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A subgroup of the group of fractional ideals prime to a modulus of a number field, containing the principal ray subgroup. It is the inverse image of a subgroup of the ray class group. The existence theorem of global class field theory realizes it as the kernel of an Artin map, and identifies its quotient with the corresponding Galois group. This is the ideal-theoretic class-field meaning of congruence subgroup.

For a positive integer $m$, the finite ray class group of $\mathbb{Q}$ is $(\mathbb{Z}/m\mathbb{Z}{)}^{\times }/⟨-1⟩$: the unit group permits either sign of a rational generator. Its ray class field is the maximal real subfield of the cyclotomic field $\mathbb{Q}({\zeta }_m)$. Including the real place requires positive principal generators and gives the full group $(\mathbb{Z}/m\mathbb{Z}{)}^{\times }$ and full cyclotomic field. Thus modulus $(5)$ gives $\mathbb{Q}(\sqrt{5})$, while $(5)\infty$ gives $\mathbb{Q}({\zeta }_5)$. The Artin reciprocity map sends a prime $\ell \nmid m$ to the exponent $\ell$, taken modulo $\{\pm 1\}$ in the finite case. Ramification at real places is possible only when those places are included in the modulus of a number field.

The ring of integers is $\mathbb{Z}[\omega ]$, where ${\omega }^2+\omega +1=0$, and its class number is one by the Minkowski bound for ideal classes. Modulo seven it is ${\mathbb{F}}_7\times {\mathbb{F}}_7$. The six roots of unity reduce to the kernel of the product map on the two residue unit groups, so the ray class group is ${\mathbb{F}}_7^{\times }$, of order six. The relative Artin symbol on $K({\zeta }_7)$ is exponentiation by the norm of a fractional ideal modulo seven. It kills the principal ray subgroup, and $[K({\zeta }_7):K]=6$, proving that this is the ray class field.

For $K=\mathbb{Q}(\sqrt{-6})$, reduction of $\mathbb{Z}[\sqrt{-6}]$ modulo $(\sqrt{-6})$ gives $\mathbb{Z}/6\mathbb{Z}$. Its unit group is covered by the units $\{1,-1\}$ of $K$. The ray class exact sequence therefore identifies the ray class group with the ordinary ideal class group, of order two. Consequently this ray class field is the Hilbert class field of Q of square root minus six. To verify its lack of ramification, the integral basis candidate $1,\sqrt{2},b,\sqrt{2}b$, where $b=(1+\sqrt{-3})/2$, has discriminant $8^2(-3{)}^2=576$. The relative discriminant formula also forces the field discriminant to be divisible by $(-24{)}^2=576$. Equality proves that the relative discriminant ideal is trivial.

For an algebraic group defined over ${\mathbb{F}}_q$, its Frobenius morphism acts as entrywise $q$th powers in a standard matrix realization. Its differential is zero, and its fixed-point group is finite. A conjugate $F^{\prime }=\phi F{\phi }^{-1}$ defines an isomorphic rational structure. This group morphism must be distinguished from a field Frobenius endomorphism alone and from the convention called geometric Frobenius in Galois theory, which is inverse to the arithmetic generator on field points.

For simply connected $X$, rationalization of a topological space is a map $r:X\to X_{\mathbb{Q}}$ to a rational space inducing ${\pi }_n(X_{\mathbb{Q}})={\pi }_n(X)\otimes \mathbb{Q}$. It is also a rational homology equivalence. Its universal property is the bijection $r^{\ast }:[X_{\mathbb{Q}},Z]\to [X,Z]$ for every rational space $Z$. A Postnikov tower constructs it by replacing each fibre group with its rational tensor product and transporting the Postnikov invariant through the resulting cohomology isomorphism, then taking the homotopy inverse limit.

On an irreducible variety, every nonempty open subset is dense, so a germ determines a rational section. For a locally free sheaf, this map is injective: a local section which vanishes on a dense open has all its regular coefficient functions zero, hence is zero. Local freeness is important; a section of a torsion sheaf can vanish generically without having zero germ.

For a locally free sheaf on an irreducible variety, the stalk inclusion identifies sections regular near $P$ inside its rational sections. The quotient records the obstruction to regularity at that point. This definition does not assume a nonsingular point or a discrete valuation parameter.

