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The Riemann-Roch theorem states, for a divisor $D$ on a smooth projective curve of genus $g$,where $\ell (D)=\dim H^0(C,\mathcal{O}(D))$ and $K_C$ is a canonical divisor, of degree $2g-2$. Negative-degree divisors have no nonzero global sections. The theorem's usual formulation here is over an algebraically closed field, as in the curve course.

For a smooth projective algebraic curve, a linear system of divisors comes from a finite-dimensional space $V\subseteq H^0(C,\mathcal{O}(D))$. A basis of sections $s_0,\dots ,s_n$ defines $f(P)=[s_0(P):\cdots :s_n(P)]$, using any local trivialization; a common scalar change does not alter the projective point. A base-point-free linear system makes this defined everywhere, and $f^{\ast }{\mathcal{O}}_{{\mathbb{P}}^n}(1)\cong \mathcal{O}(D)$. Pulling back a hyperplane gives the zero divisor of a linear combination of the sections, an element of the original system.

A system with base divisor $B$ can instead define a map after removing $B$, but its hyperplane pullbacks have class $D-B$. Thus base-point-freeness is needed for the stated original-system property. For a closed immersion, the system must also separate distinct points and tangent directions; equivalently it is a very ample linear system. On a smooth curve the latter tangent condition is separation of the length-two subscheme at each point.

In characteristic two, let $K={\mathbb{F}}_2(t)$ and $L=K(u)$ with $u^2+u=t$. The Artin–Schreier polynomial is irreducible: if $r\in K$ solved it, a finite pole of $r$ would create a pole of even order in $r^2+r$, impossible for $t$; at infinity a pole of $r$ likewise gives even order, while $t$ has order one. A constant $r$ cannot solve it either. Its derivative is one, so this degree-two extension is separable field extension. Adjoining $\sqrt{\alpha }$ in characteristic two is either trivial or purely inseparable field extension, since the derivative of $X^2-\alpha$ vanishes. Hence $\boxed{L\ne K(\sqrt{\alpha })\text{ for every }\alpha \in K}$.

For the second example take $\boxed{K={\mathbb{F}}_{11}}$. Every extension still has characteristic eleven, and $X^{11}-1=(X-1{)}^{11}$ in it. Thus the only eleventh root of unity is one, whose order is one, and there is no primitive eleventh root in any extension.

Take the Vandermonde determinant $\Delta ={\prod }_{i<j}(x_i-x_j)$. Its square is a symmetric polynomial, hence a polynomial in the elementary symmetric polynomials and belongs to $K$. It is nonzero because the variables are algebraically independent. A transposition of two variables fixes every $e_i$ and therefore fixes $K$, but sends $\Delta$ to $-\Delta \ne \Delta$ in characteristic zero. Thus $\boxed{\alpha =\Delta \notin K,{\alpha }^2\in K}$.

Here one must distinguish two conventions for a radical extension. The solvability theorem says that a finite extension with solvable normal-closure Galois group is contained in a tower of radical adjunctions. Under the convention that this containment defines a radical extension, this is the requested assertion.

Let $N/K$ be the finite normal closure and let $m$ be divisible by the orders of its Galois group elements. Adjoin a primitive $m$th root of unity, obtaining $K^{\prime }=K({\zeta }_m)$, which itself is a radical adjunction since ${\zeta }_m^m=1$. The Galois group of $N{K}^{\prime }/K^{\prime }$ embeds in the solvable group of $N/K$, so it is solvable. A composition series with cyclic prime-order quotients gives a tower of cyclic prime-degree extensions above $K^{\prime }$. Each is a radical adjunction, as the following argument shows.

For a cyclic extension $E/F$ of prime degree $p$ containing ${\zeta }_p$, let $\sigma$ generate its Galois group. The $F$-linear operator $\sigma$ satisfies $X^p-1$, a product of distinct linear factors in characteristic zero. Thus it is diagonalizable. Because $\sigma \ne 1$, it has an eigenvector $a\ne 0$ with eigenvalue ${\zeta }_p^j\ne 1$. Then $a^p\in F$ and $a\notin F$, so $E=F(a)$ by primeness of the degree. Applying this at every step produces a radical tower containing $L$.

If “radical extension” instead means that $L/K$ itself must be such a tower, the printed statement is false without this qualification. For example the real cubic subfield of $\mathbb{Q}({\zeta }_7)$ is cyclic and solvable, but contains no nonrational $a$ with $a^n\in \mathbb{Q}$: all three conjugates are real roots of the same equation $X^n=a^n$, which has at most two real roots, while any nonrational element of the cubic field has three distinct conjugates. Consequently no nontrivial radical adjunction can start a tower inside it. Containment in a radical tower is the valid general conclusion.

