For the left shift, ${\sigma }^p(s)=s$ precisely when the binary sequence is obtained by repeating its first $p$ symbols. There are exactly $2^p$ such sequences, although some have smaller least period. Hence the set of all periodic sequences is a countable union of finite sets and is at most countable. Passing from periodic points to periodic orbits cannot increase this cardinality.

To prove infinitude, for every integer $p\ge 1$ take the repeated word $(0^{p-1}1{)}^{\infty }$. Its least period is $p$, because its ones are separated by exactly $p$ positions; a smaller positive period would carry a one to a zero. Distinct least periods give distinct periodic orbits. Thus there are countably infinitely many periodic orbits.

In fact $J=R$, so the quotient module is zero. This requires an argument using left ideals: a left ideal is not generally closed under multiplication on the right.

Write $x=x_1$, $y=y_1$, so that $yx-xy=1$. The commutator product rule givesBoth $y^2{x}^2$ and $x^2{y}^2$ lie in $J=R{x}^2+R{y}^2$, so their differencebelongs to $J$. Multiplying on the left and subtracting another known element of the left ideal givesBecause the scalar $6$ is invertible in $\mathbb{C}$, it follows that $y\in J$, and hence $xy\in J$. Therefore $h-4xy=2\in J$, giving $1\in J$. We concludeThus the answer is yes for every $n\ge 1$. All ideal operations in the proof use left multiplication. In particular, simply replacing $J$ by the commutative ideal $(x_1^2,{\xi }_1^2)$ of the associated graded ring would miss the lower-order consequences that force $J=R$.

Use the quotient filtration induced from the order filtration of a Weyl algebra. The key step is to identify the entire associated graded module of the left ideal $I$, rather than merely take symbols of its displayed generators.

For every member $x^{\alpha }{y}^{\beta }$ of the ordered monomial basis of a Weyl algebra, right multiplication by the pure $y$-monomials givesNo commuting past an $x_i$ is necessary. It follows that $I$ is exactly the complex linear span of the basis monomials satisfying eitherIntersecting this linear span with each filtration piece and then taking symbols therefore yields the monomial idealBy b,

To determine its Krull dimension, compute the radical of an ideal:Indeed, $K$ lies in the displayed intersection, an intersection of prime ideals. Conversely, $({\xi }_1{\xi }_2{)}^3\in K$ and $({\xi }_1{\xi }_3{)}^3\in K$, so both generators of that intersection lie in $\sqrt{K}$. The equality of the intersection with $({\xi }_1{\xi }_2,{\xi }_1{\xi }_3)$ follows by factoring out ${\xi }_1$ and reducing modulo $({\xi }_2,{\xi }_3)$.

The order characteristic variety of a Weyl algebra module in this case has two irreducible components:They are coordinate affine spaces of dimensions $2n-1$ and $2n-2$, respectively, in the $2n$ coordinates $x_1,\dots ,x_n,{\xi }_1,\dots ,{\xi }_n$. The Krull dimension is the larger of these component dimensions. ThusIn particular, two displayed generators do not force a drop of two in dimension: their common factor creates the component ${\xi }_1=0$ of dimension $2n-1$.

We first prove that any two nonzero principal left ideals $Ra$ and $Rb$ intersect nontrivially. Suppose instead that $Ra\cap Rb=0$. The nonzero left idealsthen form a direct sum. Indeed, take a finite relationIt impliesAs $R$ is a noncommutative domain and $a\ne 0$, this gives $u_0=0$. Right cancellation of the nonzero element $b$ reduces the remaining relation to one of the same form with one fewer term. Repetition gives $u_1=\cdots =u_m=0$, proving the asserted direct sum.

Consequently its finite partial sums give a strictly ascending chain of left ideals, contradicting the ascending chain condition in a left Noetherian ring. Thus $Ra\cap Rb\ne 0$ whenever $a,b\ne 0$. This is the left Noetherian domain Ore lemma.

Now take $s\in S=R\setminus \{0\}$ and $r\in R$. If $r=0$, choose $s^{\prime }=1$ and $r^{\prime }=0$. Otherwise chooseSince $z\ne 0$, the coefficient $s^{\prime }$ is nonzero and hence belongs to $S$. This proves the left Ore set condition in all cases.

A multiplicative subset $S$ is a left Ore set if for every $s\in S$ and $r\in R$ there are $s^{\prime }\in S$ and $r^{\prime }\in R$ withEquivalently, the subsets $Sr$ and $Rs$ intersect for each such pair. This is the common-left-multiple condition needed for Ore localization with denominators on the left. In a noncommutative domain, no extra condition concerning annihilation by denominators is needed, since all members of $S$ are regular elements of a ring.

