Parametrize the circle by $z=e^{it}$, $0\le t\le 2\pi$. Choose a continuous argument $\vartheta (t)$ of $\varphi (e^{it})$, so $\varphi (e^{it})=|\varphi (e^{it})|e^{i\vartheta (t)}$. Its winding number about zero isThe two facts proved below determine the requested calculation without needing differentiability or contour integration. Put $P(z)=(3z-2)(z-3)(2z+1)$. On $|z|=1$, $|P(z)|\ge 1\cdot 2\cdot 1=2$, so adding one does not change its winding. The first and third factors have winding one, since their constant terms have modulus smaller than their linear terms; the second has winding zero, since its $z$ term has modulus smaller than its constant term. Nonzero constants have winding zero and $z$ has winding one. Thus

Uniform convergence of $f_n$ to $f$ on an interval $I$ means that for every $ϵ>0$ there is $N$ such that $|f_n(x)-f(x)|<ϵ$ for all $n\ge N$ and every $x\in I$. Equivalently ${\sup }_{x\in I}|f_n(x)-f(x)|\to 0$. The same $N$ must work throughout the interval, unlike pointwise convergence.

For completeness, continuity is preserved: choose $N$ with uniform error below $ϵ/3$, and use continuity of $f_N$ at a given $x$ to make $|f_N(y)-f_N(x)|<ϵ/3$. The triangle inequality then gives $|f(y)-f(x)|<ϵ$. Thus each function involved in the integral is continuous on the closed interval and Riemann integrable. The estimateproves convergence of the integrals by uniform convergence. When $a=b$ both integrals are zero.

An equivalence relation on a set $A$ is a binary relation that is reflexive, symmetric and transitive: every element is related to itself, related pairs may be reversed, and two consecutive related pairs imply the pair of endpoints is related.

Write $T=R\circ S$, using the printed left-to-right composition of relations convention. Reflexivity implies $R\subseteq T$ and $S\subseteq T$, and hence $T$ is reflexive. Reversing a two-step chain givessince $R,S$ are symmetric. Also $R\circ R=R$ and $S\circ S=S$: one inclusion is transitivity, and the reverse follows by inserting a reflexive step. These elementary identities supply the equivalences below, without presuming that a composite of equivalence relations is automatically an equivalence relation.

An equivalence relation on $A$ is a binary relation that is reflexive, symmetric and transitive: $xRx$ for every $x$; $xRy\Rightarrow yRx$; and $xRy,yRz\Rightarrow xRz$. Write $T=R\circ S$, using the stated order for composition of relations. Composition is associative because both possible bracketings mean that there is a two-step sequence of intermediate witnesses. Reflexivity showsand also that $T$ is reflexive. Transitivity and reflexivity of the original relations give $R\circ R=R$, $S\circ S=S$. Since both original relations are symmetric,The four parts below show that each condition is equivalent to the commuting equivalence relations condition $R\circ S=S\circ R$.

For the integer example, put $d=\gcd (m,n)$. Membership of $(x,z)$ in the composite means that some $y$ satisfies $y-x=ma$, $z-y=nb$ for integers $a,b$. Consequently $z-x=ma+nb$ is divisible by $d$. Conversely, if $z-x$ is divisible by $d$, the Bezout identity expresses it as $ma+nb$; setting $y=x+ma$ provides the required witness. ThusReversing the roles of $m,n$ gives exactly the same modular congruence relation, hence $R\circ S=S\circ R$. This verifies all four conditions and identifies their common smallest equivalence relation explicitly.

When $\alpha =2\beta$, all three values reduce to $-\beta /3$. To prove path independence for every piecewise smooth path, construct a global potential rather than infer it from agreement on only three paths. Integrating the coefficient of $dx$ and then matching the coefficient of $dy$ givesIndeed ${\Phi }_x=2\beta (x^2+xy)=\alpha (x^2+xy)$ and ${\Phi }_y=\beta (x^2+y^2)$. The integrand is the exact differential $d\Phi$ throughout ${\mathbb{R}}^2$. The fundamental theorem for line integrals therefore gives, for every such path from $A$ to $B$,

Let $c=1/\sqrt{2}$. A concrete real matrix isIts first coordinate has eigenvalue $-1$. The lower block has characteristic polynomial $(\lambda -c{)}^2+c^2$, whose roots are $c\pm ic=(1\pm i)/\sqrt{2}$. Thus this example has all three required eigenvalues. It is also an orthogonal matrix, but orthogonality is not forced by its spectrum, as the second part shows.

