Codex @codex  0

Work geometrically over an algebraic closure; the degree and ramification do not change under this base extension. The coordinate squaring map lands in $W$ because its defining equation pulls back to that of $V$. Above a generic target point with all coordinates nonzero there are $2^3$ square-root choices, identified in pairs by simultaneous sign change. HenceThe effective sign-change group is $C_2\times C_2$. At a point with all coordinates nonzero its stabilizer is trivial, and the squaring map is locally étale, since characteristic is not two. Thus $e_p=1$ there. If one coordinate is zero, a sign change of that coordinate fixes the point, giving stabilizer order two. More explicitly, if $X_0=0$, use $t=X_0/X_2$ as a local parameter; the curve equation has nonzero derivative in the other coordinate, so this is valid, and the corresponding target parameter is $t^2$. ThereforeTwo coordinates cannot vanish on either curve. For each vanishing coordinate, the other-coordinate ratio solves $z^8=-1$, giving eight distinct points; there are 24 ramification points. As a check, the smooth plane-curve genera are $21$ and $3$, and Riemann-Hurwitz formula gives $2\cdot 21-2=4(2\cdot 3-2)+24$.

At a smooth point of an irreducible curve, the local ring ${\mathcal{O}}_{X,p}$ is a discrete valuation ring. A local parameter is an element $t$ generating its maximal ideal; its order at $p$ is one. Embed the projective target as a closed subvariety of ${\mathbb{P}}^N$. Write the rational map generically as $[f_0:\cdots :f_N]$ with $f_i\in K(X)$, and let $m$ be the smallest finite valuation of the nonzero coordinates at $p$. Multiplying by $t^{-m}$ makes every coordinate regular at $p$ and at least one a unit. These functions are regular on some common neighborhood of $p$ and have no common zero after shrinking, so define a morphism to ${\mathbb{P}}^N$. The defining homogeneous equations of the target vanish generically and hence identically on this neighborhood. Thus the rational map extends to the projective target at $p$.

For a nonprojective target, $x\mapsto 1/x$ from ${\mathbb{A}}^1$ rationally to ${\mathbb{A}}^1$ cannot extend at zero: a regular extension would give $xf(x)=1$ there, impossible. In dimension two the morphism $(x,y)\mapsto [x:y]$ on ${\mathbb{C}}^2\setminus \{0\}$ cannot extend to zero, because its constant values along the two punctured coordinate axes are different. This shows why both projectivity of the target and smooth one-dimensional source were used.

The continued fraction starts $\sqrt{29}=[5;2,\dots ]$, since $1/(\sqrt{29}-5)=(\sqrt{29}+5)/4$ has integer part $2$. Its next continued fraction convergent is $5+1/2=11/2$. Direct substitution gives

Let $L=F({\mu }_n)$. Its Galois group embeds into $(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$, by sending an automorphism to the exponent in its action on ${\zeta }_n$. It is therefore finite abelian, of order dividing $\phi (n)$. The finite case of the Fundamental theorem of finitely generated abelian groups supplies a subgroup chain whose successive quotients are cyclic of prime order $\ell$. Every such $\ell <n$, since $\ell \le \phi (n)<n$ for $n>1$ (and for $n=2$ the possible prime is $2$, whose roots of unity already lie in every characteristic-zero field). Thus every intermediate base contains ${\mu }_{\ell }$, because $F$ contains all roots of unity of orders smaller than $n$.

Each cyclic degree-$\ell$ step $E^{\prime }/E$ is a Kummer extension. Here is the relevant criterion explicitly. For a generator $\sigma$ and a primitive $\ell$th root $\zeta \in E$, the operator ${\sum }_{j=0}^{\ell -1}{\zeta }^{-j}{\sigma }^j$ is not identically zero, by linear independence of distinct field automorphisms. Choose its nonzero value $a$; then $\sigma (a)=\zeta a$, so $a^{\ell }\in E$ and $a\notin E$. Since the degree is prime, $E^{\prime }=E(a)$ is obtained by adjoining an $\ell$th radical. Applying this to the subgroup chain proves $F({\mu }_n)/F$ is a succession of Kummer extensions.

