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A norm-closed subalgebra of a Banach algebra is itself a Banach algebra with the inherited norm: completeness follows from closedness, and the submultiplicative norm bound restricts to it. A unital inclusion uses the same identity as the ambient algebra. Inverse membership may still differ between the two algebras; spectrum in a closed unital subalgebra describes that difference.

A unital subalgebra $A\subseteq \mathrm{End}(V)$ closed under the adjoint operator, on a finite-dimensional real or complex inner product space, equals its double commutant of an operator algebra. Here is a proof that does not require a classification of algebras. For $N=\dim V$, choose a basis $(v_1,\dots ,v_N)$ and let $W=\{(a{v}_1,\dots ,a{v}_N):a\in A\}\subseteq V^N$. The diagonal action of $A$ preserves $W$ and $W^{\perp }$, so the orthogonal projection onto $W$ commutes with that action. Each matrix entry of this projection belongs to $A^{\prime }$. Hence the diagonal action of $b\in A^{\prime \prime }$ commutes with the projection and preserves $W$. Since $1\in A$, the tuple $(v_1,\dots ,v_N)$ belongs to $W$, and therefore $(b{v}_1,\dots ,b{v}_N)=(a{v}_1,\dots ,a{v}_N)$ for one $a\in A$. This implies $b=a$. Without the unital hypothesis the conclusion is $A^{\prime \prime }=A+\mathbb{F}1$.

Every unital $k$-subalgebra $A\subseteq k[T,T^{-1}]$ is a Noetherian ring. If $A\subseteq k[T]$ or $k[T^{-1}]$, a nonconstant element $f\in A$ makes that ambient polynomial ring a finitely generated module over $k[f]$. Otherwise $A$ contains an $f$ with both positive and negative exponents, and both $T$ and $T^{-1}$ are integral elements over $k[f]$. In either case $A$ is a submodule of a finite module over the Noetherian ring $k[f]$, and every ideal of $A$ is finitely generated over $k[f]$, hence over $A$.

Differentiate the defining relation at $\Lambda =I+tX+O(t^2)$. The linear term gives the Lie algebraSolving this equation between matrices yieldswhere a convenient basis isThe first two generate Lorentz boosts and the last a spatial rotation. Direct commutators giveThere are three independent real coefficients, so ${\dim }_{\mathbb{R}}\mathfrak{g}=3$. A Cartan subalgebra is $\mathbb{R}J$, and the rank of a semisimple Lie algebra is one. Explicitly, if $X=a{K}_1+b{K}_2+cJ$ and $Y=a^{\prime }{K}_1+b^{\prime }{K}_2+c^{\prime }J$, their commutator vanishes exactly whenFor nonzero $X$, these equations force $Y$ to be proportional to $X$, so there is no two-dimensional commuting subalgebra. The split real rank is also one, realized by the semisimple Lorentz boost generator $K_1$. ThusFor this Lorentz algebra in three spacetime dimensions, disconnected components of the full Lorentz group do not change its tangent Lie algebra at the identity.

Now $B=\overline{\mathbb{C}[1,x]}$. Let $U$ be a bounded component of ${\rho }_A(x)$ and $\lambda \in U$. Its boundary lies in ${\sigma }_A(x)$: an open disk about a boundary point in the algebra resolvent set would meet $U$ and would belong to the same component, contradicting boundary membership. The maximum modulus principle on the bounded domain $U$ therefore gives, for every complex polynomial $p$,Here spectral mapping theorem follows by factoring $p(z)-\mu$: its commuting factors are invertible exactly when every root avoids the Banach algebra spectrum.

The rule $p(x)\mapsto p(\lambda )$ is well-defined by the displayed bound, multiplicative and bounded on the dense polynomial subalgebra. It extends to a unital character ${\phi }_{\lambda }$ on $B$. Since ${\phi }_{\lambda }(\lambda 1-x)=0$, that element cannot have an inverse in $B$. Thus $U\subseteq {\sigma }_B(x)$. Together with part (iii), this proves that the Banach algebra spectrum of a polynomial-generated Banach subalgebra is exactly the original Banach algebra spectrum with all its bounded complementary components filled in.