A vector is unstable if zero lies in the Zariski closure of its orbit. For complex $S{L}_n$, this is equivalent to an algebraic one-parameter subgroup taking it to zero as $t\to 0$. In the subgroup's weight decomposition, exactly the positive-weight components can occur in such a vector.

Suppose $a_n>0$ and, for all sufficiently large $n$, $a_{n+1}/a_n\le 1-c/n$ with fixed $c>0$. Choose an initial index $N>c$. Since $\log (1-x)\le -x$ for $0<x<1$, summing logarithmic ratios gives $\log (a_n/a_N)\le -c{\sum }_{k=N}^{n-1}1/k\le -c\log (n/N)$. Thus $a_n\le a_N{N}^c{n}^{-c}$. For $c>1$, the comparison test for series with a P-series proves convergence. This supplies a useful refinement when a limiting ratio of one makes the ordinary ratio test inconclusive.

A rate function is a lower semicontinuous map $I:E\to [0,\infty ]$. It is good if every sublevel set $\{I\le M\}$ is a compact set. This property controls escapes to infinity and makes the minimizations in the contraction principle for large deviations attain their finite infima. On ${\mathbb{R}}^d$, a lower semicontinuous rate function tending to infinity with the norm is good.

A local path action integrates a nonnegative convex cost g of the path's instantaneous input rate, assigning infinity to paths that are not absolutely continuous. When $g(\mu )=0$, the typical-rate path has zero cost. On any interval with prescribed total input, Jensen inequality makes a constant derivative minimize its cost. With a queue workload constraint this leads to constant-rate burst paths and the linear workload rate for a local convex action. Such a formula alone does not establish goodness in an unspecified path topology; goodness must be checked or assumed as part of the input large-deviation principle.

For $\gamma >1$ and positive mass density and pressure, let $\xi =x/t$ in one-dimensional adiabatic compressible flow. The smooth equations are $(u-\xi ){\rho }^{\prime }+\rho u^{\prime }=0$, $(u-\xi )u^{\prime }+p^{\prime }/\rho =0$ and $(u-\xi )p^{\prime }+\gamma p{u}^{\prime }=0$. Away from a constant branch, their coefficient determinant forces $\xi =u+\sigma c$ with $\sigma =\pm 1$ and adiabatic sound speed $c=\sqrt{\gamma p/\rho }$. The entropy equation gives $p=K{\rho }^{\gamma }$, and combining continuity with $c^{\prime }/c=(\gamma -1){\rho }^{\prime }/(2\rho )$ givesThus $u=[2\xi +(\gamma -1)J]/(\gamma +1)$, $c=\sigma (\gamma -1)(\xi -J)/(\gamma +1)$, $\rho =(c^2/(\gamma K){)}^{1/(\gamma -1)}$ and $p=K{\rho }^{\gamma }$, on intervals where $c>0$. The specific entropy is constant. Together with uniform branches, these exhaust smooth positive-gas self-similar solutions; a contact discontinuity is not a smooth fan.

With ${\beta }^2=\Omega R^2/D$ and $\epsilon =2V_{s}R/(3B\Omega D)$, the first angular harmonic of the steady cell conservation equation satisfies $a^{\prime \prime }+a^{\prime }/r-a/r^2-i{\beta }^{2}a=0$ and $a^{\prime }(1)=-1$. The regular solution is $a=-I_1(qr)/[q{I}_1^{\prime }(q)]$, $q=e^{i\pi /4}\beta$. At large $\beta$ it gives the displayed layer with width $1/\beta$. The exact Modified Bessel function of the first kind expression remains regular at the axis, unlike extrapolation of the boundary-layer formula there.

Let $A,B:U\to V$ be linear maps of finite rank of a linear map. Their sum satisfiesThe image of a linear map $A+B$ lies in $\mathrm{im}A+\mathrm{im}B$, so the dimension formula for a sum of subspaces proves the upper bound. Apply that bound to $A=(A+B)-B$ and to $B=(A+B)-A$ to obtain the lower bound. Finite rank ensures that subtraction in the displayed inequality is defined.

