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.

The image of $A+B$ lies in $\mathrm{im}A+\mathrm{im}B$. The dimension formula for a sum of subspaces gives the upper bound. Applying it to $A=(A+B)-B$ and then interchanging $A,B$ gives the lower bound. Both extremes are sharp: cancelling diagonal blocks gives the difference of ranks; diagonal maps with supports covering the whole space and positive coefficients give full rank when the sum of prescribed ranks is at least the dimension.

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 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 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.

In a random transposition shuffle on a six-cycle, the shorter circular distance between two distinguished labels takes values $1,2,3$. The symmetry of the cycle and exchangeability of other labels make this a lumped Markov chain. Counting the fifteen possible transpositions gives transition matrix $P=\frac{1}{15}\left(\begin{array}{ccc}9& 4& 2\\ 4& 9& 2\\ 4& 4& 7\end{array}\right)$. Its stationary distribution is $(2/5,2/5,1/5)$ and $P=\frac{1}{3}I+\frac{2}{3}\Pi$, where each row of $\Pi$ is this stationary distribution. Consequently the probability of opposite labels after $n$ steps from adjacent labels is $(1-3^{-n})/5$.

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.

Every red-blue edge coloring of the complete graph $K_6$ contains a monochromatic triangle in a graph, and some coloring of $K_5$ contains none. Together the five-cycle Ramsey coloring and six-vertex monochromatic triangle argument giveThe threshold is exact: one carefully chosen coloring establishes the lower bound, while the upper bound must handle every coloring.

A familiar interpretation uses six people and colors each pair red if they know one another and blue if they do not. Assuming the relation is symmetric and every pair falls into exactly one case, there are either three mutual acquaintances or three mutual strangers. The conclusion concerns all three pairs within the group, not merely two relationships through a common person. There is no prediction about which color wins or which three people form the group.

The small size makes exhaustive verification practical. The complete graph $K_5$ has $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ edges, giving $2^{10}=1024$ red-blue colorings of labeled vertices; exactly 12 have no monochromatic triangle in a graph. They are the choices of a red cycle graph of length five, whose blue complement graph is another five-cycle. The complete graph $K_6$ has $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$ edges, giving $2^{15}=32768$ colorings. Checking all $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$ triples in every coloring finds none without a monochromatic triangle in a graph; in fact the minimum number is two. These counts can be reproduced by assigning one bit to each unordered pair and testing whether the three bits of each triple agree. The monochromatic-triangle counting formula gives a separate explanation of the stronger minimum-two conclusion.

Let a power series ${\sum }_{n\ge 0}{a}_n{z}^n$ have finite radius of convergence $R$. If its terms were bounded in complex modulus at a point $z$ with $|z|>R$, say $|a_n{z}^n|\le M$, choose $w$ with $R<|w|<|z|$. ThenThe comparison test for series with a convergent geometric series would give absolute convergence at $w$, contradicting its position outside the convergence disc. Hence the terms are unbounded at every such $z$. The argument also works when $R=0$ and gives a stronger obstruction than the usual term test for divergence.

Inside both convergence discs, the triangle inequality gives absolute convergence of the sum. Suppose $R<S$ and the sum had radius $T>R$. At any modulus between $R$ and $\min (T,S)$, subtracting the absolutely convergent second series from the absolutely convergent sum would give absolute convergence of the first series outside its radius of convergence, a contradiction. Interchanging the two series covers $S<R$.