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Use the characteristic polynomial in the convention $p(\lambda )=\det (\lambda I-A)$:For fixed $\alpha >0$ with $\alpha \ne 1$, all three limiting eigenvalues are simple. Expanding this polynomial at each root, or applying first-order perturbation of a simple eigenvalue, givesThese are fixed-$\alpha$ asymptotic expansions, not uniform through $\alpha =1$. At $\alpha =1/3$ the displayed correction to ${\lambda }_{\alpha }$ vanishes because $\lambda =1/3$ is an exact root for every $ϵ$; there is no second nonzero term for that branch.

The two roots near $1$ interact when the correction $ϵ/(\alpha -1)$ is comparable to their unperturbed gap $\alpha -1$. Thus the distinguished limit is $\alpha =1+d\sqrt{ϵ}$, with bounded real $d$. Set $h=\sqrt{ϵ}$ and $\lambda =1+hL+h^{2}N+\cdots$. The polynomial becomesAt orders $h^2$ and $h^3$,With $D=\sqrt{d^2+4}$, the two interacting eigenvalues areThe $h^2$ correction to the pair is included to resolve the next perturbative term as well as the leading splitting. In particular the previously excluded fixed case $\alpha =1$ is $d=0$:The half-integer powers reflect square-root splitting of a defective double eigenvalue: the limiting matrix has a nontrivial Jordan block at $1$.

For matching, write $\Delta =\alpha -1$ and take the overlap $\sqrt{ϵ}\ll |\Delta |\ll 1$. For $d\to +\infty$, $L_{+}=d+d^{-1}+O(d^{-3})$, $N_{+}=1+O(d^{-2})$, while $L_{-}=-d^{-1}+O(d^{-3})$, $N_{-}=O(d^{-2})$. Consequently the distinguished branches reduce toThe regular coefficient satisfies $(3\alpha -1)/[(\alpha +1)(\alpha -1)]=1/\Delta +2/(2+\Delta )=1/\Delta +1+O(\Delta )$, proving agreement. For $d\to -\infty$ the labels $+$ and $-$ exchange roles and the same matching holds. The isolated negative branch also matches because $-2/(\alpha +1)=-1+O(\Delta )$. As a spectral check, positive diagonal symmetrization of a matrix transforms this tridiagonal matrix to a real symmetric matrix with off-diagonal entries $\sqrt{2ϵ}$ and $\sqrt{ϵ}$; its eigenvalues are real for positive $ϵ$.

On the unit circle, $z^{-1}=\overline{z}$, soThus the trigonometric polynomial is nonnegative, with a zero at $z=-1$, andEquivalently, writing $z=e^{i\theta }$ gives $p(z)=2+2\cos \theta =4{\cos }^2(\theta /2)$. The factor has polynomial degree one as required. Multiplication of $q$ by any constant of modulus one leaves the factorization unchanged; uniqueness of $q$ is not claimed.

This is the simplest instance of the Fejér–Riesz theorem, where nonnegativity of a trigonometric polynomial on the unit circle admits a polynomial modulus-square factorization.

For $z\ne 0$, the coefficient symmetry yieldsThus $P$ equals its reversed-conjugate polynomial. If $\zeta \ne 0$ is a root of a polynomial, with $P(\zeta )=0$, this identity gives $P(1/\overline{\zeta })=0$, provingThe reciprocal-conjugate root pairing also preserves the multiplicity of a root: reversal and complex conjugation take each factor associated with $\zeta$ to the corresponding factor associated with $1/\overline{\zeta }$, with the same exponent.

Since $P(0)=p_{-d}\ne 0$, no zero root of a polynomial occurs, so the reciprocal operation is always defined on the roots of a polynomial. Roots off the unit circle are paired on opposite sides of it; roots of a polynomial on the unit circle are fixed by this operation. Pairing alone does not force even multiplicity at those fixed roots of a polynomial; that extra fact uses nonnegativity.

