}=0$. The Hamiltonian energy balance consequently givesFor a regular Lagrangian the Legendre transform in mechanics gives $H_t=-L_t$, evaluated at the corresponding velocity. In particular, a Lagrangian with no explicit time dependence has time-translation symmetry. Noether theorem associates this continuous symmetry of the action with the conserved quantity ${\sum }_i{\dot{q}}_i\partial L/\partial {\dot{q}}_i-L=H$. For an ordinary natural mechanical Lagrangian $L=T-V$, with $T$ quadratic in the velocities, this quantity is the total energy $T+V$.

In canonical coordinates on phase space, the Poisson bracket isIt treats $t$ as a parameter. Along a solution of Hamilton's equations, the chain rule givesTherefore the Hamiltonian conservation law isThe total derivative follows the trajectory; the explicit partial derivative holds $(\mathbf{q},\mathbf{p})$ fixed.

Define the canonical momenta byTo obtain an ordinary unconstrained Hamiltonian, assume a regular Lagrangian: the velocity Hessian matrix $({\partial }^{2}L/\partial {\dot{q}}_i\partial {\dot{q}}_j)$ is invertible. The inverse function theorem then lets us solve locally for $\dot{\mathbf{q}}=\mathbf{v}(\mathbf{q},\mathbf{p},t)$. The phase space is the cotangent bundle $T^{\ast }\mathcal{Q}$, with local canonical coordinates $(\mathbf{q},\mathbf{p})$ and dimension $2n$. The Legendre transform in mechanics definesThe velocities in this formula have been eliminated in favour of phase space variables. A hyperregular Lagrangian makes this construction global; regularity alone guarantees it locally. Hamilton's equations are

Now take zero torque and $\mathbf{\omega }(0)=(1,0,1)$. The axial angular velocity is ${\omega }_3=1$ and the complex transverse component obeys $\dot{z}=i\nu z$ with $z(0)=1$ and $\nu =(I_3-I_1)/I_1$. Hence the angular momentum components in the principal body frame areThe transverse projection therefore traces a circle with signed angular frequency $\nu$. Its magnitude is constant, $L=\sqrt{I_1^2+I_3^2}$.

In the inertial frame, zero torque means that $\mathbf{L}$ is fixed. Usingand the derivative of a body-fixed basis vector, the symmetry axis satisfiesThus there is uniform torque-free precession of a symmetric top about the fixed angular momentum, at a constant cone angle $\beta$ and space angular frequencyThe cone angle stays constant, with no nutation. The body-frame and space-frame frequencies of free symmetric-top precession differ: $\nu$ describes rotating components, whereas ${\Omega }_{\mathrm{space}}$ describes the symmetry axis in space. The formulas also cover $I_3=I_1$, where the angular momentum components in the principal body frame are constant.

Let $J=I_1=I_2$, ${\omega }_3=c$ andHere $c$ and $\nu$ are constant. The first two of Euler's equations give ${\dot{\omega }}_1=-\nu {\omega }_2+{\tau }_0/J$ and ${\dot{\omega }}_2=\nu {\omega }_1+{\tau }_0/J$. Combining them producesMultiplication by the integrating factor $e^{-i\nu t}$ gives, for $\nu \ne 0$,For completeness, $\nu =0$ instead gives $z(t)=z(0)+{\tau }_0(1+i)t/J$. These formulas describe a symmetric top under constant transverse body torque.

The stated initial conditions require $b\ne 0$ and $I_3\ne J$. They give $c=b/(I_3-J)$, $\nu =b/J$ and $z(0)=i{\tau }_0/b$. Substitution and simplification yieldTaking the real part and imaginary part, the angular velocity in the principal body frame isIts value at $t=0$ is the required initial condition; differentiating the first two components verifies both forced Euler's equations.