A C-star algebra is a complex Banach algebra with a conjugate-linear involution satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$ and $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. The inequality $\Vert a{\Vert }^2\le \Vert a^{\ast }\Vert \Vert a\Vert$, and then the same inequality for $a^{\ast }$, show that the involution is isometric. In a nonzero unital C-star algebra, the identity has norm one.

For a Hermitian algebra element $h$, the exponential $e^{ith}$ is unitary for real $t$ and has norm one. If $\lambda \in \sigma (h)$, then $e^{it\lambda }\in \sigma (e^{ith})$: factoring $e^{itz}-e^{it\lambda }=(z-\lambda )q(z)$ with entire $q$ proves this, since the evaluated factors commute. Hence $|e^{it\lambda }|\le 1$ for all positive and negative $t$, forcing $\mathrm{Im}\lambda =0$. Thus Hermitian algebra elements have real Banach algebra spectrum.

For a normal C-star element $a$, commutation of $a,a^{\ast }$ givesThe middle equality of norms uses the C-star identity on the Hermitian algebra element $a^{\ast }a$. Iterating along powers $2^n$ and applying the spectral radius formula proves $r(a)=\Vert a\Vert$.

Now suppose $A$ is commutative and unital. Every maximal ideal is closed: if it were dense, it would contain an element within distance less than one of the identity, hence an invertible element by the Neumann series, which is impossible for a proper ideal. Its quotient is a complex Banach division algebra and is therefore $\mathbb{C}$ by the Gelfand-Mazur theorem. Maximal ideals thus give characters. Conversely, every character $\phi$ satisfies $\phi (a)\in {\sigma }_A(a)$, because applying it to an inverse of $a-\phi (a)1$ would give a contradiction. Characters are consequently contractive. The converse spectral inclusion follows by placing the proper ideal generated by $a-\lambda 1$ in a maximal ideal. We have provedThe character space ${\Phi }_A$ is a weak-star closed subset of the dual unit ball, characterized by $\phi (1)=1$ and $\phi (ab)=\phi (a)\phi (b)$. The Banach-Alaoglu theorem makes it compact Hausdorff. Define the Gelfand transform by $\hat{a}(\phi )=\phi (a)$. For Hermitian $h$, its values are real; writing $a=h+ik$ with both terms Hermitian shows $\widehat{a^{\ast }}=\overline{\hat{a}}$. In the commutative algebra every element is normal, soThe transform is therefore an isometric unital star-homomorphism and has closed range in $C({\Phi }_A)$. Its range separates characters by their definition and is closed under complex conjugation. The complex Stone-Weierstrass theorem makes that range dense, hence equal to $C({\Phi }_A)$. This proves the Commutative Gelfand--Naimark theorem:For a nonunital C-star algebra, its C-star unitization can be constructed without assuming an operator representation on a Hilbert space. Represent $A$ by left multiplication on itself, which is isometric because testing $L_a$ on $a^{\ast }/\Vert a\Vert$ gives $\Vert L_a\Vert =\Vert a\Vert$. Adjoin the identity operator and set $\Vert z\Vert =\Vert L_z\Vert$. For $z=a+\lambda 1$ and $b\in A$, the C-star identity in $A$ gives $\Vert zb{\Vert }^2=\Vert b^{\ast }{z}^{\ast }zb\Vert \le \Vert b\Vert \Vert L_{z^{\ast }z}b\Vert$. Consequently $\Vert L_z{\Vert }^2\le \Vert L_{z^{\ast }z}\Vert \le \Vert L_{z^{\ast }}\Vert \Vert L_z\Vert$. Applying this also to $z^{\ast }$ proves that the new involution is isometric and that its C-star identity holds. The extension by the identity is complete, since the isometric image of $A$ is closed and one adjoins just one dimension. Apply the unital result to a commutative $A^{+}$; the character $\infty$ given by $a+\lambda 1\mapsto \lambda$ has kernel $A$. Under the isomorphism, this kernel consists of the continuous functions vanishing at $\infty$. Thus $A\cong C_0({\Phi }_A)$ in the nonunital case, with locally compact character space ${\Phi }_A={\Phi }_{A^{+}}\setminus \{\infty \}$.

