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Subtract the sphere equation from the last equality to obtain $x_2^2=1/2$. Thus the equality locus is the disjoint union of two sphereseach a smooth two-dimensional manifold. The two equality gradients are independent there: $x_2\ne 0$ and at least one of $x_1,x_3,x_4$ is nonzero.

On this locus the inequality becomes $1-2x_4^2\le a$, orSince $0<a<1$, this cuts out two closed caps on each sphere. At the cap boundary, $x_1^2+x_3^2=a/2>0$, so the restricted differential of $x_4$ is nonzero. The regular level set theorem for a sublevel set therefore gives a manifold with boundary. Explicitly it consists of four disks, with four boundary circles, and

A smooth $n$-dimensional manifold with boundary has charts onto relatively open subsets of the half-space ${\mathbb{H}}^n=\{u_n\ge 0\}$, with smooth transition maps extending smoothly locally across the boundary. A point is a boundary point when its last coordinate in such a chart is zero; this condition is independent of chart. One way to see the independence is that a local diffeomorphism between half-spaces preserves their interior and hence their topological boundary.

Restrict each boundary chart to $u_n=0$ and drop the last coordinate. It maps the boundary onto an open subset of ${\mathbb{R}}^{n-1}$, and the restricted transitions are smooth diffeomorphisms. These charts, with the inherited Hausdorff and countability properties, make $\partial X$ a smooth manifold without boundary. Therefore

The ideal $(f)$ is prime, since an irreducible element of the polynomial ring is prime. By the Hilbert Nullstellensatz, it is the full vanishing ideal of $Z(f)$. Its Zariski tangent space at a point $a$ is therefore the kernel of the single linear form $d{f}_a$, and has dimension $n-1$ or $n$.

At least one partial derivative $f_j$ is nonzero. In characteristic zero this is immediate for a nonconstant polynomial. In characteristic $p$, if all partial derivatives vanished, all exponents would be multiples of $p$. Since an algebraically closed field is perfect, $f$ would be a $p$th power of a nonconstant polynomial, contradicting irreducibility.

The nonzero $f_j$ cannot vanish everywhere on $Z(f)$: the Nullstellensatz would give $f_j^r\in (f)$, hence $f\mid f_j$, contradicting $\mathrm{deg}{f}_j<\mathrm{deg}f$. Thus some point has $d{f}_a\ne 0$, and the minimum tangent-space dimension is

For an irreducible affine variety, its dimension is the minimum of its Zariski tangent space dimensions:The minimum is attained on the smooth locus. Equivalently $T_{x}X$ is the dual of ${\mathfrak{m}}_x/{\mathfrak{m}}_x^2$, or the simultaneous kernel of the differentials of the defining equations at $x$. If “variety” permits reducible sets, apply this definition to each irreducible component and take the maximum; taking the minimum over an arbitrary reducible set with components of unequal dimension would be incorrect.

The parametrization $t\mapsto (t^3,t^2)$ maps ${\mathbb{A}}^1$ to the cusp. Off the origin its inverse is $t=x/y$, since $(x/y{)}^2=y$ and $(x/y{)}^3=x$ on $x^2=y^3$ with $y\ne 0$. This is an isomorphism between dense open subsets, so the cusp is birational to ${\mathbb{A}}^1$.

At the origin the differential of the defining polynomial is zero, giving a two-dimensional Zariski tangent space. The tangent space of ${\mathbb{A}}^1$ at every point is one-dimensional. An isomorphism induces tangent-space isomorphisms, so the cusp is not isomorphic to ${\mathbb{A}}^1$. The rational inverse cannot extend regularly through the singular point.

The decimal expansion gives $M={\sum }_{i=0}^s{a}_{i}10^i$. In modular arithmetic, $10\equiv -1(\mathrm{mod}11)$, and henceAn integer is divisible by eleven exactly when its congruence class is zero, proving both directions of the alternating-digit test.

