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The cofinite filter on an infinite set consists of all subsets whose complements are finite.

In a quadratic extension $L/K$, an element is integral exactly when its relative trace and norm lie in ${\mathcal{O}}_K$.

The trace form of a number field $K$ is the symmetric bilinear form

The pairing uses the field trace of the product, rather than taking a trace of each factor separately.

For a number field $K$ and an algebraic integer $\alpha \in {\mathcal{O}}_K$, the criterion isHere the absolute field norm means the field norm down to $\mathbb{Q}$; its absolute value is one. Necessity follows from $N(\alpha )N({\alpha }^{-1})=1$ with both factors integers. Conversely, multiplication by $\alpha$ on an integral basis is a matrix with integer entries and determinant $\pm 1$. Its inverse also has integer entries, and applying that inverse to $1$ gives ${\alpha }^{-1}\in {\mathcal{O}}_K$. The integrality hypothesis matters: $\alpha =(3+\sqrt{2})/(3-\sqrt{2})=(11+6\sqrt{2})/7$ has field norm one, but is not in $\mathbb{Z}[\sqrt{2}]$.

In $\mathbb{Z}[\sqrt{2}]$, the criterion becomes $a^2-2b^2=\pm 1$. Thus $(1+\sqrt{2}{)}^{-1}=\sqrt{2}-1$ and $(3+2\sqrt{2}{)}^{-1}=3-2\sqrt{2}$ are again algebraic integers. These units produce infinite families of solutions to the Pell equation.

If $x=y^2$ in a number field, then $N(x)=N(y{)}^2$; a norm that is not a square in the base field therefore obstructs square roots.

For a finite number field $L/\mathbb{Q}$ with a rational basis ${\omega }_1,\dots ,{\omega }_n$, the norm form is the homogeneous polynomial

An embedding of $K({\alpha }_1,\dots ,{\alpha }_i)$ extends across ${\alpha }_{i+1}$ by sending it to a root of its transformed minimal polynomial. There are between one and the relative degree many choices, with equality for a separable extension.

Distinct field embeddings ${\sigma }_1,\dots ,{\sigma }_n:L\to \Omega$ are linearly independent over $\Omega$ as functions on $L$. Thus a nontrivial linear combination ${\sum }_i{a}_i{\sigma }_i$ cannot vanish on every element of $L$.

For the proof, choose a vanishing relation with the fewest nonzero coefficients and normalize one coefficient to one. If ${\sigma }_1\ne {\sigma }_m$, choose $y$ with ${\sigma }_1(y)\ne {\sigma }_m(y)$. Evaluating at $xy$ and subtracting ${\sigma }_m(y)$ times the relation evaluated at $x$ eliminates its $m$th term but leaves a nonzero first coefficient, contradicting minimality.

A connected Feynman diagram has a path between every pair of its vertices. Normalized correlation functions discard disconnected vacuum bubbles, while full correlators can still factor into disconnected components carrying external insertions.

The loop order of a connected Feynman diagram is the number $L=I-V+1$ of independent momentum cycles, where $I$ is the number of internal lines and $V$ the number of vertices. Each loop introduces an unconstrained momentum integral.

A tree-level diagram has no loops and gives the leading classical contribution allowed by the interaction vertices.

Feynman rules translate each diagram into momentum-space factors for propagators, vertices, external states, momentum conservation and loop integration.

A ferromagnetic magnon is a quantum of a collective spin wave about ferromagnetic order. In a uniform isotropic exchange ferromagnet, slowly varying misalignment costs less energy than rapid variation. Its long-wavelength energy is typically quadratic, $E(\mathbf{k})\simeq D{k}^2$, and its angular frequency is $E/\hslash$.

A one-magnon state in a fully polarized magnet reduces total spin projection by one quantum while spreading that change coherently across the sample. It is therefore different from assigning each site an independent classical tilt. The ferromagnetic spin chain gives an explicit cosine dispersion relation whose small-wavenumber limit is quadratic. An applied magnetic field or symmetry-breaking anisotropy can introduce a nonzero excitation gap, while interactions between magnons matter when their population is no longer dilute. The mode picture is most useful when the ordered reference and low-excitation approximation are justified.