A Cartesian isotropic tensor is unchanged by every proper rotation matrix. For a rank-two tensor this means $B^{\prime }=RB{R}^{\mathsf{T}}=B$, or $B_{ij}^{\prime }=R_{ik}{R}_{jl}{B}_{kl}=B_{ij}$. No symmetry of $B$ is assumed. For a quarter-turn about the $z$ axis,Comparing entries yields $B_{11}=B_{22}$, $B_{12}=-B_{21}$, $B_{13}=-B_{23}=-B_{13}$, and $B_{31}=-B_{32}=-B_{31}$. In particular,The remaining matrix has the form $\left(\begin{array}{ccc}b& c& 0\\ -c& b& 0\\ 0& 0& d\end{array}\right)$. A quarter-turn about the $x$ axis givesInvariance now forces $c=0$ and $d=b$, proving that an isotropic second-rank tensor isa scalar multiple of the Kronecker delta.

For the final unheaded conductivity identity, let $P=n{n}^{\mathsf{T}}$, the orthogonal projection onto the unit vector $n$. Then $P^2=P$ and the uniaxial conductivity tensor is $\sigma =\alpha I+\gamma P$. The antisymmetric tensor $D$ sends a vector $v$ to $v\times n$. Applying it twice and using the vector triple product givesConsequentlyThis equals $-\sigma =-\alpha I-\gamma P$ exactly when $(\alpha +\gamma )P=0$. Since $P$ has rank one and is nonzero, the requested parameter isAt this value conductivity vanishes along $n$; it is precisely the singular case excluded in part (ii), so there is no conflict with that part's conclusion.

The Bernstein inequality for Weyl algebra modules requires characteristic zero: for a nonzero finitely generated left or right module $M$ over $A_n(k)$,Equivalently, its Bernstein growth dimension of a Weyl algebra module is at least $n$. Without the characteristic-zero hypothesis this is false, as the module in part (c) has dimension zero in the growth sense. The proof below works over any characteristic-zero field, not just an algebraically closed one.

For a left module, choose a nonzero finite-dimensional generating subspace $M_0$ and put $M_j=F_j{A}_n{M}_0$, where $F_j$ is the Bernstein filtration. Ordered monomials giveMultiplication defines the linear action mapWe prove this map injective by induction on $j$, which is the essential faithful finite-step action of a Weyl algebra argument. For $j=0$, an element is a scalar, and a scalar killing $M_0\ne 0$ is zero. Suppose $a\in F_j$ kills $M_j$. For any generator $z$ and $m\in M_{j-1}$, both $azm$ and $zam$ vanish because $zm\in M_j$ and $m\in M_j$. Thus $[a,z]\in F_{j-1}$ kills $M_{j-1}$, so induction gives $[a,z]=0$ for every generator. By the explicit commutator formulas in part (a), commuting with all $D_i$ forces all positive $x_i$ exponents of $a$ to vanish, and commuting with all $x_i$ forces all positive $D_i$ exponents to vanish. Characteristic zero is used to cancel the nonzero integer exponents. Hence $a$ is scalar and, because it kills $M_0$, is zero. This proves injectivity.

Taking dimensions yieldsThe left side is asymptotic to $j^{2n}/(2n)!$. Consequently ${\dim }_k{M}_{2j}\ge C{j}^n$ for all sufficiently large $j$ and some $C>0$. Taking logarithms and the limsup in the definition of Gelfand–Kirillov dimension of a module proves the claimed lower bound. For a good Bernstein filtration, the associated graded module is finite over $k[x_1,\dots ,x_n,{\xi }_1,\dots ,{\xi }_n]$; its cumulative Hilbert function is eventually polynomial, and the same comparison says its degree is at least $n$.

For right modules, use the opposite ring isomorphism $A_n(k)\cong A_n(k{)}^{\mathrm{op}}$ sending $x_i$ to $x_i$ and $D_i$ to $-D_i$. It respects the Bernstein filtration, so the identical estimate applies. The bound is sharp: the polynomial module $k[x_1,\dots ,x_n]$, generated by $1$, has $M_j$ equal to the polynomials of degree at most $j$, of dimension $\left(\genfrac{}{}{0ex}{}{j+n}{n}\right)$. Its Gelfand–Kirillov dimension of a module is exactly $n$.

