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The key is a Laurent normalization by a monomial change of variables. We prove the slightly stronger assertion that every nonzero quotient of a Laurent polynomial ring in $r$ variables is finite as a module over an embedded Laurent polynomial ring in some number $m\le r$ of variables.

Induct on $r$. If the defining ideal is zero, the ring itself works. Otherwise choose a nonzero Laurent polynomialin that ideal, with nonzero coefficients. For a sufficiently large positive integer $N$, the weightsare distinct for all $\mathbf{a}\in F$. Indeed, each difference gives a nonzero polynomial in $N$, which has only finitely many integer roots. Make the invertible monomial substitutionIts inverse is $Y_j=X_j{X}_r^{-N^j}$, so it is an automorphism of the Laurent polynomial ring. When $r=1$ there is no substitution and distinct exponents already give distinct weights.

As a Laurent polynomial in $Y_r$, the transformed relation has unique lowest and highest exponents. Their coefficients are nonzero scalar multiples of monomials in $Y_1,\dots ,Y_{r-1}$, hence units inMultiplying by a power of $Y_r$ and the inverse leading coefficient givesHere $d>0$: a single-term Laurent polynomial would be a unit and the quotient would be zero. Reversing the equation and dividing by $b_0{Y}_r^d$ gives a monic equation for $Y_r^{-1}$. Thus both $Y_r$ and $Y_r^{-1}$ are integral over the image $B$ of $B_0$ in $T$. In fact $b_0$ being a unit also expresses $Y_r^{-1}$ as a polynomial in $Y_r$, so $T$ is generated as a $B$-module by $1,Y_r,\dots ,Y_r^{d-1}$.

The ring $B$ is itself a nonzero quotient of an $(r-1)$-variable Laurent polynomial ring. By induction, it is finite over an embedded Laurent polynomial ring in $m\le r-1$ variables. Finiteness of modules is transitive, so the same is true of $T$; a finite ring extension is an integral extension, by the determinant trick applied to multiplication by an element.

Applying this to $r=3$ yields an embedded subringThe base case $m=0$ means the subring $k$; finiteness over it would make $T$ a finite-dimensional vector space. The hypothesis excludes this, so $1\le m\le 3$, as required. No infinitude assumption on $k$ is used.

Eisenstein's criterion. Let $f(X)=a_n{X}^n+\cdots +a_0\in \mathbb{Z}[X]$ be a nonconstant primitive polynomial. If a prime number $p$ satisfies $p\nmid a_n$, $p\mid a_j$ for $0\le j<n$, and $p^2\nmid a_0$, then $f$ is an irreducible polynomial over $\mathbb{Q}$.

To prove the Eisenstein criterion, suppose a nontrivial factorization exists over $\mathbb{Q}$. Gauss lemma for polynomials gives $f=gh$ with $g,h\in \mathbb{Z}[X]$ of positive degree. Their leading coefficients are not divisible by $p$, because their product is $a_n$. Reduction modulo $p$ therefore preserves both degrees, andSince the only irreducible factor of the right side is $X$, both reduced factors are nonzero monomials of positive degree. Their constant coefficients are consequently divisible by $p$. This would make $p^2\mid g(0)h(0)=a_0$, a contradiction.

For the requested polynomial, shift the variable by one. The binomial theorem givesIt is monic, all coefficients below the leading one are divisible by $p$, and the constant coefficient is $p$, not divisible by $p^2$. The Eisenstein criterion proves irreducibility of this shifted polynomial. The substitution $X\mapsto X+1$ is an automorphism of $\mathbb{Q}[X]$, with inverse $X\mapsto X-1$, so it preserves nontrivial factorizations. HenceThis is the cyclotomic polynomial for a prime index.

The Weak Hilbert Nullstellensatz states that if $k$ is an algebraically closed field, every maximal ideal of $k[Y_1,\dots ,Y_s]$ isEquivalently, every proper ideal has a common zero. We prove the more general field statement, the Zariski lemma: a field finitely generated as a $k$-algebra is a finite field extension of $k$.

Write such a field as $L=k[b_1,\dots ,b_s]$. Choose a maximal set of algebraically independent elements $t_1,\dots ,t_r$ among these generators and denote the remaining generators by $c_1,\dots ,c_h$. Each $c_j$ is an algebraic element over $K=k(t_1,\dots ,t_r)$, so $L/K$ is a finite field extension. Clearing the denominators in their monic equations, choose a nonzero $d\in k[t_1,\dots ,t_r]$ such that each $c_j$ is an integral element overBecause $L$ is a field and $d^{-1}\in L$, we have $L=D[c_1,\dots ,c_h]$. Thus $L$ is an integral extension of $D$.