For a Hermitian $h$ in a possibly noncommutative algebra, apply this result to the commutative unital C-star subalgebra $C^{\ast }(1,h)$ in the unitization. Its Banach algebra spectrum at $h$ equals the ambient Banach algebra spectrum: both Banach algebra spectra are real, the complement of the ambient compact real Banach algebra spectrum is connected, and the resolvent-component result of Question 1 retains that whole complement in the smaller algebra. Characters of this subalgebra are determined by their values at $h$, so its character space is homeomorphic to $\sigma (h)$. We obtain the continuous functional calculus $C(\sigma (h))\to C^{\ast }(1,h)$.

Interpret ${\mathbb{R}}^{+}$ here as $[0,\infty )$. If $\sigma (h)\subseteq [0,\infty )$, apply this calculus to $t\mapsto \sqrt{t}$, obtaining a Hermitian $k$ with $k^2=h$. In a nonunital algebra this element remains in $A$ because the defining scalar function vanishes at zero. Conversely, if $h=k^2$ with $k=k^{\ast }$, spectral mapping theorem and the real Banach algebra spectrum of $k$ give $\sigma (h)=\{t^2:t\in \sigma (k)\}\subseteq [0,\infty )$. Hence nonnegative Banach algebra spectrum is equivalent to being the square of a Hermitian algebra element. A strict-positive interpretation would exclude $h=0$, and would not be equivalent to the assertion about an arbitrary Hermitian square.

Finally, use the two continuous real functions $t\mapsto \sqrt{\max (t,0)}$ and $t\mapsto \sqrt{\max (-t,0)}$ on $\sigma (x)$. Their calculus values $u,v$ are Hermitian, belong to $A$ also in the nonunital case, and satisfy

A representation of a Banach algebra is a complex-linear multiplicative map $\pi :A\to {\mathrm{End}}_{\mathbb{C}}(X)$. A nonzero irreducible action of a unital algebra automatically has $\pi (1)=I$: the image of the idempotent $\pi (1)$ is a nonzero invariant subspace and hence is all of $X$. An algebraically irreducible representation of a Banach algebra is a nonzero action if it has no nonzero proper invariant submodule, whether closed or not. A normed representation of a Banach algebra has $\pi (a)$ bounded for each $a$ on a preassigned normed vector space $X$; continuity of $\pi$ as an operator-norm-valued map is a further conclusion. This must not be confused with topological irreducibility, which excludes only closed invariant subspaces. Unitizing the algebra extends a nonunital action by $\pi (a+\lambda 1)=\pi (a)+\lambda I$, so it is enough to discuss the unital case.

For $\xi \ne 0$, irreducibility gives $A\xi =X$. The annihilator $L_{\xi }=\{a:\pi (a)\xi =0\}$ is a maximal left ideal: an intermediate left ideal gives an intermediate submodule of $X$. It is closed by the Neumann-series argument for maximal left ideals. Consequently $A/L_{\xi }$ transports a complete quotient norm to $X$, and left multiplication is contractive in this norm. Every algebraically irreducible representation of a Banach algebra can therefore be realized continuously on a Banach space. This alone does not prove continuity for its preassigned norm.

The kernel of $\pi$ is the intersection of the closed left ideals $L_{\xi }$ and is a closed primitive ideal. The Jacobson radical is the intersection of all primitive ideals, equivalently the intersection of maximal left ideals. Thus these quotient representations detect semisimplicity. Equivalent algebra representations are related by a bijective module homomorphism.

