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The multiplicative field norm $N(m+n\sqrt{-7})=m^2+7n^2$ is an integer. Units have field norm one and are exactly $\pm 1$. A nontrivial factorization of 2 would force both factors to have field norm two, which the displayed quadratic form cannot represent. Thus 2 is an irreducible element. It is not a prime element, since it divides $(1+\sqrt{-7})(1-\sqrt{-7})=8$ but neither factor. This is an example in a nonmaximal quadratic order, rather than the full ring of integers of $\mathbb{Q}(\sqrt{-7})$; the precise ring matters for factorization.

With Bohr radius $a$, define $A_l=a[i{p}_r/\hslash -(l+1)/r+1/((l+1)a)]/\sqrt{2}$. For regular radial bound states the radial inner product and compatible domains give the displayed factorization. Expanding the first-order factors produces the centrifugal term $l(l+1){\hslash }^2/(2m{r}^2)$ and Coulomb term $-{\hslash }^2/(mar)$. factorization converts a second-order spectral problem into a positivity bound and first-order terminating-state equation.

For an odd prime $p$, the least quadratic nonresidue $n(p)$ is the smallest positive integer whose nonzero residue class is a quadratic nonresidue modulo $p$. For example, the nonzero quadratic residues modulo seven are $1,2,4$, so $n(7)=3$.

The least quadratic nonresidue is itself a prime number. Otherwise it has a factorization $n(p)=ab$ with $1<a,b<n(p)$. By minimality, both residue classes are quadratic residues, so $a\equiv u^2$ and $b\equiv v^2(\mathrm{mod}p)$ would give $n(p)\equiv (uv{)}^2(\mathrm{mod}p)$, a contradiction. This illustrates why multiplicative information can constrain a least exceptional integer.

An element $a$ of an integral domain is an irreducible element when it is nonzero and not a unit, and any factorization $a=bc$ has at least one unit factor. The exclusions of zero and units are part of the definition.

An element $a$ is a prime element when it is nonzero and not a unit, and $a\mid bc$ implies $a\mid b$ or $a\mid c$. If $a=bc$ and $a$ is a prime element, assume for example $b=ad$. Cancellation in $a=adc$ gives $dc=1$, so $c$ is a unit. This proves every prime element is an irreducible element.

In a Euclidean domain, each nonzero ideal has a generator of least Euclidean value: division of another member by that generator leaves a remainder in the ideal of smaller value, so the remainder must vanish. Let $a$ be an irreducible element and suppose $a\mid bc$ but $a\nmid b$. Write $(a,b)=(d)$. Since $d\mid a$, irreducibility says $d$ is a unit or is associate to $a$. The second case would give $a\mid b$, so $d$ is a unit. Hence $1=ra+sb$ for some $r,s$. Multiplying by $c$ shows $a\mid c$. Thus irreducibles in a Euclidean domain are prime.

For $R=\mathbb{Z}[i\sqrt{7}]$, use the multiplicative integer field norm $N(m+ni\sqrt{7})=m^2+7n^2$. Units have field norm one and are exactly $\pm 1$. If $2=uv$ were a product of nonunits, their norms would be integers greater than one multiplying to four, hence both would have field norm two. But $m^2+7n^2=2$ has no integer solution: $n\ne 0$ makes the left side at least seven, and $n=0$ would give $m^2=2$. Therefore 2 is an irreducible element.

On the other hand,is divisible by 2, while neither factor is, because divisibility by 2 requires both integer coefficients to be even. Thus 2 is not a prime element in this ring. It is also not a prime element in $\mathbb{Z}[i]$: $2\mid (1+i{)}^2=2i$, but $2\nmid 1+i$. In fact $2=(1+i)(1-i)$ is already a nontrivial factorization there.

For a purely imaginary candidate $z=it$, $t\in \mathbb{R}$, the imaginary part of the polynomial is $2t(t^2-4)$. The candidate $t=0$ is not a root, while $t=\pm 2$ both make the real part zero. The resulting factorization into quadratic polynomials givesThe roots of a polynomial sum to $0+2=2$ and have product $4\cdot 4=16$, as required by Vieta's formulas for a monic quartic with cubic coefficient $-2$ and constant coefficient $16$.

ideals are coprime when $A+B={\mathcal{O}}_K$, equivalently $1=a+b$ for some $a\in A$, $b\in B$. Always $AB\subseteq A\cap B$. If $x\in A\cap B$, then $x=xa+xb\in AB$, giving $\boxed{AB=A\cap B}$ in the coprime case. For nonzero integral ideals, the Chinese remainder theorem gives ${\mathcal{O}}_K/(AB)\cong {\mathcal{O}}_K/A\times {\mathcal{O}}_K/B$, and taking finite cardinalities yields $\boxed{N(AB)=N(A)N(B)}$.