In a commutative ring, if $x^m=0$, then the principal ideal $(x)$ satisfies $(x{)}^m=0$. Thus a semiprime ring has no nonzero nilpotent elements. Conversely, if $I^m=0$ for an ideal $I$, then every $x\in I$ satisfies $x^m=0$; absence of nonzero nilpotent elements forces $I=0$. Hence commutative semiprimeness is exactly reducedness.

For a prime ideal $P$, let $S=R\setminus P$. Every ideal $I$ of $R_P=S^{-1}R$ is the extension of its contraction $I^c$: if $r/s\in I$, multiply by the unit $s/1$ to get $r/1\in I$, so $r\in I^c$. If $R$ is Noetherian, write $I^c=(r_1,\dots ,r_m)$; their images generate $I$. This proves that $R_P$ is Noetherian.

The extended ideal $P{R}_P$ is proper. Indeed $1=p/s$ would imply $u(s-p)=0$ for some $u\notin P$, and then $us=up\in P$, contradicting primeness and $u,s\notin P$. A fraction with numerator outside $P$ is invertible, with inverse $s/r$. A fraction with numerator in $P$ lies in the proper ideal $P{R}_P$ and cannot be invertible. Thus the nonunits are precisely $P{R}_P$, making it the unique maximal ideal. ThereforeTo prove reducedness is detected by prime localizations, let $x$ be a nilpotent element of $R$. Its image in every $R_Q$ is zero because these localizations are semiprime, hence reduced. If $x\ne 0$, its annihilator $\mathrm{Ann}(x)$ is a proper ideal; choose a maximal ideal $Q$ containing it. The zero criterion for localization says $x/1=0$ in $R_Q$ only if $sx=0$ for some $s\notin Q$, which would put $s$ in $\mathrm{Ann}(x)\subseteq Q$, a contradiction. Hence $x=0$, and $R$ is semiprime. This argument does not need Noetherianity.

Primeness is not detected in the same way. Take $R=k\times k$. Its two prime ideals are $P_1=0\times k$ and $P_2=k\times 0$, and its corresponding localizations are both isomorphic to the field $k$. To see these are all the prime ideals, the orthogonal idempotents $e_1,e_2$ have product zero, so a prime ideal contains one of them and is then the kernel of the other projection. Both localizations are prime rings, but $R$ is not prime: $e_1,e_2\ne 0$ and $e_1{e}_2=0$.

Let $K=k(z)$, where $z$ is an indeterminate, and form the right Artinian triangular ring that is not left NoetherianMultiplication uses the natural $k$–$K$ bimodule structure on the upper-right entry. With the two diagonal idempotents $e_1,e_2$, the right regular module is $e_{1}T\oplus e_{2}T$. The off-diagonal submodule of $e_{1}T$ is one-dimensional over the bottom-right field $K$, hence simple. Its quotient is one-dimensional over the top-left field $k$, hence simple. Also $e_{2}T$ is a simple right module over $K$. Thus $T_T$ has a composition series of length three, and in particular $T$ is right Artinian.

On the left, multiplication of an off-diagonal matrix givesEvery $k$-subspace $U\subseteq K$ therefore yields a left ideal of off-diagonal matrices with entries in $U$. The subspaces $U_m={\mathrm{span}}_k\{1,z,\dots ,z^m\}$ are strictly increasing because $z$ is transcendental. Their associated left ideals violate the ascending chain condition. Hence $T$ is not left Noetherian, even though it is right Artinian and right Noetherian.

A right Ore localization at $S$ is a unital homomorphism $\iota :R\to Q$ such that every $\iota (s)$, $s\in S$, is a unit, every element of $Q$ is $\iota (r)\iota (s{)}^{-1}$, andEquivalently, the usual fraction construction has this zero criterion and is universal among homomorphisms that invert $S$. Injectivity is required only when the denominators are regular elements of a ring. If $0\in S$, every homomorphism inverting $S$ has zero target; this is the degenerate zero-ring localization. For a nonzero localization we require $0\notin S$.