The algebra of module endomorphisms is exactly $\mathbb{C}I$. Indeed, if a module endomorphism $D$ sends a fixed cyclic vector $\xi$ to $a\xi$, then $D(b\xi )=ba\xi$. It is induced by bounded right multiplication by $a$ on $A/L_{\xi }$, so is bounded for the quotient norm on an irreducible Banach module. Every nonzero module endomorphism is injective and surjective by irreducibility, and its inverse is bounded by the same argument. This commutant of an operator algebra is a closed division subalgebra of the bounded operators on this Banach module; the Gelfand-Mazur theorem makes it $\mathbb{C}I$.

The Jacobson density theorem now says that for finitely many linearly independent ${\xi }_1,\dots ,{\xi }_n$ and arbitrary ${\eta }_1,\dots ,{\eta }_n$, one element of $A$ sends all ${\xi }_j$ to ${\eta }_j$. Here is the induction behind it. The one-vector assertion is $A{\xi }_1=X$. Assuming the first $n-1$ coordinates can be prescribed, set $L=\{a:a{\xi }_j=0(j<n)\}$. Its image $L{\xi }_n$ is an invariant subspace. If it is $X$, it adjusts the last image without changing the first ones. If it is zero, the rule $(a{\xi }_1,\dots ,a{\xi }_{n-1})\mapsto a{\xi }_n$ is a well-defined module map $X^{n-1}\to X$. Each coordinate map is scalar by the module endomorphism result, forcing ${\xi }_n$ to be a linear combination of the earlier vectors, a contradiction. This proves the assertion.

We can now give Johnson's continuity argument with its crucial construction. If $X$ is finite dimensional, the closed kernel has finite codimension and $A/\ker \pi$ is finite dimensional, so $\pi$ is continuous. Suppose instead that $X$ is infinite dimensional. LetThis is an invariant submodule, because the orbit map of $\pi (b)\xi$ is the orbit map of $\xi$ composed with the bounded map $a\mapsto ab$. Thus either $Y=X$ or $Y=0$.

To rule out $Y=0$, choose linearly independent unit vectors ${\xi }_0,{\xi }_1,\dots$. There are elements $a_n\in A$ such that, with $c_n=a_n\cdots a_1$,For completeness, construct them in the canonical quotient-norm Banach module. At stage $n$ the current tail vectors are independent. In the closed left annihilator of its first vector, density permits arbitrary images of any finite list of the remaining vectors. The condition that finitely many such images are independent is open in this Banach subspace and dense: perturb any element by $t$ times an element prescribing independent images; a suitable determinant is a nonzero polynomial in $t$, with only finitely many exceptional values. The Baire category theorem intersects these open dense conditions for all initial finite tail lists. The resulting $a_n$ kills the first current vector and retains independence of the entire remaining tail. This is the countable annihilation sequence in an irreducible Banach module.

The gliding-hump continuity principle supplies the other ingredient: if $E,F$ are Banach spaces, $S:E\to F$ is linear, $T_j:E\to E$ and $U_n:F\to F_n$ are bounded, and $U_{n}S{T}_1\cdots T_m$ is continuous whenever $m>n$, then $U_{n}S{T}_1\cdots T_n$ is continuous eventually. The product form of this principle is recorded in Proposition 1.3 of Thomas's paper. Here is the contradiction mechanism. Rescale so $\Vert T_j\Vert ,\Vert U_n\Vert \le 1$ and put $P_m=T_1\cdots T_m$. If infinitely many diagonal maps are discontinuous, choose increasing $n_j$ and successively tiny $x_j$ soAlso choose $\Vert x_j\Vert \le 2^{-j}/(1+{\max }_{i<j}\Vert U_{n_i}S{P}_{n_i+1}\Vert )$. The convergent sum $x={\sum }_j{P}_{n_j}{x}_j$ has, after its $j$th term, a tail factoring through $P_{n_j+1}$. Applying that continuous composite bounds the tail contribution by one. The display forces $\Vert U_{n_j}Sx\Vert >j$, contradicting $\Vert U_{n_j}Sx\Vert \le \Vert Sx\Vert$. Only finite decompositions and the continuous tail composites are used; no unjustified interchange of a discontinuous map with an infinite sum occurs.