For a nonzero prime ideal, the chain of finite-index inclusions givesby the given hypothesis. Distinct nonzero prime ideals are maximal and coprime, as are their positive powers. Unique factorization of ideals in a number field therefore gives $N(A)={\prod }_{P}N(P{)}^{v_P(A)}$. Exponents add under ideal multiplication, proving multiplicativity for arbitrary nonzero integral ideals. This is the standard domain of the finite ideal norm used here.

For the Dedekind factorization theorem, choose an integral primitive element $\theta$ of $K$ with monic minimal polynomial of an algebraic element $f\in \mathbb{Z}[X]$, and assume the rational prime $p$ does not divide the index $[{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$. Factorover ${\mathbb{F}}_p$, with distinct monic irreducible factors. For integer polynomial lifts $f_i$, the distinct prime ideals above $p$ are $P_i=(p,f_i(\theta ))$, andThus the residue degree is $\mathrm{deg}{\bar{f}}_i$, the ramification index is $e_i$, and ${\sum }_i{e}_i\mathrm{deg}{\bar{f}}_i=[K:\mathbb{Q}]$. The index restriction is essential to this direct factorization formula.

For $K=\mathbb{Q}(\sqrt{-2})$, the ring of integers is $\mathbb{Z}[\sqrt{-2}]$, so the index is one and the theorem applies to every prime. Modulo two, $X^2+2=X^2$, giving one prime ideal of ideal norm two. For an odd rational prime, a prime ideal has ideal norm $p$ if the polynomial splits, and ideal norm $p^2$ if it is irreducible. In the former case the supplied condition forces $p\equiv 1$ or $3(\mathrm{mod}8)$; in the latter case every odd square is one modulo eight. ConsequentlyThe odd part of any ideal norm is thus one or three modulo eight, because $3^2\equiv 1$. Multiplication by a power of two leaves possible residues $1,3$ for exponent zero, $2,6$ for exponent one, $4$ for exponent two, and $0$ for exponent at least three. Hence $\boxed{N(I)\not\equiv 5,7(\mathrm{mod}8)}$ for every nonzero integral ideal.

Apply the radial momentum operator to a smooth radial test function. More generally $[f(r),p_r]=i\hslash f^{\prime }(r)$, giving the first two labelled identities. Expanding the square gives the third:The kinetic-energy decomposition is $p^2=p_r^2+L^2/r^2$. A central potential commutes with $L^2$ and $L_z$, so choose simultaneous angular eigenfunctions $Y_{lm}$, with $L^2=l(l+1){\hslash }^2$ and $l=0,1,\dots$. The radial Hamiltonian consequently has the stated centrifugal term. The radial inner product uses measure $r^{2}dr$; adjoints below are taken on compatible regular bound-state domains. The formal radial momentum is not thereby asserted to be a self-adjoint observable on a half-line.

Set $q=l+1$, $W=-q/r+1/(qa)$ and $D=i{p}_r/\hslash$. Since $D^{†}=-D$ formally and $[D,W]=W^{\prime }=q/r^2$,The hydrogen radial Hamiltonian factorization follows by matching the Coulomb coefficient, which fixes the Bohr radiusFor the second pair of labelled relations, the opposite product changes the sign of $W^{\prime }$, henceUsing $[A,A^{†}A]=[A,A^{†}]A$ in the factorization then givesThis Coulomb angular-momentum intertwining means applying $A_l$ to an energy eigenstate either gives zero or an eigenstate of $H_{l+1}$ with the same energy. For a normalized negative-energy state,At fixed $E<0$ this would become negative if raising $l$ continued indefinitely. The bound-state ladder termination by norm positivity occurs at some $L$ with $A_L\psi =0$, givingThe terminating radial function is proportional to $r^L{e}^{-r/[(L+1)a]}$, regular and normalizable; adjoint intertwiners generate lower $l$ values at fixed $n$. Positive-energy scattering states are not subject to this terminating negative-energy argument. Here $m$ is the mass in the supplied Hamiltonian; including nuclear motion replaces it by the reduced mass.

After integrating the overall orientation of a scalar-parent three-body decay, the Lorentz-invariant phase space is constant in two pairwise squared invariant masses. The two-body factorization $d{\Phi }_3=dsd{\Phi }_2(M;\sqrt{s},m_2)d{\Phi }_2(\sqrt{s};m_1,m_3)/(2\pi )$ and the relative-angle Jacobian give the displayed density. The event density includes $|\mathcal{M}{|}^2/(2M)$ and need not be flat: a resonance produces bands. Identical-particle counting supplies its usual global symmetry factor.