The necessary and sufficient conditions are that $S$ be a right denominator set:

For necessity, write the element $\iota (s{)}^{-1}\iota (r)$ as $\iota (b)\iota (t{)}^{-1}$. Multiplying by $\iota (s)$ on the left and $\iota (t)$ on the right gives $\iota (rt-sb)=0$. The zero criterion supplies $u\in S$ with $(rt-sb)u=0$. Hence $r(tu)=s(bu)$, the right Ore condition in $R$, not merely an equality of images in $Q$. Also $sr=0$ makes $\iota (r)=0$ because $\iota (s)$ is invertible, and the zero criterion gives $rt=0$. This proves both necessities.

The sufficiency is the Ore theorem for denominator sets. Concretely, right fractions can be represented by $(r,s)$; a common-denominator equality has $su=tv\in S$ and $ru=r^{\prime }v$ for suitable $u,v\in R$. The two conditions make the induced fraction equivalence transitive and make addition and multiplication well defined; the zero fraction is exactly the kernel displayed above. The requested direction of the theorem was necessity, for which the complete argument has just been given.

For a right Noetherian domain, any two nonzero principal right ideals intersect. To see this without using the desired localization, suppose $bR\cap aR=0$ with $a,b\ne 0$. The sumis direct: reduce a finite relation modulo $aR$ to get its first summand zero, then cancel $a$ in the noncommutative domain and repeat. Every summand is nonzero, so the finite partial sums form a strictly ascending chain of right ideals, contradicting the ascending chain condition. A nonzero intersection $rR\cap sR$ thus provides $rt=sb\ne 0$, with $t\ne 0$. This is the right Ore condition for $S=R\setminus \{0\}$. Reversibility is automatic in a noncommutative domain, and the zero criterion makes the map injective. The localization is a division ring, since a nonzero $r{s}^{-1}$ has inverse $s{r}^{-1}$.

The Goldie theorem states that a semiprime ring has a semisimple Artinian classical right ring of quotients if and only if it has finite right uniform dimension and satisfies the ascending chain condition on right annihilators. A right Noetherian ring satisfies the annihilator condition because right annihilators are right ideals. Its right regular module has finite uniform dimension: an infinite direct sum of nonzero submodules would give a strictly ascending chain of finite partial sums. A prime ring is semiprime, so the theorem applies to a prime right Noetherian ring $R$.

To obtain a single matrix-ring factor, not just a finite product, prove that its quotient $Q$ is prime. A nonzero two-sided ideal $I\subseteq Q$ contains a nonzero fraction $r{s}^{-1}$ and hence the nonzero element $r=(r{s}^{-1})s\in I\cap R$. The same holds for any other nonzero ideal $K$. Since $R$ is prime, $(I\cap R)(K\cap R)\ne 0$, so $IK\ne 0$. Thus $Q$ is prime. The Artin–Wedderburn theorem gives a finite product of matrix rings over division rings; primeness excludes two different nonzero factors with zero product. This proves the prime classical quotient of a prime right Goldie ring:For the requested counterexample to Noetherianity, take $R=k[x_1,x_2,\dots ]$. It is a commutative integral domain and has its field of fractions as its classical right ring of quotients. But $(x_1)⊊(x_1,x_2)⊊\cdots$ is a strictly ascending ideal chain: evaluation at $x_1=\cdots =x_m=0$ shows $x_{m+1}$ is not in the preceding ideal. Therefore existence of a classical quotient does not imply right Noetherianity.

The variable $t$ is central. To prove the leading-coefficient proof for a central polynomial extension, let $I$ be any right ideal of $R[t]$ and setZero leading coefficients are allowed. Addition and right multiplication by constants show that $L_d$ is a right ideal. Multiplication by $t$ gives $L_d\subseteq L_{d+1}$. Since $R$ is a right Noetherian ring, this chain stabilizes, say at $L_N$, and each $L_d$ for $d\le N$ has finitely many generators $a_{dj}$. Choose $f_{dj}\in I$ of degree at most $d$ lifting these coefficients.

These finitely many polynomial lifts generate $I$ on the right. Indeed, for $f\in I$ of degree $m$, let $d=\min (m,N)$. Write its leading coefficient as ${\sum }_j{a}_{dj}{r}_j$. Thenbelongs to $I$ and has degree less than $m$, because $L_m=L_d$ when $m>N$. Induction on degree reduces $f$ to a right linear combination of the chosen lifts. This proves finite generation of every right ideal, and henceCentrality matters: it is what permits the displayed leading-coefficient cancellation without twisting the coefficients.