Since $p$ is real on the unit circle,Multiply the difference by $z^d$. It is an ordinary polynomial of polynomial degree at most $2d$ vanishing at every point of the unit circle. A nonzero polynomial has only finitely many roots of a polynomial, so all its coefficients vanish. This proves the conjugate symmetry of trigonometric polynomial coefficients:For arbitrary nonzero complex $z$, coefficient substitution now gives preciselyEquivalently $p(1/\overline{z})=\overline{p(z)}$. The complex conjugation on the right applies to the whole value, including the coefficients; it cannot simply be discarded away from the circle.

The PDF additionally assumes $p_d\ne 0$. Then $p_{-d}=\overline{p_d}\ne 0$, so the ensuing polynomial $P=z^{d}p$ has polynomial degree exactly $2d$ and nonzero constant term. These clauses and this subpart are absent from the damaged TeX, but present in the PDF.

Use localization and cancellation of the assumed interval-adapted wavelet basis: at scale $j$, every detail wavelet has support diameter at most $B{2}^{-j}$, L1 norm at most $B^{\prime }{2}^{-j/2}$, and annihilates polynomials of degree less than $q$. These bounds also apply to the modified boundary wavelets; a restriction of an arbitrary whole-line wavelet without the cancellation-preserving boundary construction would not suffice. The finitely many coarse scaling functions can be treated separately.

Write $\alpha =r+\beta$ with $0<\beta \le 1$ as above. Because $\alpha <q$, we have $r<q$. Choose $x_0$ in the support of a function of ${\psi }_{j,n}$, and let $T_{x_0}$ be the degree-$r$ Taylor polynomial of $f$. The Hölder-Taylor remainder bound givesFor $r=0$ this is the Hölder seminorm bound. For $r\ge 1$, the integral Taylor remainder is bounded using $f^{(r)}(t)-f^{(r)}(x_0)$, whose magnitude is at most $[f^{(r)}{]}_{C^{0,\beta }}|t-x_0{|}^{\beta }$.

The vanishing moments remove $T_{x_0}$ from the inner product. Applying the remainder estimate and the L1 norm bound on the localized support yields the wavelet coefficient decay for Hölder functionsThusHere $C$ depends on the fixed wavelet family, the exponent and boundary construction, but not on $f,j,n$. The proof uses the assumed regularity and moments; it does not identify the minimal Daubechies wavelet order with its differentiability order, which need not coincide.

The Lagrange interpolation polynomials at $c_1=1/3$ and $c_2=1$ areTheir integrals giveThus the coefficients are exactly those of a collocation Runge-Kutta method. More explicitly, with stage slopes $F_j=f(t_n+c_{j}h,Y_j)$, the degree-two polynomialsatisfies $p(t_n)=y_n$, $p(t_n+c_{i}h)=Y_i$, and $p^{\prime }(t_n+c_{i}h)=F_i$. These are the collocation Runge-Kutta method equations, and $y_{n+1}=p(t_n+h)$. For an implicit Runge-Kutta method, the nearby stages exist uniquely for sufficiently small $h$ under Lipschitz continuity; see stage solvability of an implicit Runge-Kutta method.

Write $e=(1,1{)}^T$, $c=Ae$, and interpret powers of $c$ componentwise. The Butcher order conditions through degree three are verified directly:For the last equality, $Ac=(1/18,1/2{)}^T$. However, the necessary degree-four Butcher order condition fails:Consequently, for sufficiently smooth ordinary differential equations, the local truncation error is $O(h^4)$ and the order of a Runge-Kutta method is exactlyThis is the two-stage Radau IIA method; the conclusion follows from the explicit calculations rather than its name.

On the Dahlquist test equation, $g(y)={\lambda }^{2}y$, so the amplification polynomial of a multistep method becomesBoth roots of a polynomial, including the parasitic one, must satisfy the root condition for a multistep method.