A subring over which a field is integral must itself be a field. Indeed, for $0\ne a\in D$, an integral equation for $a^{-1}$ multiplied by $a^{m-1}$ expresses $a^{-1}$ as an element of $D$. But if $r>0$, the ring $D$ is not a field: choose an irreducible polynomial $q(t_1)\in k[t_1]$ which does not divide $d$. There are infinitely many monic irreducible polynomials in $k[t_1]$, by the usual product-plus-one argument, whereas only finitely many can divide the nonzero $d$. The prime ideal $(q)$ stays proper after inverting $d$, so $q$ has no inverse in $D$. Therefore $r=0$, proving that $L/k$ is a finite field extension, and in particular an algebraic field extension.

Apply the Zariski lemma to the residue field $k[Y_1,\dots ,Y_s]/\mathfrak{m}$ of a maximal ideal. When $k$ is an algebraically closed field, this field is $k$ itself. The images $a_i$ of $Y_i$ then give $\mathfrak{m}=(Y_1-a_1,\dots ,Y_s-a_s)$, proving the Weak Hilbert Nullstellensatz.

For the real Laurent polynomial ring, the same argument applies because it is the finitely generated algebraEvery residue field at a maximal ideal is consequently a finite field extension of $\mathbb{R}$, and so is $\mathbb{R}$ or $\mathbb{C}$. Here we use the fundamental theorem of algebra, which makes $\mathbb{C}$ the algebraic closure of $\mathbb{R}$ of degree two. The images of $X_1,X_2$ must be nonzero since these elements are units. It follows that the answer iswith $(\alpha ,\beta )$ and $(\overline{\alpha },\overline{\beta })$ defining the same maximal ideal. If both coordinates are real, evaluation has image $\mathbb{R}$; otherwise its image is $\mathbb{C}$. Thus every displayed kernel is maximal, and the Zariski lemma shows that this list is exhaustive.

For explicit generating sets of an ideal, there are three cases. For real nonzero $a,b$,If $\alpha =a\in {\mathbb{R}}^{\times }$ and $\beta =s+it$ with $t\ne 0$,If $\alpha =u+iv$ with $v\ne 0$ and $\beta =s+it$, thenThe last two quotients are $\mathbb{C}$; the nonzero coordinate condition ensures that the Laurent inverses exist in them. There are no further coincidences in the classification: equal kernels induce an $\mathbb{R}$-isomorphism between the evaluation image fields. The only $\mathbb{R}$-automorphisms of $\mathbb{C}$ are the identity and complex conjugation, so the two pairs must agree or be conjugate.

To prove the intersection assertion, fix a prime ideal $P$ and an element $f\notin P$. The integral domain $B=R/P$ has nonzero image $\bar{f}$. Its localization of a ring $B[{\bar{f}}^{-1}]$ is nonzero and still a finitely generated algebra over $\mathbb{R}$, so it has a maximal ideal $\mathfrak{n}$. By the Zariski lemma, the field $L=B[{\bar{f}}^{-1}]/\mathfrak{n}$ is finite over $\mathbb{R}$.

The image of $B$ in $L$ is a subring containing $\mathbb{R}$ and is a finite-dimensional $\mathbb{R}$-vector space. It is an integral domain, hence a field: multiplication by any nonzero element is an injective linear map of a finite-dimensional vector space and therefore surjective. The kernel of $B\to L$ is thus a maximal ideal which avoids $\bar{f}$, since $\bar{f}$ is invertible in $L$. Its inverse image in $R$ is a maximal ideal containing $P$ but avoiding $f$. As this works for every $f\notin P$,This is the Jacobson ring property for this Laurent polynomial ring.

For a nonzero finitely generated module $M$, its set of associated primes isHere ${\mathrm{Ann}}_R(m)=\{a\in R:am=0\}$ is the annihilator of the element. The Laurent polynomial ring $R$ is a Noetherian ring, so the nonempty collection of annihilators of nonzero elements of $M$ has a maximal member, say $P={\mathrm{Ann}}_R(m)$. It is proper because $m\ne 0$.

If $ab\in P$ and $b\notin P$, then $bm\ne 0$ andMaximality forces ${\mathrm{Ann}}_R(bm)=P$, hence $a\in P$. Thus $P$ is a prime ideal, proving $\boxed{{\mathrm{Ass}}_R(M)\ne \varnothing }$. This argument in fact only needs that $R$ is Noetherian, not the finite generation of $M$.

A proper ideal $Q$ is $P$-primary when its radical of an ideal is $P$ andEquivalently, $ab\in Q$ and $b\notin P$ imply $a\in Q$; equivalently, every zero divisor of $R/Q$ is nilpotent. These formulations distinguish the actual primary ideal property from the weaker assertion that there is only one minimal prime ideal.