Both extremes occur for square matrices. For prescribed ranks $r\ge s$, take $A=\mathrm{diag}(I_r,0)$ and $B=-\mathrm{diag}(I_s,0)$ to obtain matrix rank $r-s$. For the largest possible matrix rank, use diagonal supports whose union has size $\min (n,r+s)$ and assign positive real diagonal entries, obtaining that matrix rank without cancellation.

For independent $N(0,1)$ increments, a random walk has $S_j\sim N(0,j)$. If $B$ is a bounded measurable set, its occupation time probability obeys $\mathbb{P}(S_j\in B)\le |B|/\sqrt{2\pi j}$. Consequently $\mathbb{E}{\sum }_j{1}_B(S_j)/j<\infty$, and Tonelli's theorem makes this series finite almost surely. Kronecker lemma then gives the displayed zero occupation natural density. This also shows why its empirical occupation probabilities cannot converge to a probability measure simultaneously on all bounded intervals and the whole real line: continuity of probability from below would contradict the preserved total mass one.

An upward skip-free random walk has integer-valued increments bounded above by one. Arbitrarily large downward jumps are permitted, but a first crossing of a nonnegative integer level $k$ from below must land exactly at $k$. This removes the overshoot from the stopped likelihood ratio under exponential tilting and permits the exact geometric maximum of an upward skip-free random walk. For a real threshold $k$, the first crossing is the crossing of $\lceil k\rceil$, so the overshoot-free formula must use that integer level.

Place six positions on a cycle and distinguish two labels. At each step choose one of the fifteen unordered pairs of distinct positions uniformly and transpose their contents. This convention excludes identity moves. The shorter circular distance between the distinguished labels takes values $1,2,3$. Cycle symmetry and exchangeability of the other labels make this distance a Lumped Markov chain.

Counting transpositions gives the transition matrix and Stationary distributionThe stationary weights also follow by counting unordered position pairs: six pairs have distance one, six have distance two, and three are opposite. If $\Pi$ has every row equal to $\pi$, then $P=\frac{1}{3}I+\frac{2}{3}\Pi$ and $P^n=3^{-n}I+(1-3^{-n})\Pi$. Starting with adjacent labels, the Probability of opposite labels after $n$ steps is therefore $(1-3^{-n})/5$.

The parent Random transposition shuffle uses a different convention: two positions are sampled independently, allowing an identity move with Probability $1/6$. Under that convention the distance transition matrix is $\tilde{P}=\frac{1}{6}I+\frac{5}{6}P=\frac{4}{9}I+\frac{5}{9}\Pi$, and the opposite-label Probability is $(1-(4/9{)}^n)/5$. Explicitly stating the sampling convention prevents a spurious change in the convergence rate.

For a random sum of independent claims, conditioning on the count gives $M_S(t)=\mathbb{E}[M_X(t{)}^N]$ and therefore the displayed cumulant-generating function. Write $a_j={\kappa }_{N,j}$ and $b_j={\kappa }_{X,j}$. The chain rule gives ${\kappa }_{S,1}=a_1{b}_1$, ${\kappa }_{S,2}=a_1{b}_2+a_2{b}_1^2$, and ${\kappa }_{S,3}=a_1{b}_3+3a_2{b}_1{b}_2+a_3{b}_1^3$. In raw severity moments $m_j=\mathbb{E}{X}^j$, the third cumulant is $a_1{m}_3+3(a_2-a_1)m_1{m}_2+(a_3-3a_2+2a_1)m_1^3$. Positive exponential moments justify two-sided differentiation near zero; finite ordinary moments alone suffice for the corresponding left derivatives when claims are positive.

Let $Y\mid \Lambda \sim \mathrm{Poisson}(e^{\eta +\Lambda })$, with $\Lambda \sim N(0,{\tau }^2)$. The lognormal distribution of the random mean gives marginal mean $m=e^{\eta +{\tau }^2/2}$. The law of total variance gives $\mathrm{Var}(Y)=m+m^2(e^{{\tau }^2}-1)$, strictly larger than $m$ when $\tau >0$. Thus independent observation-level random effects produce overdispersion even though the conditional Poisson distributions have variance equal to their means.