The product ${\prod }_j{G}_j$ of the finite groups with their discrete topology is a topological group under coordinatewise multiplication and inversion. The compatibility equations defining $G={\underleftarrow{\lim }}_j{G}_j$ are preserved by both operations, so $G$ is a subgroup. Their restrictions to the subspace topology on $G$ are continuous. Hence $G$ with its standard inverse-limit topology is a topological group.

Give each nonempty finite set $X_j$ the discrete topology. The product space $X={\prod }_j{X}_j$ is compact by the Tychonoff theorem. For each $i\le j$, the compatibility condition $f_{ij}(x_j)=x_i$ defines a closed subset $C_{ij}\subseteq X$.

These sets have the finite intersection property. Indeed, for finitely many conditions choose an index $k$ above every index occurring in them, choose any $x_k\in X_k$, and use the transition maps from $k$ to define all required coordinates; choose the remaining coordinates arbitrarily. Compactness therefore givesThis is the nonemptiness theorem for inverse limits of finite sets.

For an inverse system of sets $(X_j,f_{ij})$ indexed by a directed set $J$, the inverse limit is the set of compatible tuplesIts projection to $X_j$ sends $(x_i{)}_i$ to $x_j$.

LetThe extension $K_2/K$ is the unramified quadratic extension. Since ${\mathbb{F}}_{p^2}^{\times }$ contains all $(p^2-1)$st roots of unity, $L/K_2$ is a cyclic, tamely and totally ramified extension of degree $p^2-1$, with automorphisms $u\mapsto \zeta u$.

The Frobenius automorphism of ${\mathbb{F}}_{p^2}/{\mathbb{F}}_p$ extends by fixing $u$ and conjugates $\zeta$ to ${\zeta }^p$. Hence $L/K$ is Galois, its inertia group isand its residue-field Galois group isThe tame ramification also gives $G_1=1$.

Let $G=\mathrm{Gal}(L/K)$ and normalize $v_L$. The lower ramification groups arewith $G_{-1}=G$. In particular $G_0$ is the inertia group, and $G_1$ is the wild inertia group. When the extension is totally ramified, the uniformizer criterion for lower ramification groups permits the equivalent test on one uniformizer.

The reduction of $P=(0,0)$ is a nonidentity point of the group $\tilde{E}({\mathbb{F}}_3)$ of prime order seven. Hence $7P$ reduces to the identity and lies in $E_1({\mathbb{Q}}_3)$. Part i shows that $P$ has infinite order, so $7P\ne O$. Every nonidentity point in $E_1({\mathbb{Q}}_3)$ has formal parameter $z=-x/y$ of positive valuation and thereforehas negative valuation. Thus $7P$ does not have integral coordinates.

Let $E_r({\mathbb{Q}}_2)$ be the $r$th term in the filtration of elliptic-curve points over a local field. Reduction givesof order four, whilehas order two. The formal logarithm is injective on $E_2({\mathbb{Q}}_2)$ and identifies it with an additive subgroup of ${\mathbb{Q}}_2$, so $E_2({\mathbb{Q}}_2)$ is torsion-free. A finite subgroup of $E({\mathbb{Q}}_2)$ therefore injects into $E({\mathbb{Q}}_2)/E_2({\mathbb{Q}}_2)$, whose order is $4\cdot 2=8$. Thus $|E({\mathbb{Q}}_2{)}_{\mathrm{tors}}|$ divides $8$.

For good reduction at $p$, the reduction of an elliptic curve gives a mapwhose kernel is its formal group of an elliptic curve, and the torsion in that kernel is $p$-primary. Hence the prime-to-$p$ part of any torsion subgroup injects into $\tilde{E}({\mathbb{F}}_p)$.

At $p=2$, the odd part of $E(\mathbb{Q}{)}_{\mathrm{tors}}$ must divide $4$, so it is trivial. The torsion group is therefore a $2$-group. At $p=3$, all of this group has order prime to $3$ and injects into a group of order $7$, so it too is trivial. Thus

After passing to a subsequence, weak compactness and the Rellich-Kondrachov compactness theorem giveFor fixed $w\in H^1(U)$, the Sobolev inequality gives $w\in L^6(U)$, soMeanwhile ${\partial }_j{u}_{n_k}\rightharpoonup {\partial }_{j}u$ in $L^2(U)$. Pairing this weak convergence with the strong convergence of the products, or equivalently using the weak-strong product convergence lemma, yields

Linearity in $u$ follows from linearity of the weak derivative and Lebesgue integration. By the Holder inequality, the three-dimensional Sobolev inequality, and the preceding $L^3$ interpolation estimate,Thus $\Phi$ is a linear functional and a continuous linear map on $H^1(U)$.

The Holder inequality interpolates between $L^2(U)$ and $L^6(U)$:Taking cube roots and applying the three-dimensional Sobolev inequality $H^1(U)\hookrightarrow L^6(U)$ givesThis is the H1 L3 interpolation inequality in three dimensions.

The boundedness in the Sobolev space $H^1(U)$ and the weak sequential compactness of bounded sequences in a reflexive Banach space give a subsequence converging weakly to some $u\in H^1(U)$. For each integer $m\ge 2$, the Rellich-Kondrachov compactness theorem makes $H^1(U)\hookrightarrow L^m(U)$ compact because the dimension is two. Repeated extraction followed by the diagonal argument gives one subsequence $u_{n_k}$ converging strongly to $u$ in every $L^m(U)$ with integral $m\ge 2$.

For any finite real $p\ge 1$, choose an integer $m\ge \max \{2,p\}$. Since $U$ has finite measure, the Lp inclusion on a finite measure space givesThe same subsequence therefore works for every finite $p$.