For the final example, set $u=X_1-1$ and $v=X_2-1$. ThenSince $R/P\cong \mathbb{R}[X_2,X_2^{-1}]$ is an integral domain, $P$ is a prime ideal. Moreover $I\subseteq P$ and $u^2\in I$, so $\sqrt{I}=P$. Consequently $P$ is the unique minimal prime ideal over $I$.

Nevertheless $uv\in I$, while $u\notin I$ and no power of $v$ belongs to $I$. For the first assertion, reduce modulo $v$: the image of $I$ is $(u^2)$ in $\mathbb{R}[X_1,X_1^{-1}]$, and $u$ is not a multiple of $u^2$ because $u$ is a nonunit in this integral domain. For the second, $v\notin P$, so $v^j\notin P\supseteq I$ for every $j\ge 1$. Thus $I$ is not a $P$-primary ideal: a unique minimal prime does not imply primary. In $R/I$, the nonzero class of $u$ is killed by $v$, a zero divisor which is not nilpotent.

An integral domain $A$ is an integrally closed domain if every element of its fraction field which is an integral element over $A$ belongs to $A$. Explicitly, if $z\in \mathrm{Frac}A$ satisfiesthen $z\in A$. Thus $\boxed{A\text{ is integrally closed exactly when its integral closure in its fraction field is }A}$.

Suppose $A$ is a unique factorization domain and $z=a/b$ is integral, with $a,b\in A$, $b\ne 0$, and with no common irreducible element as a factor. Multiplying its monic equation by $b^d$ givesHence $b$ divides $a^d$. Every irreducible element dividing $b$ is a prime element, so it would divide $a$, contrary to the choice of the fraction. Therefore $b$ is a unit, and $z\in A$. We have proved

The going-down theorem is the following. Let $A\subseteq B$ be an integral extension of integral domains with $A$ an integrally closed domain. IfthenWe first prove this for a module-finite ring extension and then remove that extra hypothesis. We use the Lying-over theorem, the going-up theorem, and the incomparability theorem for integral extensions: primes exist over every base prime, prime chains can be extended upwards, and distinct primes over the same base prime cannot contain one another. The going-up theorem follows by applying the Lying-over theorem to the quotient by the lower prime. These results concern arbitrary integral extensions; none assumes the going-down theorem.

For the finite case, let $K=\mathrm{Frac}A$ and $L=\mathrm{Frac}B$. Since $B$ is module-finite, $L/K$ is a finite field extension. Take a finite normal field extension $E/K$ containing $L$, obtained as a splitting field of the minimal polynomials of finitely many field generators. The extension $E/K$ need not be a separable field extension. Let $C$ be the integral closure of $A$ in $E$. Then $B\subseteq C$, and $C$ is integral over both $A$ and $B$.

We need one auxiliary fact: the finite group $G={\mathrm{Aut}}_K(E)$ acts transitively on the prime ideals of $C$ over any fixed prime ideal $\mathfrak{p}$ of $A$. First observe that an element $w\in E$ fixed by all of $G$ is a purely inseparable algebraic element over $K$. Indeed, all $K$-embeddings of $K(w)$ into an algebraic closure extend to $K$-automorphisms of the normal field extension $E$. Hence the minimal polynomial of an algebraic element $w$ has only one distinct root. In characteristic zero this says $w\in K$; in characteristic $p>0$ it says $w^{p^e}\in K$ for some $e\ge 0$.

Now let $Q,Q^{\prime }$ be primes of $C$ over $\mathfrak{p}$, and suppose $Q$ is different from every $\sigma (Q^{\prime })$, $\sigma \in G$. By the incomparability theorem for integral extensions, $Q$ is contained in none of these conjugate primes. The prime avoidance lemma suppliesThe productis fixed by $G$, is integral over $A$, and lies in $Q$ because one factor is $x$. Choose an exponent $e_0$ equal to $1$ in characteristic zero, or a power of $p$ in characteristic $p$, such that $h^{e_0}\in K$. Since $A$ is an integrally closed domain, $h^{e_0}\in A$. It also lies in $Q\cap A=\mathfrak{p}$, hence in $Q^{\prime }$. Because $Q^{\prime }$ is a prime ideal, some factor $\sigma (x)$ lies in $Q^{\prime }$. This means $x\in {\sigma }^{-1}(Q^{\prime })$, contradicting our choice. Thus $Q=\sigma (Q^{\prime })$ for some $\sigma$, proving transitivity. The power $e_0$ accounts for the possible purely inseparable field extension and makes the argument valid in every characteristic.

