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A martingale is an adapted Lebesgue integrable process with $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$. A stopping time $T$ has $\{T\le n\}\in {\mathcal{F}}_n$ for every $n$. A precise bounded-time optional sampling theorem says: if $S\le T\le N$ are stopping times, then $\mathbb{E}[M_T\mid {\mathcal{F}}_S]=M_S$. To prove it, writeFor $A\in {\mathcal{F}}_S$, the event $A\cap \{S\le k<T\}$ lies in ${\mathcal{F}}_k$. Every summand multiplied by its indicator therefore has expectation zero. Summing proves the conditional expectation identity. All terms are Lebesgue integrable since there are finitely many. Unbounded stopping requires an additional hypothesis such as uniform integrability of the stopped martingale; it is not justified solely by almost-sure finiteness.

The martingale convergence theorem ensures a finite almost-sure limit if ${\sup }_n\mathbb{E}|M_n|<\infty$. Uniform integrability also gives convergence in $L^1$. In particular bounded or nonnegative martingales satisfy the almost-sure convergence criterion. A symmetric random walk is a martingale but cannot have a finite limit, since each increment is $\pm 1$ and does not tend to zero. Some hypothesis is therefore necessary.

If there are $j$ pessimists, the count increases with probability $(K-j)j/[K(K-1)]$ and decreases with the identical probability. The remaining outcomes leave it unchanged, so $P_n=j_n/K$ is a bounded martingale. It converges almost surely. From any nonunanimous state, there is a fixed positive probability, bounded uniformly over the finitely many states, of reaching unanimity within $K-1$ steps by letting each other individual copy a chosen seed. Repeating blocks shows eventual absorption has probability one. Thus the finite voter-model consensus probability isThe last equality follows from bounded convergence and the constant martingale expectation. The population eventually becomes wholly pessimistic or wholly optimistic, with probabilities equal to their respective initial proportions.

A divisor on a complex manifold, specialized to a compact Riemann surface, is a finite sum $D={\sum }_p{n}_{p}p$ with integer coefficients, and $\mathrm{deg}D={\sum }_p{n}_p$. Two divisors are linearly equivalent when $D^{\prime }=D+(h)$ for a nonzero meromorphic function $h$, whose principal divisor on a complex manifold records zero orders minus pole orders. DefineMultiplication by $1/h$ gives an isomorphism $L(D)\to L(D^{\prime })$, because $(f/h)+D^{\prime }=(f)+D$. Its inverse is multiplication by $h$, proving $\ell (D^{\prime })=\ell (D)$.

On the Riemann sphere move $P$ to infinity by a Möbius coordinate. Meromorphic functions on the sphere are rational; those with no finite poles and a pole of order at most one at infinity are exactly $az+b$. Thus $\ell (P)=2$, without using Riemann–Roch.

The Riemann-Roch theorem for a compact connected genus-$g$ surface is $\ell (D)-\ell (K-D)=\mathrm{deg}D+1-g$, where $K$ is a canonical divisor. Choose $D=(g+1)P$. Since the second dimension is nonnegative, $\ell (D)\ge 2$. Constants span only one dimension, so a nonconstant $f$ exists whose only possible pole is at $P$ with order at most $g+1$. The degree of its holomorphic map to the sphere is the total pole order. Therefore

A bounded operator on a complex Hilbert space is self-adjoint when $T=T^{\ast }$, equivalently $⟨Tx,y⟩=⟨x,Ty⟩$. The spectral theorem for compact self-adjoint operators says its nonzero spectrum consists of real eigenvalues of finite multiplicity, with at most countably many and with accumulation possible only at zero. Their eigenspaces are mutually orthogonal; their closed span is the orthogonal complement of the kernel. On that span the operator has the norm-convergent spectral action $Tx={\sum }_j{\lambda }_j⟨x,e_j⟩e_j$.

Take $H={\ell }^2(\mathbb{N})$ and $T{e}_n=e_n/n$. It is self-adjoint and has infinite-dimensional range. Truncation to the first $N$ coordinates is a finite-rank operator and differs from $T$ in norm by $1/(N+1)$, so $T$ is compact.

The resolvent set comprises those $\lambda$ for which $T-\lambda I$ has a bounded everywhere-defined inverse; its complement is the spectrum, and the point spectrum consists of eigenvalues. In this exampleFor $\lambda$ outside the displayed closed set the inverse is diagonal with entries $(1/n-\lambda {)}^{-1}$ uniformly bounded, proving membership of the resolvent. Each $1/n$ has eigenvector $e_n$. Zero is not an eigenvalue, but $\Vert T{e}_n\Vert \to 0$ prevents a bounded inverse; also $(1/n{)}_n\in {\ell }^2$ has no preimage since that would be the non-square-summable constant sequence. This justifies every asserted spectral case.