Apply this principle with $E=A$, $F=\mathcal{B}(\overline{X})$, where $\overline{X}$ is the completion in the original norm, and $S(a)$ the bounded extension of $\pi (a)$. Set $T_j(a)=a{a}_j$ and $U_n(R)=R{\xi }_n$. ThenFor $m>n$ this is zero, hence continuous. For $m=n$ it is the orbit map of a nonzero vector, hence discontinuous if $Y=0$. This contradicts the principle. Therefore $Y=X$.

Finally, apply the Uniform boundedness principle on the Banach space $A$ to the continuous maps $a\mapsto \pi (a)\xi$ indexed by $\Vert \xi \Vert \le 1$. They are pointwise bounded because each individual $\pi (a)$ is bounded in the original norm. It givesThis proves Johnson's continuity theorem for irreducible normed representations, including when the original normed vector space is incomplete. The quotient norm, countable density construction, and gliding hump explain the substantive steps rather than assuming the desired continuity in advance.

First, a maximal left ideal $L$ is closed. Its norm closure is again a left ideal, since multiplication is continuous. If that closure were all of $A$, some $\ell \in L$ would satisfy $\Vert 1-\ell \Vert <1$ and would be invertible by the Neumann series. But a proper left ideal cannot contain an invertible element: ${\ell }^{-1}\ell =1$ would put $1$ in $L$. Maximality therefore forces $\overline{L}=L$.

Consequently $X=A/L$ is a nonzero quotient Banach space. It is a simple module for left multiplication by $A$, because the inverse image of any submodule under $A\to A/L$ is a left ideal containing $L$, and hence is either $L$ or $A$. Since $La\subseteq L$, right multiplication defines a well-defined bounded module endomorphismIndeed, changing $u$ by an element of $L$ changes $ua$ by an element of $L$, and taking the infimum over representatives proves the quotient norm bound. Right multiplication commutes with the left action by associativity.

Choose $\lambda$ in the nonempty spectrum of an element of $T$ in the Banach algebra $\mathcal{B}(X)$. Both the kernel and range of $T-\lambda I$ are submodules. If this module endomorphism were nonzero, simplicity would force its kernel to be zero and its range to be all of $X$. It would be a bounded bijection of Banach spaces, with bounded inverse by the bounded inverse theorem, contradicting the choice of $\lambda$. Thus $T=\lambda I$. Applying this identity to $1+L$ yields $a-\lambda 1\in L$. If also $a-\mu 1\in L$, then $(\lambda -\mu )1\in L$; since $1\notin L$, $\lambda =\mu$. ThereforeThis is the scalar right action on a maximal-left-ideal quotient; it does not assume that $L$ is a two-sided ideal or that $A/L$ is an algebra.

Now consider the center of an associative algebra $Z$. For each $u\in A$, the map $z\mapsto zu-uz$ is a continuous linear map, so its kernel is closed. Their intersection $Z$ is therefore closed. The identity belongs to $Z$, as do sums and scalar multiples of its elements. If $z,w\in Z$, then for every $u\in A$,Thus $zw\in Z$. Also $zw=wz$, since $z$ commutes with every element of $A$, including $w$. Hence the center is a closed commutative unital subalgebra, and is a Banach algebra in the inherited norm.

For every $z\in Z$, the inclusion $Lz=zL\subseteq L$ allows the first argument to apply. Let $\chi (z)$ be the unique scalar for which $z-\chi (z)1\in L$. Equivalently, the induced right multiplication on $A/L$ is $T_z=\chi (z)I$. Sincethe map $\chi :Z\to \mathbb{C}$ is a complex-linear unital character of an algebra. It is surjective because $\chi (\alpha 1)=\alpha$, and its kernel is precisely $L\cap Z$. ConsequentlyIndeed any ideal properly containing the kernel would have a nonzero image in the field $\mathbb{C}$ and thus would be all of $Z$. The algebra character is also continuous: $|\chi (z)|=\Vert \chi (z)I\Vert =\Vert T_z\Vert \le \Vert z\Vert$, although continuity is not needed for maximality.