Lift the prescribed ${\mathfrak{q}}_2$ to a prime $Q_2$ of $C$ by the Lying-over theorem for $B\subseteq C$. Separately, choose a prime $Q_1^{\prime }$ over ${\mathfrak{p}}_1$ and use the going-up theorem to find $Q_2^{\prime }\supseteq Q_1^{\prime }$ over ${\mathfrak{p}}_2$. Transitivity gives $\sigma \in G$ with $\sigma (Q_2^{\prime })=Q_2$. ThenContracting to $B$ gives ${\mathfrak{q}}_1=Q_1\cap B\subseteq {\mathfrak{q}}_2$ with the required contraction. This proves the going-down theorem for the module-finite case.

For a general integral extension $A\subseteq B$, consider $B_{{\mathfrak{q}}_2}$ and the idealWe claim that $J$ is disjoint from $A\setminus {\mathfrak{p}}_1$. Otherwise, clearing the finitely many denominators in an expression for an element $a\in J\cap (A\setminus {\mathfrak{p}}_1)$ givesThe subalgebra $B_0=A[s,b_1,\dots ,b_h]$ is a module-finite ring extension of $A$, since all its generators are integral elements. Apply the finite case to ${\mathfrak{q}}_2\cap B_0$: there is a prime $Q_1\subseteq {\mathfrak{q}}_2\cap B_0$ contracting to ${\mathfrak{p}}_1$. The displayed identity puts $sa$ in $Q_1$, but $a\notin Q_1$. Thus $s\in Q_1\subseteq {\mathfrak{q}}_2$, a contradiction.

Localize $B_{{\mathfrak{q}}_2}$ further at the multiplicative subset $A\setminus {\mathfrak{p}}_1$. The extension of $J$ is a proper ideal, by the disjointness just proved. Choose a maximal ideal containing it and contract back to a prime ideal of $B_{{\mathfrak{q}}_2}$. This prime contains $J$ and avoids $A\setminus {\mathfrak{p}}_1$. Its contraction ${\mathfrak{q}}_1$ to $B$ lies inside ${\mathfrak{q}}_2$ by the prime ideal correspondence for localization, and its contraction to $A$ is exactly ${\mathfrak{p}}_1$. This completes the proof of the going-down theorem without any finite generation assumption on $B$.

A right Noetherian ring satisfies the ascending chain condition on right ideals; equivalently, every right ideal is a finitely generated module.

Here is a noncommutative Hilbert basis theorem proof adapted to the stated hypothesis. Set $F_{-1}=0$ andThe equality follows inductively from $A+xA=A+Ax$, by moving one coefficient past one $x$ at a time. These spaces form an exhaustive filtered algebra structure on $B$, with $F_n{F}_m\subseteq F_{n+m}$. This does not assert uniqueness of the displayed expressions or the existence of a coefficient-moving automorphism.

For a right ideal $I\subseteq B$, defineEach $L_n$ is a right ideal of $A$. In fact, if $a\in L_n$ and $c\in A$, write $c{x}^n=x^{n}b+u$ with $b\in A$, $u\in F_{n-1}$; then $ac{x}^n=(a{x}^n)b+au\in I+F_{n-1}$. Also $L_n\subseteq L_{n+1}$ by right multiplication by $x$. Since $A$ is a right Noetherian ring, this ascending chain stabilizes at some $L_N$.

Choose finite generators $a_{nj}$ for each $L_n$, $0\le n\le N$, and choose $f_{nj}\in I$ with $f_{nj}-a_{nj}{x}^n\in F_{n-1}$. These finitely many elements generate $I$ as a right ideal. To see this, induct on $m$ for $f\in I\cap F_m$. Write $f=a{x}^m+u$ with $u\in F_{m-1}$; then $a\in L_m$. Put $d=\min (m,N)$ and express $a={\sum }_j{a}_{dj}{c}_j$. Write $c_j{x}^d=x^d{b}_j+u_j$ with $u_j\in F_{d-1}$. The differencelies in $I\cap F_{m-1}$, so the induction applies. At $m=0$ the remainder is zero. Hence $B$ is right Noetherian.

For the quantum torus, take $q\in k^{\times }$ and the convention $YX=qXY$. Begin with the polynomial ring $k[X]$, which is Noetherian by the Hilbert basis theorem. Adjoining $X^{-1}$ preserves the hypothesis because it commutes with $k[X]$, and gives $k[X^{\pm 1}]$. Adjoin $Y$ next. The relation $Yf(X)=f(qX)Y$ and its inverse coefficient-moving relation give $A+YA=A+AY$ for $A=k[X^{\pm 1}]$. Finally adjoin $Y^{-1}$ to $k[X^{\pm 1}][Y;\sigma ]$, where $\sigma (X)=qX$. On a monomial, $Y^{-1}{X}^i{Y}^j=q^{-i}{X}^i{Y}^{j-1}$; when $j=0$ this is in $A{Y}^{-1}$, and when $j>0$ it is in $A$. The reverse inclusion follows by the same relation. The preceding argument applies at each step, proving the quantum torus is right Noetherian. Nonzero $q$ is required for this notation.