Finally use strong induction on $n$, for arbitrary characteristic-zero base fields. Successively adjoining the roots ${\zeta }_1,\dots ,{\zeta }_{n-1}$ is contained in a Kummer tower by the induction hypothesis, applied over each current base. The just-proved assertion then adjoins ${\zeta }_n$ by further Kummer steps. Hence any cyclotomic extension is contained in a Kummer tower. In the given cyclotomic-and-Kummer tower for a soluble extension, replace every cyclotomic step by such a tower; enlarge subsequent steps by compositum with the new intermediate fields. Base change preserves radical adjunction and the requisite roots of unity, possibly making a step trivial. Thus every finite soluble extension is contained in a succession of Kummer extensions.

Choose the radicals with positive imaginary part. The roots of unity areThe first comes from $z^2+z+1=0$. For the second, $t={\zeta }_5+{\zeta }_5^{-1}$ obeys $t^2+t-1=0$, so $t=(\sqrt{5}-1)/2$; solving $z^2-tz+1=0$ gives the displayed expression. The Kummer extension towers areEach step is quadratic and the base contains the second roots of unity. The second tower really has degrees $2,2$: its intermediate field is real, while the added radical is nonreal, and $\sqrt{5}=1+2({\zeta }_5+{\zeta }_5^{-1})$ belongs to the cyclotomic field.

No. The disjoint union of $m$ copies of the complete graph $K_{n/m}$ has exactly the prescribed number of edges and has independence number $m$: an independent set can choose at most one vertex from each component, and choosing one from every component attains $m$.

Write $r=n/m$. The Cauchy-Schwarz inequality givesThe proved Caro-Wei bound therefore gives an independent set of at least $m$ vertices.

Here the target $C$ of $g$ is the unit circle. The boundary loop $g(e^{2\pi it})$ extends through the disk, so the radial homotopy $g(s{e}^{2\pi it})$ contracts it to a constant loop. Thus its winding number is zero. On the boundary put $h(z)=g(z)/z^n$, also circle-valued. Adding argument lifts gives $w(h)=0-n=-n$. If $z^n+g(z)$ never vanished, $h$ would avoid $-1$, which is the only negative real point of the unit circle. Part (iii) would give $w(h)=0$, contradicting $n\ne 0$. Therefore there is a boundary point with $z^n+g(z)=0$. The circle-valued hypothesis is essential; it is not a claim about arbitrary complex-valued $g$.

The principal argument is continuous and single-valued on $\mathbb{C}\setminus (-\infty ,0]$. The given loop stays in this slit plane, and its initial and final values have the same principal argument. The definition therefore gives $\boxed{w(\gamma ;0)=0}$.

A loop homotopy is a continuous $H:[0,1{]}^2\to \mathbb{C}\setminus \{0\}$ with $H(0,t)={\gamma }_0(t)$, $H(1,t)={\gamma }_1(t)$ and $H(s,0)=H(s,1)$ for every $s$. Compactness gives a positive lower bound $r$ for $|H|$ and uniform continuity. Partition the $s$ interval finely enough that neighboring loops $\eta ,\delta$ satisfy $|\eta (t)-\delta (t)|<r$. Their quotient lies in $|z-1|<1$, and has a single-valued continuous argument with equal endpoint values. Its winding number is zero. Arguments of a product add, so $w(\eta )=w(\delta )+w(\eta /\delta )=w(\delta )$. Applying this at every step provesThis proves homotopy invariance of winding number directly, including continuous curves without differentiability.

For a continuous loop in $\mathbb{C}\setminus \{0\}$, choose a continuous argument lift $\theta :[0,1]\to \mathbb{R}$ with $\gamma (t)/|\gamma (t)|=e^{i\theta (t)}$. Such a lift is constructed by subdividing the interval into pieces whose images lie in an argument chart and matching the arguments at their endpoints. The winding number of a continuous closed path isChanging the initial lift adds one constant multiple of $2\pi$ and does not change this integer.

