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The Riemann-Roch theorem statesOn a genus-one curve $K\sim 0$. If $D$ has degree zero, then $D+P_0$ has degree one, and Riemann--Roch gives $\ell (D+P_0)=1$. Its unique nonzero section has an effective divisor of degree one, say $P$, soIf also $D\sim Q-P_0$, then $P\sim Q$; a nonconstant function with divisor $P-Q$ would define a degree-one map to ${\mathbb{P}}^1$, forcing the genus to be zero. Hence $P=Q$.

This identifies $X$ with ${\mathrm{Pic}}^0(X)$ by $P\mapsto [P-P_0]$ and defines $P+Q=R$ throughFor the stated cubic, the line through $A=(2,1)$ and $B=(-2,-1)$ is $y=x/2$. Its third intersection is $(1/4,1/8)$, so reflection in the $x$-axis gives

Because $P_0$ is an inflection point, a line section is linearly equivalent to $3P_0$. A point $P$ is an inflection point exactly when some line has intersection divisor $3P$, equivalentlyUnder the group-law identification this is precisely $3(P-P_0)=0$, so the inflection points are exactly the three-torsion points.

In local coordinates around $p\in R$ and $f(p)\in S$, a nonconstant analytic map has the formAfter shrinking the chart, $h$ has an analytic $m$th root, so a change of coordinate makes the map $z\mapsto z^m$. It maps small discs onto neighborhoods of the origin; hence every nonconstant analytic map of Riemann surfaces is open. If $R$ is compact and $S$ connected, $f(R)$ is compact and therefore closed, and it is also nonempty and open. Thus $f(R)=S$.

For $f:\mathbb{D}\to \mathbb{D}$ with $f(0)=0$, the removable-singularity theorem makes $g=f/z$ analytic. On $|z|=r$, $|g(z)|=|f(z)|/r\le 1/r$, and the maximum principle gives the same bound throughout ${\mathbb{D}}_r$. Applying the Schwarz lemma to an automorphism $h$ and to $h^{-1}$ shows that every automorphism fixing zero is

If $F:\mathbb{C}\to \mathbb{C}$ is an analytic isomorphism fixing zero, continuity of $F^{-1}$ implies that the preimage of every closed disc is compact. Hence $|F(z)|\to \infty$ as $|z|\to \infty$, and defining $F(\infty )=\infty$ gives an analytic isomorphism of the Riemann sphere.

Consequently an automorphism of $\mathbb{D}$ fixing zero does extend to $\mathbb{C}$. A general disc automorphism need not: for $a\ne 0$,has a finite pole and cannot extend to an entire isomorphism.

For $s>n/2$ and $u\in H^s({\mathbb{R}}^n)$, Fourier inversion and Cauchy--Schwarz giveThe last integral is finite exactly when $s>n/2$. Thus the sobolev embedding theorem gives

To see why the endpoint relevant to $H^1({\mathbb{R}}^3)$ fails, choose a smooth cutoff $\chi$ supported near the origin and equal to one there, and setNear zero, $|u{|}^2$ contributes ${\int }_0^1{r}^{3/2}dr$, while $|\nabla u{|}^2$ contributes a constant multiple of ${\int }_0^1{r}^{-1/2}dr$. Both are finite, so $u\in H^1({\mathbb{R}}^3)$, but $u$ is unbounded.

The real Stone-Weierstrass theorem says that if $X$ is compact Hausdorff and $A\subset C(X,\mathbb{R})$ is a subalgebra containing the constants and separating points, then $A$ is uniformly dense in $C(X,\mathbb{R})$.

Let $\overline{A}$ be the uniform closure. Polynomial approximation to $\sqrt{t}$ on a bounded interval shows that $|f|\in \overline{A}$ whenever $f\in \overline{A}$. Hence $\overline{A}$ is closed underGiven $h\in C(X)$, separation and the constants provide, for each $x,y$, a function in $A$ agreeing with $h$ at $x$ and $y$. Compactness first combines finitely many such functions by minima to obtain one that agrees at $x$ and lies below $h+\epsilon$ everywhere; a second finite cover and maxima produces $f\in \overline{A}$ with $|f-h|<\epsilon$. Thus $h\in \overline{A}$.

The space $C_b(\mathbb{R})$ is Banach because a uniform Cauchy sequence converges uniformly to a bounded continuous function. Compactness is essential: the algebra of bounded continuous functions having finite limits at both $+\infty$ and $-\infty$ contains constants and separates points, for example using $\tanh x$, but its uniform closure has the same limiting property. It cannot uniformly approximate $\sin x$, so it is not dense in $C_b(\mathbb{R})$.

Because $K$ is a subcomplex, $\partial C_k(K)\subseteq C_{k-1}(K)$, sois well defined and squares to zero. The short exact sequence of chain complexesgives the long exact sequenceTaking $L=\partial {\Delta }^{n+1}$ and using the contractibility of ${\Delta }^{n+1}$ givesFor $0<k<n$, exactness makes $H_k(K)$ a quotient of a subgroup of $H_{k+1}(L,K)$, whose rank is at mostThe same stated inequality for $k=n$ follows from $H_n(K)\hookrightarrow H_n(L)\cong \mathbb{Z}$. Finally, an $n$-cycle in the boundary sphere has equal signed coefficients on all $n$-faces. A proper subcomplex omits an $n$-face, so that coefficient and hence every coefficient is zero. Therefore $H_n(K)=0$.

Write $N={\prod }_i{p}_i^{e_i}$ with odd primes $p_i$. The Jacobi symbol iswhere each factor on the right is a Legendre symbol. The supplementary law isFor coprime positive odd $m,n$, quadratic reciprocity givesApplying this with $m=3$ givesFinally, factor $d=2^{e}m$ with $m$ odd. The supplementary laws for $-1$ and $2$, followed by reciprocity on every odd prime factor of $m$, express $(-d/a)$ solely in terms of the residue class of the positive odd integer $a$ modulo $d$ when $d\equiv 0$ or $3(\mathrm{mod}4)$. If $a\equiv b(\mathrm{mod}d)$, all those factors, including the zero cases, agree; hence