For a noncommutative ring, a prime ideal of a noncommutative ring means a proper two-sided ideal $P$ such that $UV\subseteq P$ for two-sided ideals $U,V$ implies $U\subseteq P$ or $V\subseteq P$. Equivalently, $aAb\subseteq P$ implies $a\in P$ or $b\in P$. This definition does not require $A/P$ to be a noncommutative domain.

Retain the right Noetherian ring hypothesis for the last assertion. More generally, the ascending chain condition on two-sided ideals suffices. We claim that every proper two-sided ideal $I$ contains a product of finitely many prime ideals of a noncommutative ring, each containing $I$. If not, choose a maximal counterexample $I$. It cannot be a prime ideal of a noncommutative ring. Thus there are two-sided ideals $U,V$ strictly containing $I$ with $UV\subseteq I$: add $I$ to the two witnesses for failure of the defining condition for a prime ideal of a noncommutative ring. By maximality, both $U$ and $V$ contain products of finitely many prime ideals of a noncommutative ring containing them. Concatenating these products gives a product inside $UV\subseteq I$, a contradiction.

Apply the claim to $I=0$ in a nonzero $A$, obtaining $P_1\cdots P_r=0$. Every prime ideal of a noncommutative ring $Q$ contains one of the $P_i$, by repeated application of the definition of a prime ideal of a noncommutative ring. For the prime radical of a noncommutative ring $N$, it follows thatIndeed, $N\subseteq P_i$ gives $N^r\subseteq P_1\cdots P_r=0$, and ${\bigcap }_i{P}_i\subseteq Q$ for every $Q$ gives equality of the intersections. If $A=0$, the empty intersection is the whole zero ring and the conclusion is immediate.

The final assertion is false for arbitrary algebras without the preceding chain condition. For example, in the commutative ring $k[z_1,z_2,\dots ]/(z_i^2:i\ge 1)$ the nilradical is $(z_1,z_2,\dots )$, its only prime ideal, but the product of any number of distinct $z_i$ is nonzero. Thus this nilradical is not a nilpotent ideal.

Put $A^e=A{\otimes }_k{A}^{\mathrm{op}}$, so an $A$-bimodule is a left $A^e$-module through $(a\otimes b^{\mathrm{op}})m=amb$. The Hochschild cohomology isEquivalently, the Hochschild cochain complex has $C^n(A,M)={\mathrm{Hom}}_k(A^{\otimes n},M)$ and coboundary mapIts cohomology agrees with the displayed Ext functor because the bar resolution of an associative algebra is free over $A^e$ when $k$ is a field. The Hochschild cohomological dimension isThe supremum uses all bimodules, not merely the finitely generated one supplied in the question; an unbounded projective dimension is infinity.

An extension in this classification is a square-zero extension of an algebra: a short exact sequence $0\to M\to E\to A\to 0$, where $E$ and $A$ are unital algebras, $E\to A$ is unital, and $M$ is its two-sided ideal with $M^2=0$ and induced $A$-bimodule structure equal to the prescribed one. An equivalence of extensions is an algebra isomorphism of the middle terms commuting with the maps and inducing the identity on both $A$ and $M$. Arbitrary isomorphisms of middle algebras, or extensions without the square-zero ideal requirement, are not classified by this cohomology group.

Choose a $k$-linear map section $s:A\to E$ with $s(1)=1$. Its multiplication defectis a normalized Hochschild cocycle: $\mu (1,a)=\mu (a,1)=0$, and associativity in $E$ gives $\delta \mu =0$. Changing $s$ to $s+g$, where $g(1)=0$, changes the defect to $\mu +\delta g$, since $M^2=0$.

Conversely, for a normalized Hochschild cocycle $\mu$, put $E_{\mu }=A\oplus M$ as a vector space and defineThe Hochschild cocycle equation is exactly associativity, and $(1,0)$ is the identity. If ${\mu }^{\prime }=\mu +\delta g$, the map $E_{{\mu }^{\prime }}\to E_{\mu }$, $(a,m)\mapsto (a,m+g(a))$, is an equivalence. Conversely, every equivalence has this form after choosing sections. The normalized Hochschild cochain complex computes the same cohomology as the full complex: in the bar resolution of an associative algebra, the degenerate terms containing an inserted identity form a contractible subcomplex. Passing to the normalized bar resolution of an associative algebra, then applying the Hom functor, gives the same cohomology. Thus every class has a normalized representative. We obtain a bijection between $H{H}^2(A,M)$ and equivalence classes of square-zero extensions.