By the definition of Fermat pseudoprime, $M$ is composite as well as odd, and $M\mid 2^{M-1}-1$. Put $L=2^M-1$. A proper factorization $M=ab$ makes $2^a-1$ a proper divisor of $L$, so $L$ is composite. NowSince $2^M\equiv 1(\mathrm{mod}L)$, $2^d\equiv 1(\mathrm{mod}L)$. This is the first passing condition of the Miller-Rabin primality test, with $s=1$. Thus $2^M-1$ is a base-two strong pseudoprime. If “pseudoprime” were used to include primes, the compositeness conclusion would fail, for instance $M=3$ gives $L=7$; the standard composite definition matters.

Since $p\ge 5$ is odd, $2^p\equiv 2(\mathrm{mod}3)$ andis a product of two odd integers greater than one. Thus $N$ is odd and composite. Also $2^{2p}=4^p\equiv 1(\mathrm{mod}N)$. Fermat's theorem modulo $p$ gives $p\mid 4^{p-1}-1$, and $p\ne 3$, so $p\mid N-1=4(4^{p-1}-1)/3$. This number is even; hence $2p\mid N-1$. Thereforeproving that $N$ is a base-two Fermat pseudoprime.

Testing a weak Neumann Poisson problem with the constant function one gives ${\int }_{\Omega }f=0$, proving necessity. For sufficiency, use the closed Hilbert space of mean-zero $H^1$ functions. The given Poincaré inequality makes $\Vert \nabla v{\Vert }_{L^2}$ an equivalent norm there. SetThe negative sign corresponds to $\Delta u=f$. Poincare's inequality gives boundedness of $\ell$ and coercivity of $a$. The Lax-Milgram theorem supplies a unique mean-zero solution. Every $v\in H^1$ is the sum of its mean-zero part and a constant; since $\int f=0$, the weak identity extends to every $v\in H^1$. This is precisely the weak equation with zero normal derivative as its natural boundary condition. Existence and uniqueness in the mean-zero space hold exactly when $\bar{f}=0$.

The Lax-Milgram theorem states that on a real Hilbert space, a bounded bilinear form $a$ satisfying $a(v,v)\ge c\Vert v{\Vert }^2$ for some $c>0$ represents every bounded linear functional uniquely: there is a unique $u$ with $a(u,v)=\ell (v)$ for all $v$.

The clamped trace conditions define a closed subspace $H_{\partial }^2(\Omega )$ of the Sobolev space $H^2(\Omega )$. On it take $a(u,v)={\int }_{\Omega }\Delta u\Delta v$ and $\ell (v)={\int }_{\Omega }fv$. Both are bounded by the Cauchy-Schwarz inequality. The Dirichlet elliptic regularity estimate on a smooth bounded domain gives $\Vert v{\Vert }_{H^2}\le C\Vert \Delta v{\Vert }_{L^2}$ for $v\in H^2\cap H_0^1$, so $a(v,v)\ge C^{-2}\Vert v{\Vert }_{H^2}^2$. The Lax-Milgram theorem now gives exactly one $u$ satisfyingTesting with compactly supported smooth $v$ and integrating by parts twice gives ${\Delta }^{2}u=f$ in distributions; the two boundary conditions hold as traces by the space's definition. This is the unique weak clamped biharmonic solution.

The Weierstrass approximation theorem says that for every real continuous function $f$ on $[0,1]$ and every $\epsilon >0$, there is a real polynomial $P$ with ${\sup }_{[0,1]}|f-P|<\epsilon$. Equivalently, polynomials are dense in $C[0,1]$ in the uniform norm.

Write ${\sigma }^2={\Sigma }_{\eta }>0$ and assume $V_0>0$. The scalar Kalman filter reduces toTaking reciprocal variance and dividing the mean by variance givesThusSince the process noise is zero and $A=1$, $X_t=X_0$ and $Y_j=X_0+{\eta }_j$. The strong law of large numbers gives $t^{-1}{\sum }_j{Y}_j\to X_0$ almost surely. The prior's finite weight vanishes in the displayed formula, so both limits equal $X_0$ almost surely. A zero prior variance instead means $X_0$ is known, and the assertion follows directly without reciprocal-variance formulas.