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For a Dollar investor, the Euro bank account has Dollar value $e^{r_{E}t}/Y_t$. After Dollar discounting, it is $e^{(r_E-r_D)t}/Y_t$. Its conditional mean growth exponent is $r_E-r_D+{\sigma }^2-\mu$, so the martingale requirement givesWith equal rates, $Y^{-1}$ is a martingale under $Q_D$ by part (a), whereas $Y$ has positive drift ${\sigma }^2$ and is not a martingale when $\sigma \ne 0$. In the degenerate zero-volatility case, both exchange rates are constant and both are martingales; the asserted nonmartingale conclusion implicitly requires a genuinely risky exchange rate.

The Dollar bank account is a tradable asset whose Euro value is $Y_t{e}^{r_{D}t}$. Under the Euro equivalent martingale measure, discount this value by $e^{r_{E}t}$. The result is a martingale precisely when $\mu +r_D-r_E=0$, givingWhen the rates coincide, $\mu =0$ and part (a) proves that the exchange rate itself is a martingale under $Q_E$.

The independent Gaussian increments of Brownian motion give $E[Y_t\mid {\mathcal{F}}_s]=Y_s{e}^{\mu (t-s)}$; the exponential is integrable at every finite time. ThereforeSimilarly $E[Y_t^{-1}\mid {\mathcal{F}}_s]=Y_s^{-1}{e}^{({\sigma }^2-\mu )(t-s)}$, so $\boxed{Y^{-1}\text{ is a martingale}⟺\mu ={\sigma }^2}$. This uses actual conditional expectations, not only the local stochastic differential equation.

Combining the eigenvalue cases givesThis agrees with the interior drift: the switching phase spends proportions $\beta /(\alpha +\beta )$ and $\alpha /(\alpha +\beta )$ in $C$ and $W$, giving level drift $(\lambda \beta -\mu \alpha )/(\alpha +\beta )$. The harmonic calculation establishes recurrence even at the zero-drift boundary, rather than inferring it from drift alone.

Here $\vartheta <1$ is equivalent to $\mu \alpha <\lambda \beta$, which gives $b<1$ from its explicit formula. Any nonnegative solution is of the formsince its second component at level zero must be one. Letting $i\to \infty$ forces $a\ge 0$ for nonnegativity. The choice $a=0$ is nonnegative, and subtracting it from any other such solution gives $a[(1,1{)}^T-{\vartheta }^i(b,1{)}^T]\ge 0$. Therefore minimality selectsIn particular the return probability after leaving $W_0$ is $b<1$, so the irreducible chain is transient. These formulas are the hitting probabilities for a velocity-switching Markov ladder.

If $\vartheta >1$, the general level solution is a constant multiple of $(1,1{)}^T$ plus a multiple of ${\vartheta }^i(b,1{)}^T$. Boundedness of probabilities removes the growing component, and $h_{W_0}=1$ then fixes the constant to one. If $\vartheta =1$, the nonzero Jordan part produces linear growth unless the initial vector lies in the one-dimensional eigenspace; the same conclusion follows. ThusStarting at $W_0$, the chain leaves to $C_0$ and returns with probability one. An irreducible chain with a recurrent state has all states recurrent, so the chain is recurrent in both cases.

Since one eigenvalue is one, the other is the determinant. Direct calculation givesIndeed the first row requires $(1+\alpha /\lambda -\vartheta )b=\alpha /\lambda$, and the second row then follows from the characteristic equation. At $\vartheta =1$, $b=1$ and the eigenvectors coincide; the nonidentity matrix has a repeated eigenvalue and a Jordan block.

Condition on the first jump, putting $h_{W_0}=1$. The hitting probabilities obeyThe relevant characterization is that a hitting probability is the minimal nonnegative harmonic extension of its target-boundary value. Thus among nonnegative solutions of these equations with $h_{W_0}=1$, choose the componentwise least one. Other harmonic solutions need not be probabilities. This also identifies the solution without imposing an unjustified boundary condition at infinity.

The arrows in the PDF give $C_i\to C_{i+1}$ at rate $\lambda$, $W_i\to W_{i-1}$ at rate $\mu$ for $i\ge 1$, and switches $C_i\to W_i$ and $W_i\to C_i$ at rates $\alpha$ and $\beta$. All rates are positive. One can travel upwards in level in the $C$ phase and downwards in the $W$ phase, switching at either end. Hence any state reaches any other through a finite positive-rate path: the continuous-time Markov chain is irreducible. Its total jump rates are uniformly bounded, so it is also nonexplosive.

Let $c$ be the center of the containing disk and take arclength around the closed curve. The periodic function $\Phi (s)=|\gamma (s)-c{|}^2/2$ has a maximum. At a maximum $s_0$,By the Cauchy-Schwarz inequality and the containing radius, $1\le |\gamma (s_0)-c||{\gamma }^{\prime \prime }(s_0)|\le r|k(s_0)|$. ThereforeClosedness provides an interior maximum on the periodic parameter domain. Without it, the straight segment $\gamma (s)=(s,0)$ for $-r<s<r$ lies in the disk but has signed curvature identically zero, giving the required counterexample.