For a formal associative deformation, a completion convention is necessary. The usual star product lives on the formal power series moduleThis is the adic completion of a module applied to the ordinary tensor product, rather than literally the ordinary $A{\otimes }_{k}k[[t]]$ when $A$ is infinite-dimensional. For example, ${\sum }_{n\ge 0}{X}^n{t}^n$ lies in $k[X][[t]]$ but not in the ordinary tensor product, whose coefficient spaces have finite-dimensional span. The two agree when $A$ is finite-dimensional. We interpret the printed notation in this standard completed sense; the infinite iteration below requires that interpretation.

A star product is a $k[[t]]$-bilinear, unital product continuous for the adic topology satisfying associativity of the formIt is a trivial formal deformation if a $k[[t]]$-linear automorphism continuous for the adic topology $T=\mathrm{id}+{\sum }_{r\ge 1}{t}^r{T}_r$, with $T(1)=1$, satisfies $T(a\ast b)=T(a)T(b)$.

If $\mathrm{Dim}(A)\le 1$, then $H{H}^2(A,A)=0$. Suppose changes of coordinates have removed all coefficients below order $r$. The order-$r$ part of associativity then says $\delta {\mu }_r=0$. Hence ${\mu }_r=\delta g_r$ for a $k$-linear map $g_r:A\to A$. Its normalization gives $g_r(1)=0$, since $(\delta g_r)(1,1)=g_r(1)$. Transport the product by $T_r=\mathrm{id}+t^r{g}_r$:The order-$r$ coefficient becomes ${\mu }_r-\delta g_r=0$, and lower coefficients remain zero. Repeating constructs compatible changes of coordinates modulo every $t^N$. They converge in the adic topology to an invertible $T$ fixing $1$, with inverse obtained coefficient by coefficient. The limit product is ordinary multiplication. Therefore every star product is trivial under the completed formal-series convention. In fact, the argument only needs the vanishing of $H{H}^2(A,A)$, not all of Hochschild cohomological dimension at most one.

Use the presentation $D_6=⟨r,s:r^3=s^2=1,srs=r^{-1}⟩$. Its subgroup $⟨r⟩$ is the unique subgroup of order $3$, hence is a characteristic subgroup. An automorphism must send $r$ to $r$ or $r^{-1}$ and $s$ to one of the three reflections $r^{j}s$. Thus there are at most six automorphisms.

The center of a group of $D_6$ is trivial, so conjugation embeds $D_6$ as a group of six inner automorphisms. These exhaust the possibilities. Hence every automorphism of $D_6$ is inner.

For every $t\in T_N$, the torus action mapis an automorphism of $X_{\Sigma }$. It induces isomorphisms of local rings, so it carries smooth points to smooth points and singular points to singular points. The singular locus is consequently invariant under $T_N$. If it contains one point $x$, it contains the entire orbit $T_N\cdot x$; hence the torus-invariance of the singular locus of a toric variety proves that $X_{\Sigma }^{\mathrm{sing}}$ is a union of the orbits in the orbit-cone correspondence.

For $n\ge 2$ with $n\ne 6$, bijectivity of $\varphi$ and the previous part implyThe supplied numerical fact forces $k=1$. Thus $\varphi$ permutes the transpositions.

Suppose $n\ge 3$, and write $s_r=\varphi ((1r))$ for $r=2,\dots ,n$. These are distinct transpositions. For $r\ne t$, the product $(1r)(1t)$ has order of a group element three, so $s_r{s}_t$ also has order three. Two distinct transpositions have this property precisely when their two-element sets of moved points meet in one point: disjoint ones have a product of order two, whereas ones sharing one point have a three-cycle as product. The two-element support of a permutation associated with each $s_r$ therefore meets that of every other $s_t$.

We now show that they have a common point. Choose two; the support of a permutation is $\{a,b\}$ for one and $\{a,c\}$ for the other, where $a,b,c$ are distinct. Any further two-element set meeting both either contains $a$ or is $\{b,c\}$. If $\{b,c\}$ occurs, any two-element set meeting all three must itself be one of $\{a,b\},\{a,c\},\{b,c\}$: a set containing $a$ must also contain $b$ or $c$, and a set omitting $a$ must contain both $b,c$. Hence a pairwise-intersecting family with no common point consists of exactly those three edges of a triangle.

For $n\ge 5$, there are $n-1\ge 4$ distinct $s_r$, ruling out that triangle. For $n=3$, the two sets already have a common point. For $n=4$, the triangle alternative would make all three $s_r$ fix the fourth point. They would then generate a proper subgroup of $S_4$, contrary to surjectivity of $\varphi$ and part (a)(iv), which imply that the $s_r$ generate $S_4$. Thus in every case $n\ge 3$ there is a common point $a$, and we may writeThe $b_r$ are distinct and different from $a$, so they exhaust the remaining $n-1$ points. Define a permutation $\sigma$ byThenThe two group homomorphisms $\tau \mapsto \varphi (\tau )$ and $\tau \mapsto {\sigma }^{-1}\tau \sigma$ agree on a generating set of a group, so agree on every product of its elements, hence throughout $S_n$:Finally, $S_1$ is trivial and $S_2$ has a unique nonidentity element, so every automorphism of either is the identity and we can take $\sigma =e$. This completes all degrees $n\ne 6$ and proves that every such automorphism is an inner automorphism.