For a discrete group, one standard definition of amenability is the fixed-point property: every continuous affine action on a nonempty compact convex set in a locally convex space has a fixed point. The equivalent state formulation, which may be stated without proof here, is that there is a left-invariant invariant mean on ${\ell }^{\infty }(\Gamma )$. Equivalently, every unital left-translation-invariant star-subalgebra of ${\ell }^{\infty }(\Gamma )$ admits a left-invariant state, with continuity for the supremum norm; one may equally pass to its norm closure. A state means a positive linear functional of norm one with value one on the constant function one. Restriction proves the subalgebra formulation from the full-algebra version, and taking the full algebra proves the converse. Such a mean is finitely additive on indicator functions, not necessarily countably additive.

Write $(\lambda (g)\mu )(h)=\mu (g^{-1}h)$ on ${\ell }^1(\Gamma )$ and $(L_{g}f)(h)=f(g^{-1}h)$ on bounded functions. We prove equivalence of this mean condition, the Reiter condition, and existence of a Følner sequence.

Assume first that an invariant mean $m$ exists. Fix a finite set $K\subseteq \Gamma$. In the real Banach space ${⨁}_{g\in K}{\ell }^1(\Gamma )$ with sum norm, consider the convex setIf zero were outside its norm closure, the Hahn-Banach separation theorem would give bounded real functions $f_g$ and $\delta >0$ such thatfor every such $\mu$. Testing point masses at $h$ showsApplying the positive invariant mean makes the left side zero and the right side at least $\delta$, a contradiction. Therefore zero is in the closure: for any $\epsilon >0$ a finitely supported probability measure makes the sum of the translation defects over $K$ less than $\epsilon$. Enumerate the countable group, take the first $n$ elements for $K$ and $\epsilon =1/n$, and obtain a sequence satisfying the Reiter condition. This proves (1) implies (2).

Next assume (2). An ${\ell }^1$ probability measure can be approximated in norm by finitely supported probability measures, by restricting to a large finite set and renormalizing. Replacing $\mu$ by $\nu$ changes each defect by at most $2\Vert \mu -\nu {\Vert }_1$. Thus simultaneous small defects for finitely many translations can be achieved with finite support.

For such $\nu$, set $F_t=\{h:\nu (h)>t\}$. The elementary identity $|r-s|={\int }_0^{\infty }|1_{r>t}-1_{s>t}|dt$ for $r,s\ge 0$ givesIf the sum of defects over $K$ is less than $\epsilon$, some $t$ with $F_t\ne \varnothing$ must satisfy ${\sum }_{g\in K}|F_t△g{F}_t|<\epsilon |F_t|$; otherwise integration would contradict that strict inequality. Applying this layer-cake extraction of Følner sets to successive finite sets of translations gives nonempty finite $F_n$ with each relative boundary tending to zero. Hence (2) implies (3).

Conversely, normalized counting measures ${\mu }_n=1_{F_n}/|F_n|$ satisfyso (3) implies (2). Finally, from (2) define states $m_n(f)={\sum }_h{\mu }_n(h)f(h)$. The dual unit ball of ${\ell }^{\infty }(\Gamma )$ is weak-star compact by the Banach-Alaoglu theorem; take a convergent subnet, since this compactness need not be metrizable. Its limit $m$ is positive with $m(1)=1$, andThus $m$ is invariant, proving (2) implies (1). Together these implications establish all three equivalences. Countability is used to obtain sequences from finite tests; it is not a justification for replacing the compactness subnet by a subsequence.

Let $V$ be a finite-dimensional real or complex inner product space, and let $A\subseteq \mathrm{End}(V)$ be a unital subalgebra stable under the adjoint operator. Define its commutant of an operator algebra by $A^{\prime }=\{T:Ta=aT\text{ for all }a\in A\}$. The double commutant theorem for star-algebras asserts $\boxed{A^{\prime \prime }=A}$. The inclusion $A\subseteq A^{\prime \prime }$ follows immediately from the definition; we prove the reverse.