The planar planar isoperimetric inequality is $4\pi |\Omega |\le |\partial \Omega {|}^2$, with equality exactly for a disk. For one boundary component of length $L$, take its unit-speed periodic parametrization $\gamma =(x,y)$ and subtract its mean position. Green's area formula and the Cauchy-Schwarz inequality giveApply the periodic Wirtinger inequality to each mean-zero coordinate: $\int |\gamma {|}^2\le (L/2\pi {)}^2\int |{\gamma }^{\prime }{|}^2$. Since unit speed gives the final integral $L$, it follows that $|A_{\mathrm{signed}}|\le L^2/(4\pi )$. For several boundary components, orient each with the domain on its left. Green's formula, followed by the triangle inequality, givesThis proves the stated inequality, including disconnected domains or holes. Equality permits only one component; equality in the two integral inequalities makes its coordinates first sine-cosine harmonics with the corresponding orthogonal rotation relation. Unit speed then makes the curve a circle, so its domain is a disk. Conversely disks attain equality.

Parametrize the regular curve by arclength $s$. Its unit tangent is $T={\gamma }^{\prime }$, and its curvature is $\kappa =|T^{\prime }|$. Where $\kappa >0$, define the principal normal $N=T^{\prime }/\kappa$ and binormal $B=T\times N$. The torsion of a space curve is $\tau =-B^{\prime }\cdot N$. Orthonormality and differentiation give the Frenet-Serret formulasFor a general parameter, $\kappa =|\dot{\gamma }\times \ddot{\gamma }|/|\dot{\gamma }{|}^3$ and $\tau =\det (\dot{\gamma },\ddot{\gamma },\stackrel{...}{\gamma })/|\dot{\gamma }\times \ddot{\gamma }{|}^2$. The normal and torsion formulas require nonzero curvature; at zero-curvature points this particular Frenet frame need not be defined.

Units have norm $\pm 1$. Norm $-1$ is impossible modulo three. The least positive unit greater than one isIndeed its norm is one. A smaller positive norm-one unit would have $0<y=({\epsilon }^{\prime }-1/{\epsilon }^{\prime })/(2\sqrt{30})<2$, forcing $y=1$, but then $x^2=31$. Multiplying an arbitrary positive unit by powers of $\epsilon$ reduces it to $[1,\epsilon )$, so minimality proves that all units are $\pm {\epsilon }^n$.

The equation with right side $+5$ is impossible modulo three, since it would require $x^2\equiv 2(\mathrm{mod}3)$. For norm $-5$, let $\beta =5+\sqrt{30}$, so ${\beta }^2=5\epsilon$ and $\beta /\epsilon =\sqrt{30}-5$. For unit reduction for quadratic norm equations, change the sign of $x+y\sqrt{30}$ if necessary to make it positive, then multiply by a power of the fundamental unit to place it in $[\beta /\epsilon ,\beta )$. Its conjugate is $-5$ divided by itself, soThe maximum on this interval occurs at its endpoints. Thus $y=1$, and $x^2=25$; only $x=-5$ lies in the half-open interval. Reversing the reduction gives all solutions:Every displayed element has norm $-5$, and the reduction proves completeness, including both signs of $x$ and $y$.

Put $s=\sqrt{30}$. Since $30\equiv 2(\mathrm{mod}4)$, the ring of integers is $\mathbb{Z}[s]$ and its field discriminant is $120$. All three primes ramify:For example ${\mathfrak{p}}_p^2\subseteq (p)$ and both ideals have norm $p^2$, proving equality. The Minkowski bound for ideal classes is $\sqrt{120}/2=\sqrt{30}<6$. Thus the ideal class group is generated by the classes of these prime ideals; an ideal of norm four contributes only the already principal ${\mathfrak{p}}_2^2$.

The elements $6+s$ and $5+s$ have norms $6$ and $-5$. Their ideal factorizations give ${\mathfrak{p}}_2{\mathfrak{p}}_3=(6+s)$ and ${\mathfrak{p}}_5=(5+s)$. This also explains the printed hint: $N((6+s)/2)=3/2$, a fraction, not the integer $32$. Therefore $[{\mathfrak{p}}_3]=[{\mathfrak{p}}_2]$ and $[{\mathfrak{p}}_5]=1$, leaving at most two ideal classes. The class of ${\mathfrak{p}}_2$ is nontrivial: a principal generator would have norm $\pm 2$, but $x^2-30y^2=\pm 2$ would require $x^2\equiv 2$ or $3(\mathrm{mod}5)$, impossible. Hence

A primitive sixth root of unity satisfieswhich is irreducible over $\mathbb{Q}$ because its discriminant is $-3$. For the other polynomial choose $\alpha =i{3}^{1/6}$, so ${\alpha }^6=-3$ and ${\alpha }^3=-i\sqrt{3}$. Then ${\zeta }_6=(1-{\alpha }^3)/2\in \mathbb{Q}(\alpha )$, so this field already contains all six roots. The Eisenstein criterion at $3$ proves $x^6+3$ irreducible, giving a splitting field of degree six.