For congruence classes in $\mathbb{Z}/N\mathbb{Z}$, defineThese operations are independent of representatives. If $a^{\prime }=a+Nr$ and $b^{\prime }=b+Ns$, then $(a^{\prime }+b^{\prime })-(a+b)=N(r+s)$ and $a^{\prime }{b}^{\prime }-ab=N(rb+sa+Nrs)$. Thus changing representatives leaves both results in the same congruence class. The usual integer laws descend to make this a commutative ring with zero $[0]$ and identity $[1]$.

Directly tracking the supplied permutations givesThus every group element reduces to $a^k$ or $b{a}^k$, $0\le k<6$. These are twelve distinct elements: $a$ has order six, while $b$ is not a power of $a$ because it exchanges the two six-element supports that every power of $a$ preserves. Therefore $|G|=12$. This is the dicyclic group with parameter three; the calculations give its order-twelve dicyclic character table.

Conjugation by $b$ inverts powers of $a$, while conjugation by $a$ changes the exponent in $b{a}^k$ by two modulo six. Conjugation by $b$ preserves that exponent's parity. The complete conjugacy classes areTheir sizes are $1,1,2,2,3,3$.

The normal subgroup $N=⟨a^2⟩$ has order three. The quotient is cyclic of order four, generated by $bN$, since $(bN{)}^2=aN$. Inflating its characters gives ${\chi }_j(b)=i^j$, ${\chi }_j(a)=(-1{)}^j$, for $j=0,1,2,3$. These are all the linear characters: in any one-dimensional representation, the relation $ba{b}^{-1}=a^{-1}$ forces the image of $a$ to have square one, so $a^2$ is in the kernel.

In the displayed class order, the full character table isTo construct the last two rows, put $\zeta =e^{i\pi /3}$ and, for $k=1,2$, setThe matrices obey all three relations. The distinct eigenlines of ${\rho }_k(a)$ are exchanged by ${\rho }_k(b)$, so the representation is irreducible. Its traces on $a^m$ are $2\cos (km\pi /3)$ and on $b{a}^m$ are zero, producing the rows above. Their weighted inner products are orthonormal, and the degree-square sum $4\cdot 1^2+2\cdot 2^2=12$ proves completeness.

For a representation $\rho$ of $G/N$, its inflation of a group representation to $G$ is $g\mapsto \rho (gN)$, with character $\tilde{\chi }(g)=\chi (gN)$. Inflation preserves irreducibility because the invariant subspaces do not change.

For irreducible complex characters ${\chi }_i$, the row character orthogonality areThe column relations are ${\sum }_i{\chi }_i(g)\overline{{\chi }_i(h)}=0$ for nonconjugate $g,h$, and $|C_G(g)|$ when $g,h$ are conjugate. Class sizes convert the row sum into its usual weighted character-table form.

For $p\nmid n$, the reduction of the cyclotomic polynomial ${\Phi }_n$ has precisely the primitive $n$th roots as its roots. If $p\mid {\Phi }_n(x)$, the residue class of $x$ has multiplicative order $n$ in ${\mathbb{F}}_p^{\times }$, so Lagrange's theorem gives $n\mid p-1$. Conversely, if $n\mid p-1$, cyclicity of the multiplicative group supplies an element of order $n$, which is a root of ${\Phi }_n$; any integer representative supplies $x$. HenceThree explicit examples are ${\Phi }_3(X)=X^2+X+1$ with $x=2,p=7$; ${\Phi }_4(X)=X^2+1$ with $x=2,p=5$; and ${\Phi }_6(X)=X^2-X+1$ with $x=3,p=7$. Each prime is coprime to the relevant $n$ and has the stated congruence.