Identify the pointed torus with ${\mathbb{R}}^2/{\mathbb{Z}}^2$ with $p$ at the origin. Its fundamental group is ${\mathbb{Z}}^2$, generated by the two coordinate loops. A homeomorphism fixing $p$ induces an automorphism of this group, givingFunctoriality makes it a homomorphism. It is nontrivial, indeed onto: every integer invertible matrix defines a pointed torus homeomorphism by $x\mapsto Ax$ modulo the Euclidean lattice.

If $f_{\ast }=I$, lift $f$ to a continuous $F:{\mathbb{R}}^2\to {\mathbb{R}}^2$ with $F(0)=0$. Its deck-transformation law is $F(x+n)=F(x)+n$ for every $n\in {\mathbb{Z}}^2$, because the induced map on coordinate loops is identity. Thenis well defined: adding $n$ changes the bracketed expression by exactly $n$. It is a continuous pointed homotopy from $f$ to identity. Hence every kernel element is homotopic to the identity, although the intermediate maps need not be homeomorphisms.

The supplied integral basis differs unimodularly from $1,\zeta ,{\zeta }^2,{\zeta }^3$, since $1=-\zeta -{\zeta }^2-{\zeta }^3-{\zeta }^4$. With $f(t)={\Phi }_5(t)=(t^5-1)/(t-1)$, the field discriminant is $\mathrm{Norm}(f^{\prime }(\zeta ))$, the degree-four sign being positive. At a nonidentity fifth root,soThe Minkowski bound for ideal classes gives an integral ideal in every ideal class with norm at most $[3/(2{\pi }^2)]\sqrt{125}<2$. Its positive integer norm must be one, so it is the whole integer ring. Every class is trivial, proving all ideals are principal.

For $1\le n\le 4$, $u_n=(1-{\zeta }^n)/(1-\zeta )=1+\zeta +\cdots +{\zeta }^{n-1}$ is an algebraic integer. The numerator has the same norm five as the denominator, because its conjugates permute the nonidentity fifth roots. Thus $\mathrm{Norm}(u_n)=1$, proving it is a unit. Multiplying the four numerator factors gives five, henceThe ideal $P=(1-\zeta )$ has norm five; explicitly reduction modulo it sets $\zeta =1$ and gives the quotient ${\mathbb{F}}_5$. Thus it is prime and $(5)=P^4$: five is totally ramified, with ramification index four and residue degree one. A rational prime can ramify only if it divides the field field discriminant, so no rational prime other than five ramifies, and there are no other ramified prime ideals.

Expand the finite Euler product into convergent geometric series. Every positive integer $n\le x$ has all its prime factors at most $x$, soThe harmonic series diverges, hence $P(x)\to 0$. If the sum of reciprocal primes were finite, then the inequality $-\log (1-t)\le 2t$ for $0\le t\le 1/2$ would giveuniformly in $x$, contradicting $P(x)\to 0$. Thus the sum of prime reciprocals diverges. Unique factorization justifies the expansion; no estimate for the distribution of primes is required. This is the Euler proof that the sum of reciprocals of primes diverges.

Take $u\in L\setminus K$, whose irreducible polynomial is $t^2+ct+d$. Since the characteristic is not two, $x=2u+c$ satisfies $x^2=c^2-4d=a\in K$ and $L=K(x)$. Irreducibility makes $a$ nonzero and nonsquare. If another quadratic extension is $K(y)$ with $y^2=b$, an isomorphism sends $x$ to $s+ty$, with $t\ne 0$. Squaring gives $a=s^2+t^{2}b+2sty$. Linear independence of $1,y$ forces $s=0$, so $a=t^{2}b$ and $b/a$ is a square. Conversely, if $b/a=q^2$, sending $x$ to $y/q$ gives an isomorphism. If $b$ itself is a square, the target is just $K$ and cannot be isomorphic; the square-ratio condition also fails.

The multiquadratic field $F$ is the splitting field of ${\prod }_i(t^2-a_i)$, a product of separable quadratic polynomials: each has distinct roots because $a_i\ne 0$ and the characteristic is not two. Even if factors repeat, this splitting field is a separable normal extension, and therefore Galois. Each automorphism sends $x_i$ to $\pm x_i$, and these signs determine it, embedding its Galois group into $(\mathbb{Z}/2{)}^n$. Every subgroup of this elementary abelian group is $(\mathbb{Z}/2{)}^m$ for some $m\le n$.

For a quadratic intermediate field, its fixing subgroup $H$ has index two and is the kernel of a nontrivial one-dimensional character $\chi :G\to \{\pm 1\}$. The coordinate sign characters span the dual of $G$: a linear functional on the subspace $G\subseteq {\mathbb{F}}_2^n$ extends to the ambient space by extending a basis. Hence $\chi ={\prod }_{i\in I}{\chi }_i$ for some subset $I$. Set $y={\prod }_{i\in I}{x}_i$. Then $\sigma (y)=\chi (\sigma )y$, so $y$ is fixed by $H$ but not by all of $G$, and $y^2={\prod }_{i\in I}{a}_i\in K$. Thus $K(y)$ is quadratic and equals the fixed field of $H$. This constructs all quadratic subfields of a multiquadratic extension: