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The multiplicative norm first rules out zero divisors: if $xy=0$ and $x\ne 0$, then $0=|x||y|$ forces $y=0$. If the dimension is zero the claimed bound already holds, so suppose it is positive.

It is worth checking that the printed description of a ring with an underlying real vector space really permits differentiation of multiplication. This follows from the norm identity even if compatibility with real scalars is not separately stated. For fixed $x$, the map $L_x(y)=xy$ is additive by distributivity, andIt is therefore continuous. Additivity gives rational linearity. Approximating a real number by rationals and using continuity proves $L_x(ty)=t{L}_x(y)$ for every real $t$. The same argument holds in the other variable. This is automatic real bilinearity from a multiplicative norm, so multiplication is a bilinear map between finite-dimensional vector spaces, and hence is smooth.

Let $S=\{x\in A:|x|=1\}$ be the unit sphere of the given positive-definite bilinear form. It is a copy of $S^{n-1}$. The squaring mapis well defined because $|x^2|=|x{|}^2=1$, and is smooth. Commutativity gives its derivativeFor $x\in S$ and $v\ne 0$, the norm identity gives $|D{Q}_x(v)|=2|v|$, so the derivative is injective, including on the tangent space $T_{x}S=x^{\perp }$. It takes tangent vectors into $T_{x^2}S$ since $Q$ maps the sphere to itself. Alternatively, polarization of $|xy{|}^2=|x{|}^2|y{|}^2$ gives $⟨xy,xz⟩=|x{|}^2⟨y,z⟩$, so $⟨x^2,2xv⟩=2⟨x,v⟩=0$ directly.

The only identifications under squaring are antipodal:The factorization uses commutativity, and the last implication uses the absence of zero divisors. Conversely $Q(-x)=Q(x)$. It follows that $Q$ induces an injective smooth mapThe antipodal quotient map is a local diffeomorphism, so the injectivity of $DQ$ implies injectivity of $Dq$. Since Real projective space is compact, the stated compact injective immersion is an embedding theorem makes $q$ a closed embedding.

For $n\ge 2$ its source and target have the same dimension $n-1$. The inverse function theorem makes its image open as well as closed in the connected sphere $S^{n-1}$. It is nonempty, so the image is the whole sphere; consequently $q$ is a diffeomorphism. But for $n\ge 3$ this is impossible:Indeed, the antipodal two-sheeted covering map $S^{n-1}\to {\mathbb{RP}}^{n-1}$ has simply connected total space in this range, and is its universal cover, with deck group $\mathbb{Z}/2$. A sphere of dimension at least two is simply connected. These fundamental groups cannot be those of homeomorphic spaces.

We have obtained the dimension bound for a commutative Euclidean normed algebra:The real numbers and complex numbers with their usual products and Euclidean norms attain dimensions one and two. In dimension two, the induced map ${\mathbb{RP}}^1\to S^1$ is entirely compatible with the argument: the circle squaring map identifies exactly the two antipodal preimages. We neither assume a multiplicative identity nor use a classification of real division algebras; the squaring argument proves the bound from the supplied hypotheses.

Every Brauer class has a central division algebra representative $D$, and every central simple algebra is a matrix algebra over such a $D$. Put $d=\mathrm{deg}D$; its reduced norm is a homogeneous degree-$d$ polynomial on the $d^2$-dimensional vector space $D$.

If $d>1$, then $d^2>d$, so the $C_1$ property supplies a nonzero $x\in D$ with $\mathrm{Nrd}(x)=0$. But every nonzero element of a division algebra is invertible, and its reduced norm cannot be zero. Thus $d=1$, and a central division algebra of degree one is $K$ itself. Every central simple algebra over $K$ is therefore a matrix algebra over $K$, givingThis proves the trivial Brauer group of a C1 field. The essential ingredient is the anisotropy of a division algebra's reduced norm, not the norm polynomial of an arbitrary matrix algebra, which certainly can vanish on nonzero singular matrices.

Let $G=\mathrm{Gal}(L/K)$. A semilinear action on an $L$-vector space $V$ is an additive action with $\sigma (av)=\sigma (a)\sigma (v)$. The Galois descent of vector spaces theorem says that $V^G$ is a $K$-vector space and the natural mapis an isomorphism. In particular ${\dim }_K{V}^G={\dim }_{L}V$. Morphisms also descend: an $L$-linear equivariant map is the scalar extension of its restriction to invariant subspaces.

We prove that the invariant vectors span $V$ over $L$, without dividing by $|G|$. For $a\in L$ and $v\in V$, the vectoris invariant, because every $\tau \in G$ merely permutes the terms. Suppose their $L$-span were proper. There would be a nonzero $L$-linear functional $\lambda$ annihilating that span. For each fixed $v$, we would haveThe Artin independence theorem makes all coefficients zero, in particular $\lambda (v)=0$ for every $v$, a contradiction. To recall its elementary proof, take a nontrivial linear relation between distinct field automorphisms with the fewest nonzero coefficients. Replace the input $a$ by $ab$ and subtract one automorphism's value on $b$ times the original relation. Choose $b$ on which two participating automorphisms differ. The resulting nonzero relation has fewer terms, contradicting minimality.

Choose an $L$-basis $v_1,\dots ,v_m$ consisting of invariant vectors. If $v={\sum }_i{a}_i{v}_i$ is invariant, uniqueness of coordinates gives $\sigma (a_i)=a_i$ for every $\sigma$, so every $a_i$ lies in $K$. These vectors are therefore a $K$-basis of $V^G$, proving the displayed map is an isomorphism. Restriction and scalar extension are visibly inverse on equivariant morphisms, completing the descent theorem. The argument remains valid when the characteristic divides $|G|$.

Now put $B=M_n(L)$, and let ${\sigma }_0$ act on $B$ by applying $\sigma$ to each entry. The group ${\mathrm{Aut}}_L(B)$ is a $G$-group under $\sigma (h)={\sigma }_{0}h{\sigma }_0^{-1}$. Its nonabelian first cohomology consists of cocycles $c_{\sigma }$ satisfyingmodulo $c_{\sigma }^{\prime }=b{c}_{\sigma }\sigma (b{)}^{-1}$ for $b\in {\mathrm{Aut}}_L(B)$. This convention is equivalent to the usual one after replacing $b$ by its inverse. The matrix-automorphism calculation above identifies this coefficient group with ${\mathrm{PGL}}_n(L)$.

For a central simple algebra $A$ of degree $n$ split by $L$, choose $\varphi :A{\otimes }_{K}L\to B$. Transport the natural semilinear action to $T_{\sigma }=\varphi (1\otimes \sigma ){\varphi }^{-1}$ and setIt is $L$-linear and multiplicative, and $T_{\sigma }{T}_{\tau }=T_{\sigma \tau }$ gives the cocycle identity. Replacing $\varphi$ by $b\varphi$ changes $c$ by the displayed coboundary equivalence. Isomorphic $K$-algebras give the same class, so this defines the descent classification of central simple algebras map naturally.

Conversely, a cocycle defines a semilinear algebra action $T_{\sigma }=c_{\sigma }{\sigma }_0$. LetThe vector-space descent theorem gives $A_c{\otimes }_{K}L\cong B$, with the natural map also respecting multiplication and the identity. Hence ${\dim }_K{A}_c=n^2$. If $I$ is a nonzero two-sided ideal of $A_c$, then $I{\otimes }_{K}L$ is a nonzero ideal of $B$, so it is all of $B$. Dimensions force $I=A_c$, proving simplicity. If $z$ is central in $A_c$, it commutes with its $L$-span $B$, and thus is a scalar matrix with scalar $a\in L$. Twisted invariance of a scalar matrix means $\sigma (a)=a$, so $a\in K$. The center is exactly $K$. Thus $A_c$ is central simple, of degree $n$, and split by $L$.

Equivalent cocycles have conjugate semilinear actions: $T_{\sigma }^{\prime }=b{T}_{\sigma }{b}^{-1}$, so $b$ restricts to a $K$-algebra isomorphism between the fixed algebras. In the other direction, an isomorphism of fixed algebras extends to an $L$-algebra automorphism of $B$ and intertwines the two actions, giving exactly that equivalence relation. Starting from $A$ recovers the fixed algebra of $A\otimes L$, namely $A$; starting from a cocycle recovers its action. ThereforeThis is a bijection of pointed sets, with the split algebra corresponding to the trivial cocycle. Fixed-degree algebra classes do not carry the Brauer group's tensor-product group law, so a pointed-set interpretation is the appropriate meaning of the isomorphism here.

We use Zorn's lemma, and no basis or other existence theorem from linear algebra. A subset $B\subseteq V$ is linearly independent if every finite relation ${\sum }_{b\in F}{\lambda }_{b}b=0$ forces every coefficient to be zero.

Order the linearly independent subsets by inclusion. This partially ordered set is nonempty, because the empty subset is independent. An empty chain has the empty subset as an upper bound. For any nonempty chain, its union is independent: every finite subset of that union lies in one member of the chain. Indeed choose one chain member containing each of its finitely many vectors, then take the largest among that finite collection of comparable members. Every finite relation in the union is therefore a relation in an independent chain member. The union is an upper bound.

Zorn's lemma gives a maximal independent subset $B$. If $v$ were outside its span, then $B\cup \{v\}$ would still be independent. In a finite relationa nonzero $\lambda$ could be inverted in the field $k$, placing $v$ in the span of $B$. If $\lambda =0$, independence of $B$ forces the remaining coefficients to be zero. This contradicts maximality, so $B$ spans $V$.

Define the free module on $B$ as the direct sum $k^{(B)}={⨁}_{b\in B}k$: its elements are coefficient families with finite support. The mapis a module homomorphism. Spanning proves surjectivity; independence proves injectivity. Hence the vector-space freeness from maximal independence givesFinite support is essential; this is a direct sum, not an unrestricted product. The zero vector space uses $B=\varnothing$ and is free of rank zero. The proof for arbitrary vector spaces explicitly uses the axiom of choice through Zorn's lemma.

A set is linearly dependent if some finite selection of distinct vectors admits a relation ${\sum }_j{c}_j{v}_j=0$ with coefficients not all zero. For a finite set this is just such a nontrivial relation using its vectors.

Prove dependence of $n+1$ vectors in ${\mathbb{R}}^n$ by induction, starting with the zero-dimensional case. If all first coordinates vanish, select $n$ of the vectors and apply induction in ${\mathbb{R}}^{n-1}$. Otherwise relabel so the last vector has first coordinate $a\ne 0$. For $1\le j\le n$, put $w_j=v_j-(v_{j,1}/a)v_{n+1}$. These $n$ vectors have zero first coordinate, so induction supplies ${\sum }_{j=1}^n{c}_j{w}_j=0$, with some $c_j\ne 0$. Expanding gives a nontrivial linear dependence among the original vectors. The same proof works over any field.

Now express vectors of $V$ in coordinates relative to its given $n$-element basis. Any other basis has at most $n$ elements: more would contain $n+1$ linearly dependent vectors. If it has $m<n$ elements, expressing the original basis in its coordinates makes the original $n$ vectors linearly dependent, a contradiction. Thus every basis has exactly $n$ elements.

Finally, the functions $\sin (jx)$, $j=1,2,\dots$, are bounded and continuous. Any finite relation among them can be multiplied by $\sin (rx)$ and integrated over $[0,2\pi ]$. The orthogonality identityforces every coefficient to vanish. Arbitrarily large linearly independent finite sets exist, so the vector space of bounded continuous functions is infinite-dimensional.

A quadratic form on a finite-dimensional real vector space is a function $q(v)=B(v,v)$ for a symmetric bilinear form $B$. Conversely $B(u,v)=\frac{1}{2}(q(u+v)-q(u)-q(v))$ recovers that bilinear form. Two quadratic forms are congruent if a linear isomorphism $T$ satisfies $q_2(v)=q_1(Tv)$; in coordinates this is matrix congruence, $A_2=P^T{A}_{1}P$, with $P$ invertible.

Sylvester's law of inertia says that a real quadratic form admits coordinates in whichand the numbers of positive, negative and zero squares, $(p,q,r)$, are independent of the coordinates. Here $p+q+r=\dim V$; $p$ and $q$ are the maximal dimensions of subspaces on which the quadratic form is positive definite and negative definite, while $r$ is the dimension of its radical of a bilinear form. These numbers are the inertia of a bilinear form. The rank of a quadratic form is $p+q$, and the signature of a quadratic form, in the signed-difference convention, is $p-q$.

Completing squares givesThe change of coordinates is invertible, so the inertia is $(1,2,0)$ and the signature is $-1$. If signature is recorded as an ordered pair instead, that pair is $(1,2)$.

The point to retain in the direct-image theorem is quasi-compactness of inverse images of affine opens and of overlaps; no separation assumption has been supplied. The underlying space of a Noetherian scheme is Noetherian, so every open subset of $X$ is quasi-compact. Fix an affine open subscheme $V=\mathrm{Spec}A$ of $Y$, put $W=f^{-1}V$, and choose a finite open cover by affine open subschemes $W={\bigcup }_i{U}_i$. For each pair $i,j$, choose a finite affine cover $(W_{ij\ell }{)}_{\ell }$ of $U_i\cap U_j$. Empty overlaps contribute no terms.

Write $M=\Gamma (W,\mathcal{F})$. The sheaf gluing axiom gives the exact sequencewhere $\delta$ is the difference of the two restrictions to each overlap chart. All terms are $A$-modules through $f$. For $a\in A$, the inverse image of $D(a)$ cuts each $U_i$ and $W_{ij\ell }$ by a principal open subscheme. Because $\mathcal{F}$ is a quasi-coherent sheaf, sections on these smaller affine charts are the corresponding module localizations at $a$. Exactness of localization and its commutation with finite products now identify the localized equalizer with the equalizer for the restricted cover. ThusThe isomorphisms respect restrictions, proving $(f_{\ast }\mathcal{F}){|}_V\cong \tilde{M}$ on the basis of principal open subschemes. Since $V$ was arbitrary, $f_{\ast }\mathcal{F}$ is a quasi-coherent sheaf. The finite covers of overlaps are what allow the proof to work for nonseparated Noetherian schemes; this is the quasi-coherence of direct image under a quasi-compact quasi-separated morphism.

For a quasi-coherent sheaf which is not coherent but has coherent direct image, use $f:{\mathbb{P}}_k^1\to \mathrm{Spec}k$ andThis infinite negative-twist sum with zero global sections is quasi-coherent: on each standard affine chart it is the sheaf associated with a direct sum of free rank-one modules. Its stalk at every point, modulo the maximal ideal, is an infinite-dimensional vector space. A finitely generated module would have a finite-dimensional quotient, so $\mathcal{F}$ is not a coherent sheaf.

Nevertheless, $\Gamma ({\mathbb{P}}^1,\mathcal{O}(-1))=0$. This follows from cohomology of twisting sheaves on projective space, or directly by gluing on the two standard affine charts: if $z=t_1/t_0$, a section is a polynomial $p(z)$ on the first chart and $q(z^{-1})$ on the second with $p(z)=z^{-1}q(z^{-1})$, which forces both to vanish. Global sections commute with this direct sum: they are the kernel of the difference map for the two-chart cover, and direct sums commute with that finite equalizer of modules. Hencewhich is coherent on $\mathrm{Spec}k$. Both schemes in this example are Noetherian.

For a finite morphism, such an example is impossible. On an affine open subscheme $V=\mathrm{Spec}A$ of the target, its inverse image is $\mathrm{Spec}B$, with $B$ a finite $A$-module. Write $\mathcal{F}{|}_{f^{-1}V}=\tilde{M}$. Its direct image corresponds to $M$ considered as an $A$-module. If the direct image is coherent, $M$ is finitely generated over $A$. The same generators also generate it over $B$, because $A$ acts through $B$. Since $B$ is Noetherian, $\tilde{M}$ is coherent. Conversely, a finite set of $B$-generators combined with a finite set of $A$-generators of $B$ gives finitely many $A$-generators of $M$. Thus coherence reflected by finite direct image gives the stronger equivalence

Finally, let $Y=\mathrm{Spec}k[x,y]$, let $X=Y\setminus \{(x,y)\}$ be the punctured affine plane, and let $j:X\hookrightarrow Y$ be the open immersion. Both are integral schemes, but $j$ is not an isomorphism of schemes because it omits a point. The cover $X=D(x)\cup D(y)$ givesinside the field of fractions $k(x,y)$. Indeed, in a reduced fraction, membership in the first localization forces every denominator factor to be associated to $x$, while membership in the second forces it to be associated to $y$. Unique factorization and coprimality force the denominator to be a unit. By the theorem just proved, $j_{\ast }{\mathcal{O}}_X$ is quasi-coherent on the affine $Y$, hence determined by this module of global sections. The natural map ${\mathcal{O}}_Y\to j_{\ast }{\mathcal{O}}_X$ corresponds to the identity of $k[x,y]$, soa coherent sheaf. This is a concrete case of codimension-two extension of regular functions on a normal variety.

An $n$-dimensional smooth manifold is a Hausdorff space with a countable topological base, equipped with a smooth atlas of homeomorphisms from open subsets onto open subsets of ${\mathbb{R}}^n$, whose overlap maps are smooth diffeomorphisms. The smooth structure is the maximal smooth atlas compatible with these manifold charts. Here manifolds have no boundary unless specified otherwise.

For a space with all the requested topological properties but no such atlas, take the topological tripod: three closed intervals joined at one endpoint. It is a compact connected subspace of the plane with its induced metric. A countable base of planar rational balls restricts to a countable base, the metric makes it Hausdorff, and compactness gives a finite subcover of every open cover, hence a locally finite refinement and paracompactness.

At an interior point of any arm, arbitrarily small neighborhoods are intervals. A coordinate ball in dimension at least two would remain connected after deleting its center; an interval does not. Dimension zero would make the space discrete. Thus any possible connected manifold structure would have dimension one. But a sufficiently small neighborhood of the junction, minus the junction, has three components, whereas an interval chart has two. This contradiction rules out even a topological manifold structure, and hence any smooth one.

The product smooth structure uses manifold charts $(\varphi ,\psi ):U\times V\to \varphi (U)\times \psi (V)\subset {\mathbb{R}}^{m+n}$. Transition maps act separately in the two coordinate blocks and are smooth with smooth inverses. Products of countable bases give a countable base, and the product remains Hausdorff. Therefore $\boxed{\dim (M\times N)=\dim M+\dim N}$ with this natural smooth structure.

Define the tangent space by point derivations: a tangent vector at $x$ is an $\mathbb{R}$-linear map $D$ on germs of smooth functions at $x$, satisfying $D(ab)=a(x)D(b)+b(x)D(a)$. Addition and scalar multiplication preserve this rule, so these derivations form a vector space. In coordinates $u^1,\dots ,u^n$, the local identityfollows by integrating the derivative of $h$ along the coordinate line segment. Derivations annihilate constants, so it gives $Dh={\sum }_{i}D(u^i){\partial }_{i}h(u(x))$. The coordinate derivations ${\partial }_i{|}_x$ are independent since they evaluate the coordinate functions as ${\delta }_i^j$. ThusFor a smooth map, define the differential of a smooth map intrinsically by $(f_{\ast }D)(h)=D(h\circ f)$. It is again linear and satisfies the derivation rule at $f(x)$, so it maps $T_{x}M$ into $T_{f(x)}N$. Its coordinate matrix is the Jacobian of the coordinate expression of $f$, independently of the charts by the intrinsic definition and chain rule.

Finally, a zero differential forces each target coordinate function to have all derivatives zero on a small connected source coordinate ball mapping into one target chart. Integration on straight segments makes those functions constant there. Hence $f$ is locally constant. Each nonempty fiber is both open and closed, so connectedness gives the zero-differential constancy theorem, $\boxed{f\text{ is constant}}$.

Choose a positive system of a root system and the corresponding triangular decomposition of a Lie algebra $L={\mathfrak{n}}^{-}\oplus H\oplus {\mathfrak{n}}^{+}$. A primitive element of a Lie algebra representation of weight $\omega \in H^{\ast }$ is a nonzero vector $v$ with $hv=\omega (h)v$ for every $h\in H$ and ${\mathfrak{n}}^{+}v=0$. Thus it is a highest-weight vector; the choice of positive roots is part of this definition. A highest-weight representation is generated by such a vector.

Let $\mathfrak{b}=H\oplus {\mathfrak{n}}^{+}$ be the corresponding Borel subalgebra. Define its one-dimensional Lie algebra representation ${\mathbb{C}}_{\omega }$ by the scalar $\omega (h)$ on $H$ and zero on ${\mathfrak{n}}^{+}$. This respects the Lie bracket, since $[\mathfrak{b},\mathfrak{b}]\subseteq {\mathfrak{n}}^{+}$. The induced Verma moduleis nonzero: the Poincaré-Birkhoff-Witt theorem identifies its underlying vector space with $U({\mathfrak{n}}^{-})$, and $v=1\otimes 1$ is a primitive element of weight $\omega$. Its weight spaces are finite dimensional, its top weight space is the line $\mathbb{C}v$, and all other weights are $\omega -\sum n_i{\alpha }_i$ with $n_i\ge 0$ and at least one positive coefficient.

A proper submodule cannot contain $v$, because $v$ generates $M(\omega )$. More strongly it has no component of weight $\omega$: any finite sum of distinct weight vectors can be projected onto its individual components by a polynomial in a generic element of $H$. Therefore the sum $N$ of all proper submodules still misses the top weight line and is proper. It contains every proper submodule, so the irreducible quotient of a Verma moduleis irreducible and retains the nonzero primitive element $v+N$. This constructs the requested representation for every $\omega \in H^{\ast }$. It is not asserted to be finite dimensional for arbitrary $\omega$.

For the sl2 Lie algebra, use $h=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$, $e=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ and $f=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$, with $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. The classification of finite-dimensional sl2 representations gives one irreducible $V_n$ for every integer $n\ge 0$. In a basis $v_0,\dots ,v_n$ its action iswhere $v_{-1}=v_{n+1}=0$. These formulas satisfy the three Lie brackets. Any nonzero invariant subspace contains a weight vector by polynomial projection using $h$; repeated application of $e$ gives $v_0$, and applications of $f$ then give the entire basis. Thus $V_n$ is irreducible.

Conversely, in any finite-dimensional irreducible module, start with an eigenvector of $h$ and apply $e$ until reaching a nonzero vector $v$ with $ev=0$. This process terminates because $e$ increases the eigenvalue by two and only finitely many eigenvalues occur. Write $hv=av$. The sl2 highest-weight lowering formula givesLet $f^{n}v\ne 0$ and $f^{n+1}v=0$, which again follows from the finite set of weights. Applying $e$ to the latter identity yields $(n+1)(a-n)f^{n}v=0$, so $a=n$. The resulting $n+1$ distinct weight vectors span an invariant subspace, hence the entire irreducible module. Therefore $\dim V_n=n+1$ and its primitive weight is ${\omega }_n(h)=n$, with $n=0,1,2,\dots$; its weights are $n,n-2,\dots ,-n$.

Use $T=P_{V^{\perp }}{|}_U:U⟶V^{\perp }$. Positivity of the directed subspace angle cosine givesso $T$ is injective. The two vector spaces have the same finite dimension, $n$. By the rank-nullity theorem, $T$ is also surjective. There is no need to assume a second positive directed subspace angle cosine in this finite-dimensional case.

For any $f\in H$, find $u\in U$ with $Tu=P_{V^{\perp }}f$. Then $f-u\in V$. If $u\in U\cap V$, then $Tu=0$, and injectivity gives $u=0$. HenceThe direct sum is again bounded: $u=T^{-1}{P}_{V^{\perp }}f$. If $n=0$, then $U=\{0\}$ and $V^{\perp }=\{0\}$, so closedness of $V$ gives $V=H$ and the conclusion directly; no angle of an empty unit sphere is needed. Equal finite dimensions are essential to the surjectivity argument, whereas mere injectivity between infinite-dimensional Hilbert spaces is insufficient.

Assume first that $S\ne 0$, as required for the irreducible-module conclusion. If $B\to {\mathrm{End}}_{\mathbb{C}}(S)$ is surjective, any nonzero invariant subspace contains every image of one of its nonzero vectors under all endomorphisms, hence is all of $S$. Thus $S$ is irreducible.

Conversely an irreducible $B$-module is finite dimensional: for $s\ne 0$, $S=Bs$ is a quotient of the finite-dimensional vector space $B$. Let $D$ be the image of $B$ in ${\mathrm{End}}_{\mathbb{C}}(S)$. It acts faithfully and irreducibly. The subspace $J(D)S$ is a submodule, so is zero or all of $S$. The latter alternative would imply $J(D{)}^{k}S=S$ for every $k$, contradicting nilpotence of the Jacobson radical. Therefore $J(D)S=0$, and faithfulness gives $J(D)=0$.

The Artin–Wedderburn theorem makes $D$ a product of matrix algebras. A faithful simple module forces there to be just one factor, because all other factors would annihilate that module. Thus $D\cong M_d(\mathbb{C})$ and $S\cong {\mathbb{C}}^d$, so the action is the full endomorphism algebra. This is the Burnside matrix-algebra theorem.

For nonzero $S$, surjectivity is equivalent to irreducibility. The zero module is a literal exception if it is admitted: its endomorphism algebra is zero, so the action map is surjective, whereas the zero module is not irreducible.

A separated dual pair consists of real vector spaces $E,F$ and a bilinear form $⟨x,y⟩$ that separates both variables: each nonzero $x\in E$ pairs nontrivially with some $y\in F$, and conversely. Thus $F$ embeds in the algebraic dual of $E$, and $E$ embeds in the algebraic dual of $F$.

The weak topology $\sigma (E,F)$ is the coarsest topology making all maps $x\mapsto ⟨x,y⟩$, $y\in F$, continuous. It makes $E$ a Hausdorff locally convex space, with topology generated by the seminorms $p_y(x)=|⟨x,y⟩|$. A neighbourhood base at zero consists of finite intersectionsSeparation ensures that the common zero set of all these seminorms is just zero, giving Hausdorffness.

We determine the continuous dual of a weak topology. Suppose a linear functional $\ell$ is continuous. A basic neighbourhood as above is contained in $\{x:|\ell (x)|<1\}$. If $x$ is in the common kernel of the finitely many evaluations, every scalar multiple of $x$ lies in that neighbourhood, so $\ell (x)=0$. Hence $\ell$ factors through the image ofExtend the resulting linear functional on $T(E)$ to ${\mathbb{R}}^m$. It has the form $z\mapsto {\sum }_j{a}_j{z}_j$, giving $\ell (x)=⟨x,{\sum }_j{a}_j{y}_j⟩$. Conversely every such evaluation is continuous by the definition of the weak topology. Since the pairing separates $F$, this identification is injective, andFor the remaining parts every topological assertion about $E^{\prime }$ or its subsets uses the stipulated weak-star topology $\sigma (E^{\prime },E)$.

Write the finite abelian group additively. A character of a finite abelian group is a group homomorphism $\chi :G\to \{z:|z|=1\}$. We first prove that there are exactly $|G|$ such characters of a finite abelian group, without assuming a structure theorem.

Use extension of a character across a cyclic quotient. Given a subgroup $H$ and $a\notin H$, let $k$ be the least positive integer with $ka\in H$. The subgroup $K=H+⟨a⟩$ has $k$ cosets of $H$. If $\varphi$ is a character of a finite abelian group on $H$, choose any of the $k$ roots ${\lambda }^k=\varphi (ka)$ and defineThis is well-defined: two representations differ by an integer multiple of $ka$, and the root equation exactly cancels that difference. It is a character of a finite abelian group, and every extension arises from one of the $k$ choices of $\lambda$. Build a chain from the trivial subgroup to $G$ by adjoining elements. The character of a finite abelian group count multiplies by the same factor as the subgroup order at every step, so $|\hat{G}|=|G|$.

For a nontrivial character of a finite abelian group $\eta$, choose $a$ with $\eta (a)\ne 1$. Translating the group sum showsso the sum is zero. Applied to $\chi \overline{\psi }$, this provesThe $|G|$ orthogonal nonzero characters of a finite abelian group therefore form a basis of all complex functions on $G$, a vector space of dimension $|G|$.

With the unnormalized Fourier coefficients $\hat{f}(\chi )={\sum }_{x\in G}f(x)\overline{\chi (x)}$, expansion in that basis gives the Fourier inversion on a finite groupUnder a normalized forward-transform convention the prefactor would instead be one. Thus the unspecified constant is determined by the convention, and the inversion itself follows directly from character of a finite abelian group counting and orthogonality.

The Steinitz exchange lemma says that if $u_1,\dots ,u_r$ are linearly independent and $v_1,\dots ,v_s$ span the same vector space, then $r\le s$ and, after reordering the $v_j$, the list $u_1,\dots ,u_r,v_{r+1},\dots ,v_s$ still spans it.

To prove this, suppose $u_1,\dots ,u_k$ have already replaced $k$ of the spanning vectors. Express $u_{k+1}$ in that spanning list. At least one coefficient of an unreplaced $v_j$ is nonzero: otherwise $u_{k+1}$ lies in the span of $u_1,\dots ,u_k$, contradicting linear independence. Solve this expression for that $v_j$ and replace it by $u_{k+1}$. The new list still spans. If all $s$ vectors were replaced before the $u_i$ were exhausted, the next $u_i$ would be dependent on its predecessors. Hence $r\le s$, and the replacement can continue through $u_r$.

For two bases of sizes $r,s$, apply the Steinitz exchange lemma in both directions to get $r\le s\le r$. Thus every basis has the same size. Given a linearly independent set and any finite basis, the replacement procedure produces a finite spanning list containing that set. Successively delete any remaining vector dependent on the others; none of the prescribed independent vectors need be deleted, since one can instead delete a dependent supplementary vector. Equivalently, greedily retain only supplementary vectors outside the span already built. The resulting list is a basis containing the prescribed set.

For the three-dimensional example, the columns have matrixTherefore the three vectors are a basis. For the four-dimensional example,This is a nontrivial linear dependence, so the four vectors are not a basis. The distinction is an instance of the adjacent sums of a cyclic basis criterion: alternating coefficients close consistently around an even cycle, but not an odd one.

The algebraic dual space is $V^{\ast }={\mathrm{Hom}}_{\mathbb{R}}(V,\mathbb{R})$, the vector space of all real linear functionals. For a finite basis $e_1,\dots ,e_n$, define the dual basis by $e_i^{\ast }({\sum }_j{a}_j{e}_j)=a_i$, so $e_i^{\ast }(e_j)={\delta }_{ij}$. Every linear functional satisfiesThis equality follows by evaluating both sides on each vector of the basis. It proves spanning directly. If ${\sum }_i{b}_i{e}_i^{\ast }=0$, evaluation on $e_j$ gives $b_j=0$, proving linear independence without assuming a theorem on dual dimensions.

For a vector subspace $U$, its annihilator of a vector subspace isExtend a basis $e_1,\dots ,e_r$ of $U$ to a basis of $V$. Then $U^{\circ }$ has basis $e_{r+1}^{\ast },\dots ,e_n^{\ast }$, so $\boxed{\dim U^{\circ }=n-\dim U}$.

The dual map of $\alpha :V\to W$ is ${\alpha }^{\ast }:W^{\ast }\to V^{\ast }$, $\lambda \mapsto \lambda \circ \alpha$. In the corresponding dual bases its matrix is the transpose of the matrix for $\alpha$. A matrix and its transpose have the same rank, giving $\mathrm{rank}{\alpha }^{\ast }=\mathrm{rank}\alpha$. Directly,Thus $\boxed{\ker {\alpha }^{\ast }=(\mathrm{im}\alpha {)}^{\circ }}$. Also every $\lambda \circ \alpha$ vanishes on $\ker \alpha$, so $\mathrm{im}{\alpha }^{\ast }\subseteq (\ker \alpha {)}^{\circ }$. Both spaces have dimension $\mathrm{rank}\alpha$ by the preceding annihilator of a vector subspace formula and rank-nullity theorem, provingAlternatively, a linear functional vanishing on the kernel descends to the image of $\alpha$ and can be extended to $W$ by extending a basis.

For the infinite-dimensional polynomial ring, the coefficient functionals $L_i$ are linearly independent: evaluate a finite relation on each monomial. However, they do not span the algebraic dual space. The linear functional $p\mapsto p(1)$ takes value one on every monomial, while any finite combination of the $L_i$ vanishes on every sufficiently high-degree monomial. Thus the $L_i$ do not form a basis of $V^{\ast }$. An infinite formal sum of coefficient functionals is not a finite linear combination.

Linearity of the integral in each argument gives a bilinear form, and $fg=gf$ gives symmetry. Also $(f,f)={\int }_{-1}^{1}f(x{)}^{2}dx\ge 0$. If a polynomial $f$ is nonzero somewhere on this interval, continuity makes $f^2$ strictly positive on a smaller interval, so the integral is positive. A polynomial vanishing on the whole interval is the zero polynomial. Thus the form is positive-definite and is an inner product.

Apply the Gram-Schmidt process to $1,x,x^2$. Odd functions have zero integral, so $x$ is already orthogonal to $1$, and the third orthogonal polynomial is $x^2-1/3$. Their squared norms areTherefore an orthonormal basis isThese are the first three normalized Legendre polynomials. Their distinct degrees also prove that they span the three-dimensional vector space.

A ring is Artinian when its ideals satisfy the descending chain condition: every chain $I_1\supseteq I_2\supseteq \cdots$ eventually becomes constant. It is Noetherian when its ideals satisfy the ascending chain condition, equivalently when every ideal is finitely generated. We first prove finite length of a commutative Artinian ring; this gives the stronger structural reason for its Noetherian property.

If $\mathfrak{p}$ is a prime ideal of an Artinian ring, the quotient $R/\mathfrak{p}$ is an Artinian integral domain. For a nonzero element $x$ of that domain, the chain $(x)\supseteq (x^2)\supseteq \cdots$ stabilizes. Thus $x^n=b{x}^{n+1}$ for some $b$, and cancellation gives $bx=1$. The quotient is a field, so every prime ideal is a maximal ideal.

There are only finitely many maximal ideals. Otherwise, choose distinct ones ${\mathfrak{m}}_1,{\mathfrak{m}}_2,\dots$. The intersections ${\mathfrak{m}}_1\cap \cdots \cap {\mathfrak{m}}_n$ form a strictly descending chain. To see strictness, for each $i\le n$ choose $a_i\in {\mathfrak{m}}_i\setminus {\mathfrak{m}}_{n+1}$; their product belongs to the first $n$ ideals but not to the next one, since ${\mathfrak{m}}_{n+1}$ is prime. This contradicts the descending chain condition.

Put $J={\bigcap }_i{\mathfrak{m}}_i$. This Jacobson radical is also the nilradical, since all primes are maximal. We need the stronger conclusion that $J$ is nilpotent, without assuming Noetherianity. Its powers stabilize, say $K=J^n=J^{n+1}=JK$. Suppose $K\ne 0$. By the descending chain condition, choose an ideal $L$ minimal subject to $KL\ne 0$. Some $x\in L$ has $Kx\ne 0$, so minimality gives $L=Rx$. Moreover $K(JL)=KL\ne 0$, and minimality gives $JL=L$. Hence $x=jx$ for some $j\in J$. But $1-j$ is a unit: it cannot lie in any maximal ideal, because $j$ lies in all of them. Thus $x=0$, a contradiction. Therefore $J^n=0$.

The Chinese remainder theorem gives $R/J\cong {\prod }_{i}R/{\mathfrak{m}}_i$, a finite product of fields. Each quotient $J^i/J^{i+1}$ is an Artinian module over this product, and each field component must be a finite-dimensional vector space; an infinite-dimensional vector space admits a strictly descending chain of subspaces. Consequently every layer has finite composition length. The finite filtrationshows that $R$ itself has finite composition length. A strict inclusion of submodules strictly increases length, so an ascending chain cannot continue indefinitely. Every commutative Artinian ring is therefore Noetherian. This is the Artinian rings are Noetherian result.

For the formal power series ring, let $B=R[[X]]$ with $R$ Noetherian, and let $H$ be any ideal of $B$. For $n\ge 0$ define a coefficient idealThese coefficient ideals of a formal power series ideal satisfy $A_n\subseteq A_{n+1}$, by multiplication by $X$. The ascending chain condition gives $A_n=A_N$ for all $n\ge N$. For each $0\le n\le N$, choose finitely many series $f_{n,j}\in H\cap X^{n}B$ whose coefficients at $X^n$ generate $A_n$.

We claim that these finitely many series generate $H$ as an ordinary ideal. Given $f\in H$, cancel its coefficient at $X^k$ successively. After coefficients below $k$ have vanished, its coefficient at $X^k$ lies in $A_k$. If $k\le N$, use an $R$-linear combination of the $f_{k,j}$. If $k>N$, use a combination of $X^{k-N}{f}_{N,j}$, since $A_k=A_N$. The remainder then belongs to $H\cap X^{k+1}B$.

Collect all the cancellations against each fixed generator. For $n<N$ its multiplier is a polynomial, while the multipliers of the $f_{N,j}$ are well-defined formal power series: at any fixed degree, only finitely many cancellation steps contribute. The remainder has every coefficient zero. ThusThis is a finite sum of ideal generators, rather than merely a topological closure assertion. Since $H$ was arbitrary,This coefficient-cancellation argument proves Noetherianity of a formal power series ring.

The corresponding Artinian assertion is false. For any nonzero ring $R$, the idealsin $R[[X]]$ are strictly decreasing, since $X^n$ has a nonzero coefficient in degree $n$ and no multiple of $X^{n+1}$ does. In particular, a field $k$ is Artinian, but $k[[X]]$ is not. The zero ring is the harmless exception.