For $\alpha =1/7$, multiply by seven and set $a=7-6z+2z^2$. The polynomial is $a{\xi }^2-8\xi +1$. The complex quadratic Schur criterion says that its roots of a polynomial are strictly in the unit disk ifHere is a direct verification on the entire left half-plane. Put $z=-r+iy$, $r\ge 0$, $t=y^2$, and $a_0=7+6r+2r^2$. Thenand expansion of the squared inequality givesEvery term is nonnegative, and at least one is positive unless $z=0$. Since $q-1>0$, taking square roots proves the strict complex quadratic Schur criterion for $z\ne 0$. At $z=0$, the roots of a polynomial are the simple values $1$ and $1/7$. The leading coefficient cannot vanish because $|a{|}^2\ge 49$. Thus the method at $\alpha =1/7$ is A-stable. Its order three is compatible with the Second Dahlquist barrier: that barrier concerns first-derivative linear multistep methods, whereas this method uses a second derivative.

For $\alpha =1/2$, choose $z=-4$, strictly in the left half-plane. The amplification polynomial of a multistep method reduces toOne root of a polynomial has modulus $(3+\sqrt{17})/4>1$. Thus the method at $\alpha =1/2$ is not A-stable.

A Runge-Kutta method advances $y^{\prime }=f(t,y)$ through stagesThe Butcher tableau records $A,b,c$, normally with $c=Ae$. Linear absolute stability tests the amplification of a linear decay mode; nonlinear B-stability compares distances between two solutions of a dissipative vector field. Both describe propagation of perturbations, but they impose different conditions.

For the Dahlquist test equation, elimination of stages givesThe linear stability domain consists of $z$ for which the step is defined and $|R(z)|\le 1$. Poles or singular stage systems must be excluded, even if an output formula has a formal cancellation. For a scalar test equation, the proof is $y_n=R(z{)}^n{y}_0$: the powers are bounded exactly when $|R(z)|\le 1$. For $y^{\prime }=Ly$, the update is $R(hL)$. If $L$ is a normal matrix, unitary diagonalization of a normal matrix makes the operator norm equal to the largest scalar amplification modulus. For a diagonalizable matrix, powers are bounded by the condition number of the eigenvector matrix times the maximal scalar power. These constants must remain uniform for mesh-dependent systems. A defective unit-modulus eigenvalue can instead produce a growing Jordan block; the scalar spectral test alone does not control a general matrix.

An A-stable method accepts the entire closed left half-plane. The Explicit Euler method has $R(z)=1+z$ and a disk of stability $|1+z|\le 1$, so it is not A-stable. No nonconstant polynomial stability function can be A-stable, since it is unbounded on the negative real axis; this excludes all consistent explicit Runge-Kutta methods. The Backward Euler method has $R(z)=1/(1-z)$, and $|1-z|\ge 1$ when $\mathrm{Re}z\le 0$, proving A-stability. It is also L-stable, because $R(z)\to 0$ for large left-half-plane $z$.

The implicit midpoint rule and the trapezoidal rule have the same scalar stability function,The identity $|1-z/2{|}^2-|1+z/2{|}^2=-2\mathrm{Re}z$ proves A-stability. On the imaginary axis their amplification has unit modulus, and along the negative real axis it tends to $-1$. Thus neither is L-stable; very stiff decay can persist as alternating numerical values. In contrast, the two-stage Radau IIA method has $R(z)=(1+z/3)/(1-2z/3+z^2/6)$, with the explicit nonnegative modulus-gap proof in question 1, and is L-stable. This illustrates why L-stability matters beyond A-stability for a stiff differential equation.

For nonlinear problems, assume a real or complex Euclidean inner product and the dissipativity hypothesisfor all relevant states at each common time. Exact solutions satisfyA B-stable Runge-Kutta method reproduces this nonexpansion of distances whenever its stages exist. Put $d=y_n-{\tilde{y}}_n$, $D_i=Y_i-{\tilde{Y}}_i$, and $G_i=f(t_n+c_{i}h,Y_i)-f(t_n+c_{i}h,{\tilde{Y}}_i)$. Then $D_i=d+h{\sum }_j{a}_{ij}{G}_j$. Expansion of the output squared norm, followed by substituting $d=D_i-h{\sum }_j{a}_{ij}{G}_j$ in its linear terms, proves the Runge-Kutta contractivity identityIf $b_i\ge 0$ and $M=(m_{ij})$ is a positive semidefinite matrix, the first sum is nonpositive by dissipativity, and the second is nonnegative: expand each coordinate of the stage vectors to express it as a sum of nonnegative quadratic forms. These are exactly the conditions for algebraic stability of a Runge-Kutta method. Hence algebraic stability implies B-stability. Applying B-stability to the linear dissipative problem $y^{\prime }=\lambda y$, or its two-dimensional real form, also proves A-stability whenever stages are well defined.

The Backward Euler method has $b=A=1$, hence $M=1$; the implicit midpoint rule has $b=1$, $A=1/2$, hence $M=0$. Both are algebraically stable and B-stable. For the two-stage Radau IIA method,a positive semidefinite matrix, while both weights are positive. It therefore combines third-order accuracy, B-stability, and L-stability. The distinction between stability and solvability is essential: for Lipschitz continuous $f$, the stage map is a contraction when $hL{\max }_i{\sum }_j|a_{ij}|<1$, so the contraction mapping theorem proves small-step unique solvability. The preceding algebraic stability proof is an estimate for existing stages; it is not a theorem asserting stage existence for arbitrary step sizes and domains.

Linear A-stability is insufficient for nonlinear B-stability. The trapezoidal rule fails B-stability, even though its scalar stability function is identical to that of the B-stable implicit midpoint rule. Its Runge-Kutta method coefficients areFailure of this sufficient algebraic stability test alone would not prove failure of B-stability. An actual counterexample does: take $f(y)=-y^3$ and $h=2$. Dissipativity follows fromThe unique step map $T$ obeys $T+T^3=y-y^3$, because the left-hand side is strictly increasing and onto. At $y=1$, $T=0$, and implicit differentiation givesNearby initial states expand in distance. This proves a true nonlinear failure rather than merely failure of a coefficient criterion.

Finally, stability of a numerical method on finite intervals connects these contractivity properties to accuracy. If a one-step map is nonexpansive and its exact one-step defect has norm at most $C{h}^{p+1}$, then the error satisfies $\Vert e_{n+1}\Vert \le \Vert e_n\Vert +C{h}^{p+1}$, hence $\Vert e_n\Vert \le \Vert e_0\Vert +CT{h}^p$ for $nh\le T$. A Lipschitz step bound $1+Kh$ gives the same order using the discrete Gronwall inequality. For a stiff differential equation, the value of B-stability is that the propagation estimate need not grow with a large negative dissipative rate. The accuracy constant still needs appropriate smoothness and uniform derivative bounds; stability alone does not establish a mesh-uniform error order.

Take the current to be dilute, vertically well mixed and Boussinesq approximation, with constant channel width $w$, a horizontal absorbing bed, negligible ambient fluid entrainment, negligible bed friction, and no resuspension. The shallow-water approximation neglects vertical inertia and gives hydrostatic pressure. Particle loss alters reduced gravity but does not alter the leading-order carrier-fluid volume or inertia; otherwise additional mass and momentum terms would be necessary. The current occupies depth $h(x,t)$, has depth-uniform speed $u(x,t)$, and has particle volume fraction $c(x,t)$, with $g^{\prime }=g_{o}c$ and constant $g_o>0$.

Volume conservation of a short channel segment gives $h_t+(uh{)}_x=0$. Its suspended particle amount per unit bed area is $hc$, and the downward particle deposition flux is $v_{s}c$. Thus $(hc{)}_t+(uhc{)}_x=-v_{s}c$, which becomes $(h{g}^{\prime }{)}_t+(uh{g}^{\prime }{)}_x=-v_s{g}^{\prime }$ on multiplying by $g_o$.