Let $L$ be any maximal left ideal. If $I⊈L$, maximality gives $L+I=R$, because the two-sided ideal $I$ is in particular a left ideal. Thus $1=\ell +i$ for some $\ell \in L$ and $i\in I$. Butis a unit by hypothesis. A proper left ideal cannot contain a unit, since left multiplication by its inverse would put $1$ in the left ideal. This contradiction proves $I\subseteq L$ for every maximal left ideal $L$, and thereforeTogether with ii, this is the ideal form of the unit criterion for the Jacobson radical.

The Jacobson radical isHere the second intersection ranges over isomorphism classes of nonzero simple modules. For the equality, every nonzero $v$ in a simple module gives a surjective module homomorphism $R\to V$, $r\mapsto rv$, whose kernel is a maximal left ideal. Conversely, every maximal left ideal $L$ gives the simple module $R/L$. This description by annihilators shows that $J(R)$ is a two-sided ideal.

Let $j\in J(R)$ and put $a=1+j$. If the left ideal $Ra$ were proper, it would lie in a maximal left ideal $L$. Both $a$ and $j$ would belong to $L$, giving $1\in L$, a contradiction. Therefore there is $u\in R$ with $ua=1$.

A left inverse alone need not be a two-sided inverse in a general ring, so one more step is necessary. The equality $u(1+j)=1$ gives $u=1-uj\in 1+J(R)$. Applying the same argument to $u$ supplies $v\in R$ with $vu=1$. Thenand hence $au=vu=1$. ThusFor the zero ring the conclusion holds directly; the maximal left ideal argument concerns the nonzero case.

A left-primitive ideal is a two-sided ideal of the formfor a nonzero simple module $V$ on the left. Equivalently, the quotient ring $R/P$ has a faithful module that is a simple module, and is therefore a primitive ring.

To prove that $P$ is a prime ideal of a noncommutative ring, let $A,B$ be two-sided ideals with $AB\subseteq P$. Since $B$ is two-sided, $BV$ is a submodule of $V$: for $r\in R$, $b\in B$ and $v\in V$, one has $r(bv)=(rb)v\in BV$. If $B⊈P$, then $BV\ne 0$, so simplicity gives $BV=V$. Consequentlyand $A\subseteq P$. Thus $AB\subseteq P$ implies $A\subseteq P$ or $B\subseteq P$. Also $P\ne R$, since $1$ acts as the identity element on the nonzero module $V$. These are precisely the required prime ideal of a noncommutative ring conditions.

Yes. An explicit holomorphic surjection is the rational functionIts denominator never vanishes inside the unit disc. To prove surjectivity, take any $w\in \mathbb{C}$ and choose a square root $r$ of $w$ with $\mathrm{Re}r\ge 0$; one of the two opposite roots always has this property. Define $z=r/(1+r)$. Then $1+r\ne 0$ andso $|z|<1$. Also $z/(1-z)=r$, and consequently $F(z)=r^2=w$.

This is a holomorphic surjection from the unit disc onto the complex plane. It does not contradict part (i): it is not injective. For example, $F(i/(1+i))=F(-i/(1-i))=-1$, with two distinct arguments of modulus $1/\sqrt{2}$.

Suppose $F:\mathbb{D}\to \mathbb{C}$ were a holomorphic bijection. By the allowed nonvanishing derivative result, $F^{\prime }(z)\ne 0$ throughout $\mathbb{D}$. The holomorphic inverse function theorem gives a holomorphic inverse locally at every image point. Since $F$ is globally bijective, these local inverses agree with its global inverse $H:\mathbb{C}\to \mathbb{D}$.

Thus $H$ is an entire function with $|H|<1$. The Liouville theorem forces $H$ to be constant, contradicting its being an inverse. ThereforeThis is the analytic obstruction behind the fact that the complex plane and unit disc are homeomorphic but not conformally equivalent.

Let $V=f(U)$, the right half-disc from part (i). Write $w=r{e}^{it}$ with $0<r<1$ and $-\pi /2<t<\pi /2$. Squaring doubles the angle and squares the radius, giving a holomorphic bijectionIts inverse is the unique square root with positive real part, and its derivative $2w$ is nonzero on $V$. The cut includes zero, because $0\notin V$.

Next, the Möbius transformation $T(s)=i(1-s)/(1+s)$ is a conformal bijection from $\mathbb{D}$ onto $\mathbb{H}$: it is the inverse Cayley transform between the half-plane and disk composed with $s\mapsto -s$. For $s=-t$, $0\le t<1$, it givesConsequently $T((-1,0])=\{iy:y\ge 1\}$. Since $T$ is a bijection on the whole disc, deleting this interval deletes exactly that ray in the image.

The required function composition is thereforeEach component is a conformal bijection between the indicated domains. The only possible denominator zeros of the simplified expression are $z=\pm 1$, outside $U$, andsince $i$ is on the excluded semicircle and $-i$ is outside the upper half-plane. This also checks the local conformality of the conformal map from the exterior half-disc to a slit half-plane.