In polar coordinates, setaway from the origin, assigning any value at the origin. This is unbounded as $r=\sqrt{x^2+y^2}\downarrow 0$. It belongs to $L^2(U)$ becauseMoreoverand henceThus $u\in H^1(U)$ but $u\notin L^{\infty }(U)$, exhibiting the failure of first-order Sobolev embedding into Linfinity in two dimensions.

The initial line $x=0$ has conormal $dx$, and the principal symbol on this conormal is $1$. It is therefore a non-characteristic hypersurface at every $(0,t_0)$. The equation's coefficients and the prescribed Cauchy data $u(0,t)=\cos t$ and $u_x(0,t)=0$ are real analytic functions. The Cauchy-Kovalevskaya theorem consequently gives a unique analytic solution in a neighbourhood of $(0,t_0)$ for every

Along $x=\sin s$ and $t=s^2$, the characteristic expression isIt vanishes exactly when

The principal symbol ofis $p(x,t;\xi ,\tau )=(1-x^2){\xi }^2$. For a regular curve $s\mapsto (x(s),t(s))$, the characteristic curve equation isAway from $x=\pm 1$ this gives $t=\text{constant}$. The two degenerate lines $x=1$ and $x=-1$ are also characteristic. These are the characteristic curves, apart from reparametrization and pieces joined at the degenerate lines.

Let $N={\bigcap }_{n\ge 0}{I}^{n}M$. Apply Artin--Rees to $N\subseteq M$. Since $N\subseteq I^{n}M$ for every $n$, stability givesfor all sufficiently large $n$, and in particular $N=IN$. The module $N$ is finitely generated because $M$ is Noetherian. The determinant trick applied to a finite generating set of $N=IN$ produces $a\in I$ withTaking $x=-a\in I$ yieldswhich is Krull intersection theorem.

If the filtration is stable from degree $\ell$, then $M^{\ast }$ is generated over $R^{\ast }$ by finite generating sets for $M_0,\dots ,M_{\ell }$. Conversely, let homogeneous elements of degrees at most $\ell$ generate $M^{\ast }$. In every degree $n>\ell$, each expression for an element of $M_n$ uses a positive-degree coefficient from $R^{\ast }$, so $M_n=I{M}_{n-1}$. Hence finite generation is equivalent to stability.

For $M^{\prime }\subseteq M$, takeThen $M^{\ast }$ is a graded submodule of the finite Rees module $⨁I^{n}M$. Since $R$ is Noetherian and $I$ is finitely generated, $R^{\ast }$ is Noetherian; hence $M^{\ast }$ is finite and the filtration is stable. Therefore, for some $\ell$ and all $n\ge \ell$,This is the Artin-Rees lemma.

The formal power series ring $P=k[[X]]$ is Noetherian, so the finite product $R=P\times P$ is Noetherian. Its maximal ideals areso there are exactly two.

The idealis principal and prime because $R/\mathfrak{p}\simeq k$. The prime chainshows that it has height one, and no longer chain exists because $\dim P=1$. Yetwith both factors nonzero, so $R$ is not a domain. This shows why locality is essential in part i.

Write $\mathfrak{p}=(x)$. The principal ideal theorem and the hypothesis give $\mathrm{ht}\mathfrak{p}=1$. We use the standard principal prime in a Noetherian local ring lemma: a principal prime of positive height in a Noetherian local ring is generated by a nonzerodivisor and is the unique minimal prime above zero. The lemma follows by applying the associated-prime description of zero divisors and Nakayama's lemma to $\mathrm{ann}(x)$; if a nonzero annihilator or another minimal component existed, the principal prime would have height zero.

Here is the needed argument directly. For every $n\ge 1$,Indeed, if $y{x}^n=0$ with $y\notin \mathfrak{p}$, then $y$ is a unit in $R_{\mathfrak{p}}$, so $x^n=0$ there. The maximal ideal $\mathfrak{p}{R}_{\mathfrak{p}}=(x)$ would then be nilpotent, making $R_{\mathfrak{p}}$ zero-dimensional, contrary to $\mathrm{ht}\mathfrak{p}\ge 1$.

Now suppose $xy=0$. The displayed containment gives $y=x{y}_1$, then $x^2{y}_1=0$ gives $y_1=x{y}_2$, and inductivelyBecause $R$ is Noetherian, the ascending chain $\mathrm{ann}(x)\subseteq \mathrm{ann}(x^2)\subseteq \cdots$ stabilizes, say at $n=N$. Then $y_N\in \mathrm{ann}(x^{N+1})=\mathrm{ann}(x^N)$, and hence $y=x^N{y}_N=0$. Thus $x$ is a nonzerodivisor.

Let $N={\bigcap }_{n\ge 0}{x}^{n}R$. It is a finitely generated ideal. If $y=xz\in N$, then $xz\in x^{n+1}R$ for every $n$; cancellation of the nonzerodivisor $x$ gives $z\in x^{n}R$ for every $n$. Hence $N=xN$, and Nakayama lemma gives $N=0$ because $x$ lies in the maximal ideal.

Every nonzero element consequently has a finite $x$-adic order. If nonzero $a,b$ satisfied $ab=0$, write $a=x^r{a}^{\prime }$ and $b=x^s{b}^{\prime }$ with $a^{\prime },b^{\prime }\notin (x)$. Cancelling $x^{r+s}$ gives $a^{\prime }{b}^{\prime }=0$, contradicting primality of $(x)$. Equivalently, $(0)$ is prime, so