Assume first $n\ge 2$. All transpositions form a single conjugacy class $T_1$. A group automorphism preserves order of a group element, so the image of a transposition is a nonidentity involution. By the order formula, its disjoint cycle decomposition contains only one-cycles and two-cycles, with at least one two-cycle. Thus it lies in some $T_k$, where $1\le k\le \lfloor n/2\rfloor$.

Moreover an automorphism maps the entire conjugacy class onto the conjugacy class of the image: for a fixed transposition $t$,and $\varphi (g)$ runs over all of $S_n$ by surjectivity. Part (a)(i) identifies that latter conjugacy class as $T_k$. Hence $\boxed{\varphi (T_1)=T_k}$, with equality rather than just inclusion. If $n=1$, $T_1$ is empty and the equality holds trivially with $k=1$.

The cyclic subgroup $A=⟨a⟩$ has order four and index two. Every index-two subgroup is normal: its complement is both the only nontrivial left coset and the only nontrivial right coset. Thus conjugation by $b$ is an automorphism of $A$. It must send its generator $a$ to another element of order four, namely $a$ or $a^3$. Since $b^{-1}=b$,equivalently $ab=ba$ or $ab=b{a}^3$, as required. Also $b\notin A$, since otherwise $a,b$ could generate at most four elements.

In the commuting case, every word reduces uniquely to $a^j{b}^{\epsilon }$ with $0\le j<4$ and $\epsilon \in \{0,1\}$. The eight expressions are distinct because $A$ and $Ab$ are disjoint. Multiplication is addition of the two exponents modulo four and two, respectively. ThereforeThis gives the multiplication table independently of any other properties of $G$, proving uniqueness up to group isomorphism.

In the inversion case, the relation $ba=a^{-1}b$ gives the same eight unique normal forms, now with multiplicationwith exponents reduced modulo four and two. Thus this case also has a uniquely determined group structure. A faithful geometric realization is the dihedral group of a square, with $a$ a quarter-turn and $b$ a reflection in a line through opposite edge midpoints. For example, on the plane,These preserve the square with vertices $(\pm 1,\pm 1)$ and satisfy $a^4=b^2=I$, $bab=a^{-1}$. They generate eight distinct transformations, so

Use the univalent functions on $D$, namely the univalent functions $F:D\to \mathbb{C}$. Each has nonzero derivative, so its image $F(D)$ is a domain conformally equivalent to $D$, and is therefore simply connected. This image is proper: otherwise $F^{-1}:\mathbb{C}\to D$ would be a bounded nonconstant entire function, contrary to the Liouville theorem.

Conversely, every proper simply connected domain $G\subset \mathbb{C}$ is $F(D)$ for some such $F$, by taking the inverse of a Riemann mapping theorem bijection. Thus these functions parameterize all proper simply connected domains by their images. The complex plane itself is the exceptional simply connected domain and must be listed separately.

The parameterization has redundancy: two univalent functions $F_1,F_2$ have the same image if and only if $F_2=F_1\circ \varphi$ for an automorphism of the unit disk $\varphi$. All such automorphisms have the formIf an interior point $w_0\in G$ is prescribed, the normalization $F(0)=w_0$ and $F^{\prime }(0)>0$ real removes this redundancy. Equivalently one can describe these maps through their convergent Taylor series on $D$, imposing univalence on the resulting holomorphic functions.

Every automorphism fixes zero and one, and preserves sums, products, negatives and multiplicative inverses of nonzero elements. Consequently elements fixed by every member of $G$ are closed under these operations and form a field. Artin fixed-field theorem gives $\boxed{[L:L^G]=|G|}$ for a finite automorphism group.

Here $\mathbb{Q}(t)$ is the fraction field of $\mathbb{Q}[t]$, with $t$ transcendental; its elements are fractions of rational-coefficient polynomials, with nonzero denominator and the usual equality by cross-multiplication. Substitution of $1-t$ or $1/t$ defines an automorphism, each its own inverse. With $\sigma \tau$ meaning first apply $\tau$ to a field element and then $\sigma$, its image on $t$ is $1/(1-t)$. Iterating gives $(t-1)/t$ and then $t$, so its order is three. The relations reduce every word to at most six possibilities, and the six resulting images areThey are distinct rational functions, so the group has six elements. Its faithful permutation action on $\{0,1,\infty \}$ identifies it with $\boxed{S_3}$.