At vertical coordinate $y$ above the bed, excess hydrostatic pressure divided by reference mass density is $g^{\prime }(h-y)$. Integrating across the depth gives the depth-integrated force flux $g^{\prime }{h}^2/2$. The horizontal momentum per unit bed area is $hu$ and the advected kinematic momentum flux is $h{u}^2$. The resulting variable-buoyancy shallow water equations areThe derivative of the pressure term acts on both $h$ and $g^{\prime }$; treating reduced gravity as horizontally constant while particles settle would lose part of the force.

For a steady interior, let $q=Q/w=uh>0$ denote volume flux per unit width. Then the particle balance is a first-order linear differential equation,Here $g_0^{\prime }=g^{\prime }(0)>0$, $v_s\ge 0$ is constant, and deposited particles are not returned to the current. In particular the decay length is $q/v_s=Q/(w{v}_s)$ for $v_s>0$, and the cancellation of depth follows from $uh=q$: dilution and transit time compensate in the deposition balance.

The steady momentum equation also givesThe inlet depth selects $K=q^2/h_0+g_0^{\prime }{h}_0^2/2$. Depth and speed at every downstream position are determined by the continuous positive root of a polynomial inThis implicit expression also makes clear why the inlet condition is needed: there are generally two possible positive depths.

The depth branches of a steady depositing gravity current are distinguished by the Froude number ${\mathrm{Fr}}^2=q^2/(g^{\prime }{h}^3)$. Differentiating the momentum invariant, and using the exponential decay of reduced gravity, givesFor $v_s>0$, supercritical flow has $\mathrm{Fr}>1$ and decreasing depth, with $h\to q^2/K$ and $u\to K/q$ as $x\to \infty$. Subcritical flow has $\mathrm{Fr}<1$ and increasing depth, with $h\sim \sqrt{2K/g^{\prime }(x)}\propto e^{v_{s}x/(2q)}$. This indefinitely deepening branch eventually violates the assumed shallow or confined-channel geometry. Hence the depth does not universally decrease: that conclusion needs a supercritical inlet.

For fixed $g^{\prime }>0$, $q^2/h+g^{\prime }{h}^2/2$ has its unique minimum at $h_c=(q^2/g^{\prime }{)}^{1/3}$, with $K_c=3q^{4/3}(g^{\prime }{)}^{1/3}/2$. The two positive roots when $K>K_c$ lie below and above $h_c$. Since $K_c$ decreases downstream while $K$ is constant, an initially strict branch remains separate from the other branch and cannot cross a critical point. A critical inlet with $v_s>0$ has no finite depth derivative: at $\mathrm{Fr}=1$ the first differentiated equation would equate zero to a positive value. It requires a singular entrance adjustment or physics beyond this steady model. If $v_s=0$, both reduced gravity and the continuous depth branch are constant. These steady solutions describe the supplied current interior, not a moving finite-release front.

For two particle populations, write $g^{\prime }=g_1^{\prime }+g_2^{\prime }$ and $g_i^{\prime }=g_{0i}{c}_i$, allowing different constant particle-density conversion factors. The bidisperse particle-laden gravity current equations areFor the steady case,Thus the depth is obtained from the same cubic polynomial invariant with the sum of the two exponentially decreasing reduced gravity contributions.

The local deposit consists of the two particle deposition fluxes $v_1{c}_1$ and $v_2{c}_2$. For a positive total deposition rate, the bidisperse gravity-current deposition fraction by particle volume isIf the particles have a common material density, this is also their local mass fraction. If they have common $g_o$, it can be written in the source's buoyancy notation asFor unequal $g_{0i}$ replace each $g_{i0}^{\prime }$ by $g_{i0}^{\prime }/g_{0i}$. The exponent sign matters: with both populations present and $v_1>v_2>0$, the faster-settling population has decreasing local deposit fraction downstream, since $f_1^{\prime }=(v_2-v_1)f_1(1-f_1)/q<0$. Equal positive settling speeds give a constant fraction. A nonsettling population contributes no deposit; if the entire local deposition rate is zero the deposit fraction is undefined.