Write $N=\dim V$, fix a basis $v_1,\dots ,v_N$, and act diagonally on $V^N$. The linear subspaceis invariant under every diagonal $a$. Since $A$ is closed under adjoints, its orthogonal complement $W^{\perp }$ is invariant too: $⟨aw,z⟩=⟨w,a^{\ast }z⟩=0$ for $w\in W^{\perp }$, $z\in W$. Hence the orthogonal projection $P$ onto $W$ commutes with every diagonal $a$.

Writing $P$ as an $N$ by $N$ block matrix $(P_{ij})$, this says $P_{ij}a=a{P}_{ij}$, so each $P_{ij}\in A^{\prime }$. If $b\in A^{\prime \prime }$, it commutes with every $P_{ij}$, and its diagonal action therefore commutes with $P$ and preserves $W$. Because $1\in A$, $(v_1,\dots ,v_N)\in W$. Therefore there is one $a\in A$ such thatEquality on the basis gives $b=a$, proving the theorem over either scalar field.

The identity hypothesis is necessary. For a possibly nonunital adjoint-stable subalgebra, adjoining the ambient identity does not change its commutant of an operator algebra, so the exact statement is $A^{\prime \prime }=A+\mathbb{F}I$. For example $A=\{0\}$ has $A^{\prime \prime }=\mathbb{F}I$ on a nonzero $V$. Thus the usual exam formulation takes the algebra to contain the identity.

The real Stone-Weierstrass theorem states that if $X$ is compact Hausdorff and $A\subseteq C(X,\mathbb{R})$ is a subalgebra containing the constants and separating points, then $A$ is uniformly dense in $C(X,\mathbb{R})$. Let $B$ be its uniform closure, still an algebra over a field. If $h\in B$ and $M\ge \Vert h{\Vert }_{\infty }$, scale the permitted polynomial approximation of absolute value on $[-1,1]$ to $[-M,M]$. Applying these polynomials to $h$ proves $|h|\in B$. Thereforebelong to $B$: it is a lattice.

Fix $f\in C(X,\mathbb{R})$ and $\epsilon >0$. For every $x,y\in X$, point separation and an affine adjustment give $g_{x,y}\in A$ with $g_{x,y}(x)=f(x)$ and $g_{x,y}(y)=f(y)$; if $x=y$, a constant suffices. For fixed $x$, the open sets where $g_{x,y}>f-\epsilon$ cover $X$. Select finitely many and take their maximum $h_x\in B$. Then $h_x>f-\epsilon$ everywhere and $h_x(x)=f(x)$. By continuity, $h_x<f+\epsilon$ on some neighborhood $V_x$ of $x$. Choose finitely many $V_x$ covering $X$ and take the minimum of the corresponding $h_x$. The result $h\in B$ satisfies $f-\epsilon <h<f+\epsilon$ everywhere. Since $B$ is closed and $\epsilon$ arbitrary, $f\in B$. Thus $\overline{A}=C(X,\mathbb{R})$, proving the theorem.

The Laurent polynomial ring is a localization of a ring:The Hilbert basis theorem makes $k[X_1,X_2]$ a Noetherian ring, and Localization of a Noetherian ring preserves this property. Thus $\boxed{R\text{ is Noetherian}}$.

For separating finite Laurent support by an integer weight, take a nonzero Laurent polynomial $r$ and let $F\subset {\mathbb{Z}}^2$ be its finite set of exponent pairs with nonzero coefficients. Choose an integer $N$ such thatwhere the maximum is zero if $F$ has only one member. Set $(a_1,a_2)=(1,N)$. If $m_1+N{m}_2=m_1^{\prime }+N{m}_2^{\prime }$, thenso $m_2=m_2^{\prime }$ and then $m_1=m_1^{\prime }$. Consequently all powers of $T$ in ${\theta }_{(1,N)}(r)$ are distinct. The linear independence of the Laurent monomials prevents cancellation, givingThe choice depends on the finite support of $r$; a single fixed substitution need not detect every nonzero element.