For $z=x+iy\in \mathbb{H}$, direct calculation givesConversely, solving $w=(z-i)/(z+i)$ givesThe denominator cannot vanish inside the unit disc, so these inverse formulas prove $\boxed{f(\mathbb{H})=\mathbb{D}}$. This is the Cayley transform between the half-plane and disk.

For $z\in U$, we also require $|z|>1$. In inverse coordinates this becomesThe equivalence holds in both directions, soThus the exterior of the unit semicircle maps to the right half of the unit disc. The circular arc maps to its imaginary diameter, which is excluded because the source inequality is strict.

At $z_0$, complex differentiability givesWriting $f^{\prime }(z_0)=\lambda e^{i\alpha }$, the assumption $f^{\prime }(z_0)\ne 0$ gives $\lambda >0$. Its real derivative is therefore the similarity $h\mapsto \lambda e^{i\alpha }h$, a rotation through $\alpha$ followed by scaling by $\lambda$.

More explicitly, for two differentiable curves ${\gamma }_1,{\gamma }_2$ with ${\gamma }_j(0)=z_0$ and nonzero tangent vectors, the chain rule givesTaking the complex argument of this ratio shows that the oriented angle is preserved, while nonvanishing ensures both image tangents remain regular. Thus

A conformal map at a point $z_0$ is a map that preserves oriented angles between regular differentiable curves through $z_0$. In the differential formulation, it is real differentiable at $z_0$ and its derivative is an orientation-preserving similarity: a rotation followed by multiplication of lengths by a positive constant. The map is conformal on $U$ when it has this property at every point:Here $R_{\alpha }$ is a planar rotation. Conformality is a local property and does not itself require global injectivity. We use the oriented convention standard in complex analysis; a reflection reverses oriented angles.

A function has a pole of order $n$ at $a$ when, on a punctured neighbourhood,Equivalently, $(z-a{)}^{n}f(z)$ has a removable singularity whose extension is nonzero at $a$. After shrinking the neighbourhood so $h\ne 0$, its reciprocal $1/f$ extends with a zero of order $n$. In particular, $|f(z)|\to \infty$ at a pole.

A holomorphic function has a zero of order $n$ at $a$ when it is holomorphic on a neighbourhood of $a$ and factors asEquivalently, $f(a)=f^{\prime }(a)=\cdots =f^{(n-1)}(a)=0$ and $f^{(n)}(a)\ne 0$. The order of a zero of a holomorphic function is the degree of the first nonzero term of its Taylor series.

An isolated singularity at $a$ is a removable singularity when there is a holomorphic function $F$ on a full disc $B(a,r)$ agreeing with $f$ away from $a$. Equivalently,The Riemann removable singularity theorem also characterizes this by boundedness of $f$ on a sufficiently small punctured disc. A preassigned incorrect value at $a$ is replaced by $L$.

A function has an isolated singularity at $a$ if, for some $r>0$, it is defined and holomorphic on the punctured disc $0<|z-a|<r$, while a holomorphic extension through $a$ has not been assumed. In particular, its domain omits $a$, or its assigned value there fails to give such an extension. The key local condition isThe convention includes removable singularities, even when a suitable extension would eliminate the missing point.

Choose $n\ge 1$ as in part (i). Since distances between iterates are nondecreasing, the triangle inequality givesLetting $ϵ\downarrow 0$ and combining with the given reverse bound proves that $\varphi$ is an isometry:For surjectivity, fix $x\in X$ and use part (i) with $y=x$ and $ϵ=1/k$ to choose $n_k\ge 1$ satisfying $d(x,{\varphi }^{(n_k)}x)<1/k$. Put $z_k={\varphi }^{(n_k-1)}x$, interpreting ${\varphi }^{(0)}$ as the identity. By sequential compactness, a convergent subsequence satisfies $z_{k_j}\to z\in X$. The assumed continuity yieldsThus every $x$ has a preimage, and $\boxed{\varphi \text{ is surjective}}$. If $X$ is empty, both conclusions hold vacuously. This proves the distance-increasing self-map theorem for sequentially compact metric spaces.

By the permitted sequential compactness of the product metric space $X\times X$, the sequence of pairs $({\varphi }^{(k)}x,{\varphi }^{(k)}y)$ has a convergent subsequence, and hence a Cauchy subsequence. Choose two of its indices $r<s$ sufficiently late thatThe distance-increasing assumption iterates: $d({\varphi }^{(r)}u,{\varphi }^{(r)}v)\ge d(u,v)$. With $n=s-r\ge 1$, use $u=x,v={\varphi }^{(n)}x$, and then $u=y,v={\varphi }^{(n)}y$, to obtainBoth summands are nonnegative, soThe use of paired iterates ensures that the same positive return time works for both points.