If “fraction of particles” means number fraction rather than particle-volume or mass fraction, let $a_i$ be a single particle's volume in population $i$. The correct weights are then $v_i{c}_i/a_i$, soSettling speeds alone do not determine $a_i$ without particle properties and a drag law. Finally, if the intended deposit is cumulative over the bed from $0$ to $x$, rather than local at $x$, integrating particle deposition flux gives the species volumes per unit channel width $D_i(x)=q{c}_{i0}(1-e^{-v_{i}x/q})$. The cumulative particle-volume fraction isFor both $v_i>0$, its limit at infinity is the incoming particle-volume fraction, whereas its limit as $x\downarrow 0$ is the local deposition fraction at the inlet. Local sorting and the cumulative inventory are different observables.

For any vector $y\in {\mathbb{F}}_2^n$, use its Hamming weight $w(y)=\#\{j:y_j=1\}$. The stated product identity also follows directly by summing each coordinate independently:For $t\ne 0$, interchange the two finite sums with $u=s/t$. Character orthogonality evaluates the inner sum over codewords, whereas the product identity evaluates the sum over all binary vectors. ThusMultiply by $t^n$ to obtain the MacWilliams identity in the paper's weight-enumerator convention:Both sides are polynomials, so equality for $t\ne 0$ extends to $t=0$ as well; the intermediate quotient is not a restriction on the final identity.

A linear code of length $n$ and rank $k$ is a $k$-dimensional linear subspace of ${\mathbb{F}}_2^n$. The Hamming weight of a vector counts its nonzero coordinates. Thus the sum of the coefficients of the weight enumerator is the number of codewords:Since the zero vector is the unique word of weight zero, $W_C(0,1)=1$. Symmetry of the enumerator therefore implies $W_C(1,0)=1$. Conversely $W_C(1,0)=1$ means that the all-one vector $1$ belongs to $C$. Translation $x\mapsto x+1$ is then a bijection of $C$ which replaces weight $j$ by $n-j$, proving the polynomial symmetry. For length zero the same conclusion holds with the unique empty vector.

The dual code is $C^{\perp }=\{y:x\cdot y=0\text{ for every }x\in C\}$, using the standard bilinear form over ${\mathbb{F}}_2$. If $y\in C^{\perp }$, each summand of the character sum equals one. Otherwise choose $x_0\in C$ with $x_0\cdot y=1$; replacing $x$ by $x+x_0$ permutes the summands and reverses their sign. The sum equals its own negative and must be zero. ConsequentlyThis is character orthogonality for the additive group of the code.

A splitting field of a nonzero polynomial $f$ over $K$ is an extension in which $f$ splits into linear factors and which is generated over $K$ by all its roots. For a constant nonzero polynomial the splitting field is $K$; the zero polynomial does not have a root-generated splitting field in this standard definition.

Given two splitting fields $L,L^{\prime }$, enumerate the roots generating $L$ and extend a $K$-embedding one root at a time. If an embedding is defined on $K({\alpha }_1,\dots ,{\alpha }_j)$, the minimal polynomial of the next root divides $f$ over that field. Its image under the embedding therefore divides $f$ over the corresponding subfield of $L^{\prime }$, and has a root in $L^{\prime }$. The quotient construction $K(\alpha )\cong K[t]/(m_{\alpha })$ extends the embedding to that root. The polynomial need not be a separable polynomial.

At the end this gives a $K$-embedding $L\to L^{\prime }$. It sends the distinct roots of $f$ injectively to its roots, hence permutes the finite set of distinct roots and contains every root in its image. Since these generate $L^{\prime }$, the embedding is onto. Thus the splitting field is unique up to a (K)-isomorphism.

