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

Going up lifts every prime chain in $A$ to one in $B$, so $\dim B\ge \dim A$. Conversely, contracting a strict chain of primes of $B$ gives a chain in $A$, and the incomparability theorem for integral extensions ensures that no strict inclusion contracts to equality. Thus $\dim A\ge \dim B$, and

A chain of length $s$ in $R/\mathfrak{p}$ lifts to a chainin $R$. Since $R$ is a domain and $\mathfrak{p}\ne 0$, prepending $(0)⊊\mathfrak{p}$ gives a chain of length $s+1$. Therefore $s+1\le r$ and

The Krull dimension $\dim R$ is the supremum of lengths $n$ of strict chainsof prime ideals. The transcendence degree ${\mathrm{trdeg}}_{k}L$ is the cardinality of a transcendence basis of $L/k$.

By Noether normalization lemma, there are algebraically independent $y_1,\dots ,y_d\in A$ such that $A$ is finite, hence integral, over $k[y_1,\dots ,y_d]$. Their fraction field has transcendence degree $d$, and $L$ is algebraic over it, so ${\mathrm{trdeg}}_{k}L=d$. Going up and incomparability show that an integral extension preserves Krull dimension, while a polynomial ring in $d$ variables over a field has dimension $d$. Hence

When every $k_i=1$, cancel common factors to writewhere $d$ is the pole order at $t=1$. Sincethe coefficient of $t^n$ in $P(M,t)$ is, for all sufficiently large $n$, a fixed linear combination of shifted binomial polynomials. It is therefore a polynomial in $n$ of degree exactly $d-1$ unless $P=0$, in which case its degree is $-1$.

For an additive length function $\lambda$ finite on the graded pieces, define the Poincare series of a graded moduleThe Hilbert-Serre theorem states thatfor a Laurent polynomial $f(t)\in \mathbb{Z}[t,t^{-1}]$.

Induct on $s$. For the last generator $x_s$ of degree $k_s$, multiplication gives an exact sequence whose kernel is the $x_s$-torsion and whose cokernel is $M/x_{s}M$. Additivity of $\lambda$ yieldsBoth modules on the right are finite graded modules over the algebra generated by $x_1,\dots ,x_{s-1}$. The induction hypothesis gives the asserted denominator. The case $s=0$ is a finite Laurent polynomial because $M$ is finitely generated over $A_0$.

The positive-degree part $A_{+}={⨁}_{n>0}{A}_n$ is an ideal and $A/A_{+}\simeq A_0$, so $A_0$ is Noetherian. Since $A$ is Noetherian, $A_{+}$ has finitely many homogeneous generators $x_1,\dots ,x_s$. Induction on degree shows that every positive-degree homogeneous element is a polynomial in the $x_i$ over $A_0$. Thus

Put $S=A\setminus \mathfrak{p}$. Primes of $B_{\mathfrak{p}}=S^{-1}B$ correspond to primes $\mathfrak{q}$ of $B$ whose contractions are contained in $\mathfrak{p}$. By going up, each such $\mathfrak{q}$ is contained in a prime lying over $\mathfrak{p}$, and this prime is uniquely ${\mathfrak{p}}^{\prime }$. Hence ${\mathfrak{p}}^{\prime }{B}_{\mathfrak{p}}$ is the unique maximal ideal of $B_{\mathfrak{p}}$.

Localizing this already local ring at its unique maximal ideal changes nothing, soFinally, localization preserves integral extensions; therefore $B_{{\mathfrak{p}}^{\prime }}=B_{\mathfrak{p}}$ is integral over $A_{\mathfrak{p}}$.

If $x\in A$ is a unit in $B$, then $x^{-1}$ is integral over $A$:Multiplication by $x^{n-1}$ expresses $x^{-1}$ as an element of $A$, so $x$ is a unit in $A$.

Use the characterization $a\in J(A)$ exactly when $1-ra$ is a unit for every $r\in A$. If $a\in J(B)\cap A$, then $1-ra$ is a unit in $B$ and hence in $A$, proving $a\in J(A)$. Conversely, if $a\in J(A)$, every maximal ideal $\mathfrak{n}$ of $B$ contracts under the integral extension to a maximal ideal of $A$, which contains $a$. Thus every $\mathfrak{n}$ contains $a$, so $a\in J(B)$. Therefore