First make the irreducible polynomial monic; for a nonmonic polynomial restore its leading coefficient in front of the product below. If $\alpha$ is one root, $K={\mathbb{F}}_q(\alpha )$ has degree $n$ and cardinality $q^n$. Thus every element satisfies $x^{q^n}=x$. The Frobenius automorphism $x\mapsto x^q$ fixes ${\mathbb{F}}_q$ and permutes the roots. If the orbit of $\alpha$ had size $d<n$, its orbit product would be a degree-$d$ polynomial fixed by Frobenius, hence with coefficients in ${\mathbb{F}}_q$, contradicting irreducibility. ThereforeThe roots are distinct, since $X^{q^n}-X$ has derivative $-1$. They all lie in $K$ and include its generator $\alpha$, so the splitting field is ${\mathbb{F}}_{q^n}$. Frobenius has order $n$ there; a degree-$n$ extension has at most $n$ automorphisms, so

TakeIt is a nonzero polynomial, but every element of the finite field ${\mathbb{F}}_q$ satisfies ${\alpha }^q=\alpha$. Thus it vanishes on every point of ${\mathbb{F}}_q^n$. Formal nonzero polynomials and nonzero polynomial functions need not coincide over a finite field.

The diagonal diagonal Ramsey number $R(s)$ is the least $n$ such that every red-blue edge coloring of $K_n$ contains a monochromatic $K_s$. More generally let $R(s,t)$ require a red $K_s$ or blue $K_t$. The base cases are $R(1,t)=R(s,1)=1$. Given $R(s-1,t)+R(s,t-1)$ vertices, select one vertex. Either it has at least $R(s-1,t)$ red neighbors or at least $R(s,t-1)$ blue neighbors. Applying the corresponding smaller problem and, when needed, adjoining that selected vertex givesInduction and Pascal's identity now prove existence and

Write ${\hat{u}}_0(k)=\hat{u}(k,0)$ and retain the left-to-right orientation of $L$. Substituting the previous relation into the deformed inverse Fourier transform gives the data-only representationThe formally remaining integral is ${\int }_L{e}^{ikx}\hat{u}(-k+i\beta ,t)dk$. Above $L$, its reflected argument lies strictly in the lower half-plane, where the transform is analytic. At large radius, $L$ approaches the rays of angles $\pi /4$ and $3\pi /4$. Close it by the upper arc between these rays. Along that arc $\mathrm{Im}k$ is bounded below by a positive multiple of its radius, whereas the reflected transform is bounded by ${\int }_0^{\infty }|u(s,t)|ds$. Thus the arc integral is bounded by a constant times $R{e}^{-cRx}$ and tends to zero for $x>0$. There are no enclosed singularities, so Cauchy integral theorem provesThis is the needed closure argument; it does not assume the unknown transform vanishes. The representation holds for smooth decaying compatible data, and extends to weaker data by the usual limiting heat-kernel interpretation.

For $k=a+ib$, $\mathrm{Re}\omega =a^2-b^2+\beta b$. The upper contour is therefore the graphoriented from left to right. Between the real axis and $L$, $b\ge 0$ and $\mathrm{Re}\omega \ge 0$. Hence $e^{ikx-\omega (t-s)}$ is bounded by $e^{-bx}$ for $0\le s\le t$. The boundary transforms are entire in $k$; with the standard smooth, compatible boundary data, integration by parts in the time integral gives the required polynomial bounds. Closing the intervening sectors and using exponential decay for $x>0$ deforms their real-line integrals onto $L$. Small endpoint truncations, followed by dominated limits, handle the sector boundaries.

For $k^{\prime }=-k+i\beta$, $\omega (k^{\prime })=\omega (k)$ and $\mathrm{Im}{k}^{\prime }\le 0$ on $L$. Thus the already valid transformed equation at $k^{\prime }$ saysIt eliminates the unknown boundary derivative through

Rapid decay allows two integrations by parts in the half-line Fourier transform. They giveSolving this scalar inhomogeneous equation by an integrating factor yieldsThe initial transform is holomorphic for $\mathrm{Im}k<0$ and continuous on the real axis under the given decay. The time integrals are entire functions of $\omega$ for fixed $t$ and smooth boundary data.

This identity is the half-line advection-diffusion global relation.