For the first fixed field put $u=t(1-t)$. It is fixed by $\sigma$, and $t$ satisfies $X^2-X+u=0$ over $\mathbb{Q}(u)$. The degree is exactly two because the nontrivial automorphism $\sigma$ fixes that field. The fixed-field degree is also two, soFor the inversion put $v=t+1/t$. It is fixed and $t$ satisfies $X^2-vX+1=0$. The same degree argument givesThe source requests polynomial generators in both cases. The second requirement is impossible. If a nonconstant polynomial $p(t)=a_d{t}^d+\cdots +a_0$ were fixed by inversion, $p(t)=p(1/t)$ would equate a Laurent polynomial with highest exponent $d>0$ to one with no positive exponent, a contradiction. Thus the only polynomial invariants of inversion are constants, which cannot generate its nonconstant fixed field. The fixed field of inversion in a rational function field therefore has the second displayed rational-function generator.

A finite extension is Galois when it is both normal and separable; over $\mathbb{Q}$ separability is automatic, so it is equivalently a splitting field of a polynomial with distinct roots. Its Galois group consists of the field automorphisms fixing $\mathbb{Q}$. If $G=\mathrm{Gal}(L/\mathbb{Q})$, the Fundamental theorem of Galois theory gives mutually inverse inclusion-reversing bijectionsThey satisfy $[L:L^H]=|H|$ and $[L^H:\mathbb{Q}]=[G:H]$. The intermediate extension $K/\mathbb{Q}$ is Galois exactly when $H=\mathrm{Gal}(L/K)$ is normal, and then $\boxed{\mathrm{Gal}(K/\mathbb{Q})\cong G/H}$ via restriction.

Every permutation of the four roots permutes the three unordered partitions into two pairs and hence permutes the three ${\beta }_j$. Thus each coefficient of $g(X)={\prod }_{j=1}^3(X-{\beta }_j)$ is a symmetric polynomial in the four roots. The Fundamental theorem of symmetric polynomials expresses it in the elementary symmetric functions, which are rational coefficients of $f$. Hence $g\in \mathbb{Q}[X]$. Each product ${\alpha }_i{\alpha }_j$, $i<j$, occurs exactly twice in ${\beta }_1+{\beta }_2+{\beta }_3$, so that sum is $2q$. Therefore the quadratic coefficient is $\boxed{-2q}$.

For the given cubic write $g(X)=(X-2{)}^3+12$. The translated cubic $Y^3+12$ is Eisenstein at three, hence irreducible. It has one real root and a nonreal conjugate pair. Its Galois group is transitive on three roots, so its order is divisible by three; complex conjugation is a transposition. A subgroup of $S_3$ with both a three-cycle and a transposition is all of $S_3$. HenceLet $G=\mathrm{Gal}(L/\mathbb{Q})\le S_4$. Irreducibility of the quartic makes the root action transitive, so four divides $|G|$. The cubic splitting field $K$ lies inside $L$ by the definitions of the ${\beta }_j$, and restriction onto its Galois group gives a quotient of order six. Thus six also divides $|G|$. Since $|G|$ divides 24, it is 12 or 24. If it were 12, the supplied index-two result would give $G=A_4$. The restriction kernel would then be a normal subgroup of $A_4$ of index six, contrary to the other supplied result. Thus

For the basis $w_1,\dots ,w_d$, its discriminant isEquivalently it is the square of the determinant of the matrix of the $d$ embeddings applied to the basis vectors. It is nonzero because number fields are separable. An integral basis is a basis of the ring of integers as a free abelian group; in particular it is a rational basis of the field and expresses every algebraic integer uniquely with integer coefficients.

To prove existence, start with any rational basis and multiply each vector by a suitable nonzero integer to make it integral. This is possible by clearing coefficients in its monic rational minimal polynomial of an algebraic element and rescaling the variable. Let $M$ be the lattice spanned over $\mathbb{Z}$ by these integral vectors, and let $D\ne 0$ be its discriminant. For any algebraic integer $\alpha ={\sum }_j{c}_j{w}_j$, each trace $\mathrm{Tr}(\alpha w_i)$ is an integer. The resulting linear system has integral coefficient matrix with determinant $D$. Its adjugate matrix shows $c_j\in D^{-1}\mathbb{Z}$. ThusA subgroup of a finitely generated free abelian group is free and finitely generated. Here its rank is $d$, since it contains $M$ and is contained in a rank-$d$ lattice. Its free generators are the required integral basis.

For squarefree $d\ne 1$, the quadratic integral basis isThe source allows squarefree nonzero integers, so the degenerate value $d=1$ gives $K=\mathbb{Q}$ and basis $\{1\}$ instead of a two-element basis.