A one-dimensional continuous group representation is a continuous group homomorphism $\rho :S^1\to {\mathbb{C}}^{\times }$. Its image is compact. If some $|\rho (z)|$ differed from one, its positive or negative powers would be unbounded, so the image lies in $S^1$. Lift $t\mapsto \rho (e^{it})$ to a continuous real argument $a(t)$ with $a(0)=0$. Then $a(s+t)-a(s)-a(t)$ is a continuous integer multiple of $2\pi$ and vanishes at $(0,0)$, hence is always zero. The continuous Cauchy functional equation gives $a(t)=ct$. Periodicity forces $e^{2\pi ic}=1$, so $c=n\in \mathbb{Z}$. Consequently $\boxed{\rho (z)=z^n}$; conversely every such map is a continuous one-dimensional group representation.

Represent a function by its graph of Kuratowski ordered pairs. For $f:\omega \to {\omega }^2$,The bound is attained by $f(n)=\omega n$, which is a cofinal function into ${\omega }^2$. Taking the rank of the set of all such functions adds one at this attained maximum. Thus $\boxed{\mathrm{rk}(({\omega }^2{)}^{\omega })={\omega }^2+1}$. The extra successor is essential: unlike the previous product, this set actually contains a member of rank ${\omega }^2$.

Use the Kuratowski ordered pair $(m,n)=\{\{m\},\{m,n\}\}$. Its rank of a set is $\max (m,n)+2$. Every pair in the Cartesian product has finite rank, but those ranks are unbounded in $\omega$. Therefore $\mathrm{rk}(\omega \times \omega )={\sup }_{m,n<\omega }(\max (m,n)+3)=\boxed{\omega }$.

An ordinal has rank equal to itself. Thus the rank of a set here is $\sup \{\omega +2,\omega +3,\omega +4\}=\boxed{\omega +4}$.

In principal axes, $2E=\sum I_i{\omega }_i^2$ and $L^2=\sum I_i^2{\omega }_i^2$. Substitute the Euler equations for a torque-free rigid body:For the squared angular momentum,Thus both $E$ and $L^2$ are conserved, even though the body-frame components of $L$ generally change.

In a frame centered at the fixed point or centre of mass, a rotating material point has velocity $\dot{r}=\omega \times r$. The identity $|\omega \times r{|}^2=|\omega {|}^2|r{|}^2-(\omega \cdot r{)}^2$ converts the rotational kinetic energy into $T=\frac{1}{2}{\omega }_i{I}_{ij}{\omega }_j$, where the inertia tensor isLikewise integrating $r\times (\omega \times r)$ gives the angular momentum $\boxed{L=I\omega }$. These expressions refer to rotational motion about the chosen origin.

At the stationary mean,The $(1,1)$ entry of the Continuous Lyapunov equation gives $-2\gamma C_{11}+2\lambda =0$, so $C_{11}=\lambda /\gamma$. Its $(1,2)$ entry gives $\gamma C_{11}-(\gamma +\beta )C_{12}-\lambda =0$, hence $C_{12}=0$. Finally $(2,2)$ gives $2\gamma C_{12}-2\beta C_{22}+2\lambda =0$. ThusBoth variances equal their means; the absence of stationary covariance does not mean that the maturation jumps are independent events in the two populations.

The three jump vectors and their rates are $(1,0)$ at rate $\lambda$, $(-1,1)$ at rate $\gamma x_1$, and $(0,-1)$ at rate $\beta x_2$. Summing rate times jump gives the Fokker-Planck drift vector, while summing rate times the outer product of the jump gives the Fokker-Planck diffusion matrix:The maturation event simultaneously decreases larvae and increases adults; this accounts for the negative off-diagonal entries. Since $A$ is affine, its expectation is its value at the mean. Setting both mean derivatives to zero yields $\boxed{⟨x_1⟩=\lambda /\gamma ,⟨x_2⟩=\lambda /\beta }$.

Let $\xi =x-⟨x⟩$ and $C_{ij}=⟨{\xi }_i{\xi }_j⟩$. In the linear noise approximation, $A_i=a_{ik}{\xi }_k$ and $B_{ij}=b_{ij}$. At the stationary mean, the preceding moment identity gives ${\dot{C}}_{ij}=a_{ik}{C}_{kj}+a_{jk}{C}_{ik}+b_{ij}$. Hence the stationary covariance matrix solves the Continuous Lyapunov equationwhich is precisely the requested component formula because $C$ is symmetric.

For a smooth test function $f$, integration by parts in the Fokker-Planck equation givesWe use the decay and absence of boundary flux intended in the question, sufficiently strong to remove the moment-weighted boundary terms. For $f=x_k$, only ${\partial }_{i}f={\delta }_{ik}$ survives, giving $d⟨x_k⟩/dt=⟨A_k⟩$. For $f=x_k{x}_l$, the first derivative is ${\delta }_{ik}{x}_l+{\delta }_{il}{x}_k$ and the second is ${\delta }_{ik}{\delta }_{jl}+{\delta }_{il}{\delta }_{jk}$. The Fokker-Planck diffusion matrix is symmetric, soThe factor one-half cancels the two equal diffusion contributions, including when $k=l$.