For Noetherianity of subalgebras of a one-variable Laurent polynomial ring, let $A$ be a unital $k$-subalgebra of $k[T,T^{-1}]$. If $A=k$, it is already a Noetherian ring. Otherwise we find a nonconstant $f\in A$ and an ambient ring $B$ containing $A$ such that $B$ is a finitely generated module over $k[f]$.

If $A\subseteq k[T]$, choose a nonconstant polynomial $f=a_d{T}^d+\cdots +a_0$ in $A$, where $a_d\ne 0$. Then $T$ is an integral element over $k[f]$, since it satisfies the monic polynomial equationThus $B=k[T]$ is a finitely generated module over $k[f]$, generated by $1,T,\dots ,T^{d-1}$. The case $A\subseteq k[T^{-1}]$ is identical after replacing $T$ by $T^{-1}$.

In the remaining case $A$ is contained in neither of these two polynomial rings. There is an element with a positive exponent and an element with a negative exponent. If either already has both, use it as $f$. Otherwise their sum has both, since the positive powers of the first and the negative powers of the second cannot cancel. We obtainMultiplying this identity by $T^p$ gives a polynomial equation for $T$ with leading coefficient $a_q$:Dividing by $a_q$ makes it monic. Similarly, multiplication by $T^{-q}$ gives a monic equation for $T^{-1}$ after division by $a_{-p}$. Both generators are therefore integral elements over $k[f]$, so $B=k[T,T^{-1}]$ is a finitely generated module over $k[f]$.

In all three cases $k[f]$ is a Noetherian ring, as a quotient of a one-variable polynomial ring. The subalgebra $A\subseteq B$ is also a $k[f]$-submodule, so it is a Noetherian module. Every ideal of $A$ is a $k[f]$-submodule of $A$ and is therefore finitely generated over $k[f]$, hence over $A$. This proves

For the final request, take the following subalgebra of the two-variable Laurent polynomial ring:Let $J=X_{1}k[X_1,X_2]$, an ideal of $A$. We haveas $k$-vector spaces. In particular, the classes of $X_1{X}_2^j$, for $j\ge 0$, are linearly independent. If $J$ had finitely many generators as an $A$-module, their classes would span $J/J^2$ over $A/J=k$, contradicting its infinite dimension. Hence $J$ is not finitely generated and $A$ is not a Noetherian ring. This illustrates why the preceding result is special to one Laurent variable.

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 $B$.

Since a subalgebra is closed under addition and scalar multiplication, the given absolute-value property yieldsThe same identity with a minus sign gives $\min (f,g)\in A$. These are pointwise operations on real-valued continuous functions.

The real Stone-Weierstrass theorem says that a real subalgebra $A\subset C(K,\mathbb{R})$ which contains the constants and separates points is dense in $C(K,\mathbb{R})$ for the uniform norm, when $K$ is compact Hausdorff space. If $A$ is also closed, $A=C(K,\mathbb{R})$. The complex version needs closure under complex conjugation; that extra condition is unnecessary here.

For a closed unital subalgebra $B$ of a unital Banach algebra $A$ and $x\in B$, the smaller-algebra algebra resolvent set is relatively open and closed in the larger one. Openness follows from the Neumann series; closedness follows from continuity of inversion in $A$ and norm closedness of $B$. Every larger-algebra resolvent component is therefore either retained whole or absorbed into the smaller-algebra Banach algebra spectrum. The unbounded component is always retained, since inverses at large modulus are norm-convergent series in $x$.

The algebra resolvent set consists of those scalars for which $\lambda 1-x$ is invertible in the algebra. It is open by the Neumann series, and the inverse-valued map is the resolvent of an element. A closed subalgebra can have a smaller algebra resolvent set because an ambient inverse need not lie in it.

A linear subalgebra closed under the involution. In bounded operators, this means closure under the adjoint operator; neither norm closure nor operator-topology closure is part of the definition. These distinctions matter in the Von Neumann double commutant theorem and the Kaplansky density theorem.

An associative algebra with a two-sided multiplicative identity. For a subalgebra of bounded operators, the term unital here means that it contains the identity of the represented Hilbert space.