Easy-plane anisotropy makes spin orientations within one plane energetically preferable to orientations normal to it. For example, a classical single-site term $D(S^z{)}^2$ with $D>0$ penalizes out-of-plane components and leaves the $xy$ plane degenerate. At sufficiently low energies, an in-plane angle can then be the main orientational degree of freedom.

The resulting approximate angular description connects a magnet to the XY model, but the microscopic form matters. For a quantum spin one-half, $(S^z{)}^2$ is proportional to the identity, so that single-site term alone does not generate orientational anisotropy. Anisotropic exchange interactions provide another route. Stating the spin size and Quantum Hamiltonian prevents transferring a classical mechanism to a quantum model where it is ineffective. Residual in-plane anisotropy can further select a finite set of directions rather than a continuous circle.

A Fermi gas is an idealized collection of noninteracting fermions. In equilibrium, each single-particle state has mean occupation $f(E)=[e^{(E-\mu )/(k_{B}T)}+1{]}^{-1}$. At zero temperature, states below the chemical potential are filled and those above it are empty.

For a fixed density, the Pauli exclusion principle forces occupation of nonzero momenta even at zero temperature, giving a Fermi energy and degeneracy pressure. A dilute high-temperature limit approaches classical statistics when occupations are small. Electrons in a simple metal and weakly interacting trapped fermionic atoms can motivate the model, while interactions, band structure and confinement determine how accurately it applies. A Fermi gas supplies a reference distribution, rather than a claim that every fermionic material is noninteracting.

For a Fermat prime $p=2^{2^k}+1$, the multiplicative group modulo $p$ has order $2^{2^k}$. Its $\phi (p-1)=(p-1)/2$ primitive roots are all quadratic nonresidues; since there are also $(p-1)/2$ nonresidues, the two sets coincide.

Lenz's law expresses the minus sign in Faraday's law of induction: the induced electromotive force drives a passive resistive loop's electric current in the direction whose own magnetic field opposes the imposed change in magnetic flux. It opposes the change of flux, not invariably the external magnetic field. Choose an oriented surface and its positive boundary direction first, find the sign of $d{\Phi }_{\mathrm{ext}}/dt$, and then use the right-hand rule to translate the signed electric current into a physical direction. ^[ref].

In the fixed-loop example in Faraday's law of induction, out-of-page magnetic flux increases at $0.0060Wb/s$. With counterclockwise positive, $I=-0.0030\mathrm{A}$ is clockwise, producing an in-page magnetic field through the loop. Reverse the ramp to $B(t)=0.40\mathrm{T}-(0.30T/s)t$ on the same one-second interval: now $\mathcal{E}=+0.0060\mathrm{V}$ and $I=+0.0030\mathrm{A}$. The counterclockwise electric current produces an out-of-page magnetic field, supporting the decreasing external magnetic flux. A steady out-of-page magnetic field with fixed loop geometry gives no induced electromotive force in either direction.

For the increasing-area example in motional electromotive force, clockwise electric current runs down the right-hand rod. Its magnetic force on a current-carrying wire points left, opposing the imposed motion to the right. Writing $I_{\mathrm{mag}}=B\ell v/R$ for the current magnitude, the power required to maintain constant speed isThus the sign agrees with conservation of energy: the external drive supplies the Joule heating. The magnetic forces do no work on individual charges; mechanical work by the drive supplies the electrical energy. This also explains the usefulness of electromagnetic braking of a sliding rod and the terminal speed of a falling conducting loop. ^[ref].

The sign of induced electromotive force alone does not determine an arbitrary circuit's instantaneous electric current. Retaining constant self-inductance $L$ gives $LdI/dt+RI=-d{\Phi }_{\mathrm{ext}}/dt$, so a pre-existing electric current can persist while the induced drive opposes its change. The distinction matters in the lossless limit of magnetic loop braking. The resistive toy examples use $L=0$ as an approximation, not a general consequence of Lenz's law.

For a film of thickness $h$ flowing down a slope of angle $\alpha$, an upslope surface stress of magnitude $\tau$ givesThe surface velocity, total flux, and basal shear reverse atrespectively.

The extremal number $\mathrm{ex}(n,H)$ is the largest number of edges in an $n$-vertex graph containing no subgraph isomorphic to $H$.

A differential form $\alpha$ is closed when $d\alpha =0$.