The quartic is irreducible by Eisenstein criterion at two, so its Galois group is transitive on its four roots. Its discriminant is $1616=16\cdot 101$, so five and seven are unramified primes for the polynomial. The squarefree factorizations areThe cubic modulo five has no root in ${\mathbb{F}}_5$, hence is irreducible. The quadratic modulo seven has discriminant five, a nonsquare there. The Frobenius cycle type theorem therefore supplies a three-cycle and a transposition in the group.

Transitivity implies that its order is divisible by four, and the three-cycle implies divisibility by three. Its order divides $24$, so it is $12$ or $24$. The only subgroup of index two in $S_4$ is $A_4$: an index-two subgroup is the kernel of a group homomorphism of a nontrivial homomorphism to $C_2$, and all transpositions are conjugate and generate $S_4$, forcing this homomorphism to be the sign. The exhibited transposition excludes $A_4$. Hence

Choose a nonzero $\alpha \in \mathfrak{a}$. It is an algebraic integer, with monic minimal polynomialIts constant coefficient is nonzero, since otherwise the polynomial would have the root zero and be divisible by its variable, contradicting irreducibility and $\alpha \ne 0$. Rearranging gives $c_0=-\alpha ({\alpha }^{d-1}+\cdots +c_1)$. The expression in parentheses lies in the ring of integers of a number field, so the ideal property gives $c_0\in \mathfrak{a}$. Hence

We prove the conjugate product of a quadratic ideal identity $\mathfrak{a}\overline{\mathfrak{a}}=(N(\mathfrak{a}))$, which applies to the given two-generator ideal without assuming its displayed generators form an integral basis.

The quadratic ring of integers of a number field has a basis $1,\omega$ over $\mathbb{Z}$: the vector 1 is primitive, since a rational algebraic integer must be an integer, and extends to an integral basis. Write ${\omega }^2-T\omega +U=0$, with $T,U\in \mathbb{Z}$. Let $B$ be the least positive integer in $\mathfrak{a}$ and let $s>0$ generate the set of coefficients of $\omega$ in elements of $\mathfrak{a}$. Its lattice basis can be chosen as $B,r+s\omega$, with $r\in \mathbb{Z}$. Since $B\omega \in \mathfrak{a}$, comparison of coefficients shows $s\mid B$ and $s\mid r$. Write $B=sd$, $r=se$ and $\theta =e+\omega$. ThenThe scaled ideal $\mathfrak{j}=s^{-1}\mathfrak{a}$ remains an ${\mathcal{O}}_L$-ideal. From ${\theta }^2-T^{\prime }\theta +U^{\prime }=0$ we get $U^{\prime }\in \mathfrak{j}\cap \mathbb{Z}=d\mathbb{Z}$, so $d\mid U^{\prime }$. The product $\mathfrak{j}\bar{\mathfrak{j}}$ is generated as an ideal by $d^2,d\theta ,d\bar{\theta },U^{\prime }$ and is contained in $(d)$.

Moreover $\gcd (d,T^{\prime },U^{\prime }/d)=1$. If a prime $\ell$ divided all three, then $\ell \mid T^{\prime }$ and ${\ell }^2\mid U^{\prime }$, so $\theta /\ell$ would satisfy the monic integer polynomial $z^2-(T^{\prime }/\ell )z+U^{\prime }/{\ell }^2$. It would lie in ${\mathcal{O}}_L$, impossible because its $\omega$ coefficient in the integral basis is $1/\ell$. Bezout identity now expresses one as an integer combination of $d,T^{\prime },U^{\prime }/d$. Multiplying by $d$, the terms are $d^2$, $d{T}^{\prime }=d\theta +d\bar{\theta }$, and $U^{\prime }$, all in the product. Hence $d$ belongs to it, proving $\mathfrak{j}\bar{\mathfrak{j}}=(d)$.

The determinant of the lattice basis $sd,s(e+\omega )$ gives the ideal norm $N(\mathfrak{a})=s^{2}d$. ConsequentlyIn particular $⟨b,\alpha ⟩⟨b,\bar{\alpha }⟩$ is principal. This does not require the original $b$ to be nonzero or the ideal itself to be principal.