Put $r(\alpha )={\zeta }_6^2\alpha$ and $s(\alpha )={\zeta }_6\alpha$. The first automorphism fixes ${\zeta }_6$ and has order three. The second sends ${\alpha }^3$ to its negative, so sends ${\zeta }_6$ to ${\zeta }_6^{-1}$; hence $s^2=1$ and $srs=r^{-1}$. These six distinct transformations exhaust the Galois group. ThereforeThe presence of sixth roots of unity inside the splitting field is why this particular binomial has degree six rather than a generic twelve-degree splitting field.

An automorphism of $L({\zeta }_n)$ fixing $K({\zeta }_n)$ fixes $K$. Because $L/K$ is a normal extension, it preserves $L$, so restriction gives a homomorphismIts kernel fixes both $L$ and ${\zeta }_n$, hence every element of their compositum. Therefore restriction is injective, proving the asserted subgroup identification. Normality is the key reason the restriction maps into automorphisms of $L$ rather than just embeddings into a larger field.

If $\theta =0$, the splitting field is $K$ and its Galois group is trivial, hence cyclic. Otherwise choose $\alpha$ with ${\alpha }^n=\theta$. The roots are $\alpha {\zeta }_n^j$, so the splitting field is $K(\alpha )$ because ${\zeta }_n\in K$. The roots are distinct since the characteristic does not divide $n$, making this a Galois extension. Each automorphism satisfies $\sigma (\alpha )=c_{\sigma }\alpha$ with $c_{\sigma }\in ⟨{\zeta }_n⟩$. Because that root-of-unity group is fixed by all automorphisms, $c_{\sigma \tau }=c_{\sigma }{c}_{\tau }$. Thus $\sigma \mapsto c_{\sigma }$ is an injective homomorphism into a cyclic group. The splitting-field Galois group is cyclic, of order dividing $n$.

The intended closed-term witness property for quantifier-free existential sentences needs at least one closed term. Under this hypothesis, suppose the set $\Gamma =\{\neg \varphi (t):t\text{ a closed term}\}$ were consistent. By the Godel completeness theorem, it has a model $M$. Take the substructure $M_0$ whose elements are the interpretations of closed terms. It is nonempty and closed under every function symbol. Atomic formulas and their negations have the same truth values in $M_0$ as in $M$, and induction extends this to every quantifier-free formula.

The provable sentence $\exists x\varphi (x)$ is true in $M_0$ by soundness. Its witness is $t^{M_0}$ for some closed term $t$, so $M\models \varphi (t)$, contradicting $M\models \Gamma$. Hence $\Gamma$ is inconsistent. The compactness theorem gives a finite inconsistent subset, so completeness and classical propositional reasoning yieldWithout a constant or nullary function, the printed assertion is false as written: in the equality-only language there are no closed terms, but $\exists x(x=x)$ is provable. Then $\Gamma$ is empty and consistent. One may repair the usual theorem by assuming a nonempty set of closed terms or adjoining a fresh constant; the latter proves a witness disjunction in the enlarged language, not the missing closed terms in the original one.

For each $n$, let ${\sigma }_n$ assert the existence of $n$ distinct elements. If a theory $T$ has arbitrarily large finite models, every finite subset of $T\cup \{{\sigma }_n:n\ge 1\}$ has a model. By the compactness theorem, this extended theory has a model, necessarily infinite. Since fields of size $2^j$ are arbitrarily large, no first-order theory can have exactly the finite fields of characteristic two as its models.

Suppose instead a finite conjunction $\chi$ axiomatized exactly the infinite fields of characteristic two. Let $T_2$ be the field axioms with $1+1=0$. Then $T_2\cup \{\neg \chi \}$ would have exactly the finite characteristic-two fields as models, contradicting the preceding conclusion. Equivalently, compactness of $T_2\cup \{{\sigma }_n\}\models \chi$ would force some finite size threshold, but arbitrarily large finite fields still fail $\chi$. The infinite characteristic-two fields are not finitely axiomatizable.

The Godel completeness theorem for first-order logic states that $\Gamma \models \varphi$ implies $\Gamma ⊢\varphi$; together with soundness, semantic consequence and derivability coincide. Equivalently, every syntactically consistent first-order theory has a model. The compactness theorem states that a set of first-order sentences has a model if and only if every finite subset has a model. These statements use nonempty first-order structures.