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Positivity of the linear functional gives monotonicity. For any $g\in C(K)$, the pointwise bounds $-\Vert g{\Vert }_{\infty }1\le g\le \Vert g{\Vert }_{\infty }1$ therefore implyThus the positive linear functional is bounded, hence continuous, and its operator norm is at most $\varphi (1)$. Evaluating on the constant function $1$, which has supremum norm one when $K$ is nonempty, gives the reverse inequality. HenceFor empty $K$ the function space and functional are zero, so the same conclusion holds. This is the norm of a positive functional on C(K) identity.

Put $c=\Vert \varphi \Vert =\varphi (1)\ge 0$. If $c=0$, the linear functional is zero and is positive. Otherwise, for a nonzero $g\ge 0$ let $h=g/\Vert g{\Vert }_{\infty }$. Then $0\le h\le 1$ and $\Vert 1-h{\Vert }_{\infty }\le 1$, so boundedness givesPositive rescaling gives $\varphi (g)\ge 0$. Thus $\varphi$ is a positive linear functional. This is the scalar form of the unital contraction positivity criterion.

We construct the Jordan decomposition of a bounded functional on C(K) directly, without assuming a measure representation. For $f\ge 0$, defineThis is finite, nonnegative, and positively homogeneous, since $0$ is allowed and $|\varphi (g)|\le \Vert \varphi \Vert \Vert f{\Vert }_{\infty }$. If $f,h\ge 0$, separately chosen approximate maximizers add to an admissible function for $P(f+h)$, giving $P(f+h)\ge P(f)+P(h)$. For the converse, any $0\le g\le f+h$ splits into the continuous functionswith $0\le g_1\le f$ and $0\le g_2\le h$. Hence $\varphi (g)\le P(f)+P(h)$ and $P$ is additive on the positive cone.

Extend it to a linear functional by ${\varphi }^{+}(u-v)=P(u)-P(v)$ for $u,v\ge 0$. This is well defined: two decompositions satisfy $u+v^{\prime }=u^{\prime }+v$, and additivity gives the same difference. Every continuous real function has such a decomposition, for example into its pointwise positive and negative parts. The extension is positive. Define ${\varphi }^{-}={\varphi }^{+}-\varphi$; for $f\ge 0$, the choice $g=f$ shows $P(f)\ge \varphi (f)$, so ${\varphi }^{-}$ is also positive. Part (a) makes both positive linear functionals continuous.

Their operator norms satisfyThe last equality uses symmetry of the real unit ball. Therefore

A character of an algebra is a nonzero multiplicative complex linear functional $\varphi :A\to \mathbb{C}$. Nonzeroness and multiplicativity force $\varphi (1)=1$, so it is onto $\mathbb{C}$. Its kernel is an ideal, and the quotient is the field $\mathbb{C}$, making the kernel maximal.

Continuity is automatic. The element $a-\varphi (a)1$ cannot be invertible, since applying $\varphi$ to an inverse equation would give $0=1$. On the other hand $a-\lambda 1$ is invertible whenever $|\lambda |>\Vert a\Vert$, by the Neumann series. Hence $|\varphi (a)|\le \Vert a\Vert$, the automatic continuity of characters bound.

Conversely let $M$ be a maximal ideal. Its closure is an ideal, because multiplication is continuous. That closure is proper: otherwise there would be $m\in M$ with $\Vert 1-m\Vert <1$, making $m$ invertible by the Neumann series and forcing $M=A$. Maximality gives $\overline{M}=M$. The quotient $A/M$ is consequently a Banach algebra. It is a division algebra: for any $a\notin M$, the larger ideal $M+Aa$ equals $A$, so the class of $a$ has an inverse. By the Gelfand-Mazur theorem, $A/M\cong \mathbb{C}$, and its quotient map gives a continuous character of an algebra with kernel $M$.

If two algebra characters have the same kernel, then $a-\varphi (a)1$ lies in the second kernel, giving $\psi (a)=\varphi (a)$ for every $a$. Thusis a bijection. This proves that maximal ideals of a commutative complex unital Banach algebra are character kernels.

There is a qualification in the printed hypotheses: quadratic ellipticity does not bound a nonsymmetric coefficient matrix. The intended standard existence proof needs $a_{ij}\in L^{\infty }(\Omega )$, or more generally a bounded principal bilinear form. To see why this does not follow from the displayed inequalities, take $\Omega =(-1,1{)}^2$ andThen ${\zeta }^{T}A(x)\zeta =|\zeta {|}^2$ for every $x,\zeta$. However, choose $u_0,{\phi }_0\in C_c^{\infty }(\Omega )$ equal to $x_2,x_1$, respectively, on a small rectangle about the origin. There $(AD{u}_0)\cdot D{\phi }_0=k(x_1)$, whose positive part has infinite integral. Thus the standard principal pairing is not a finite Lebesgue integral even for these smooth test functions. This demonstrates the missing form hypothesis, rather than assuming that all unbounded skew coefficients prevent solvability.

Under the intended bounded-coefficient hypothesis the proof is as follows. On $H=H_0^1(\Omega )$ use the Hilbert space norm $\Vert Dw{\Vert }_2$, equivalent to the $H^1$ norm by the Poincaré inequality on a bounded domain. The form from part (i) is bounded, sinceIt is a coercive bilinear form even without symmetry:Here the nonpositive sign of $q$ is essential. With $u=\psi +w$, the required equation becomesThe Cauchy-Schwarz inequality, bounded coefficients and the Poincaré inequality show that $\ell$ is a bounded linear functional on $H$. The Lax-Milgram theorem states that a bounded coercive bilinear form on a real Hilbert space represents each bounded linear functional in this way with a unique $w\in H$. It therefore yields $u=\psi +w$ with the required boundary condition and weak equation. For completeness, if $u_1,u_2$ are solutions, $z=u_1-u_2\in H_0^1$ satisfies $B(z,z)=0$; coercivity and the Poincaré inequality imply $z=0$.

With the missing bounded-form hypothesis, the weak solution exists and is unique. For nonsymmetric matrices the printed quadratic inequalities omit the hypothesis needed for this standard formulation and proof. No maximum principle is used. In part (iii), symmetry makes the quadratic upper bound a bound on the whole matrix, so no such qualification is needed there.

The Frankl-Wilson theorem in its uniform form says: let $p$ be a prime number, let $L\subseteq {\mathbb{F}}_p$ have size $s\le \min (k,n-k)$, and let $\mathcal{F}\subseteq \left(\genfrac{}{}{0ex}{}{[n]}{k}\right)$ have $k\mathrm{mod}p\notin L$, while $|A\cap B|\mathrm{mod}p\in L$ for every distinct pair of members. Then

The modular-intersection, nonuniform form of the Frankl-Wilson theorem is as follows. Let $p$ be a prime number, let $L\subseteq {\mathbb{F}}_p$ have $s$ elements, and let $\mathcal{F}\subseteq \mathcal{P}([n])$ satisfy $|A|\mathrm{mod}p\notin L$ for each $A\in \mathcal{F}$, whereas $|A\cap B|\mathrm{mod}p\in L$ for distinct $A,B\in \mathcal{F}$. ThenThis nonuniform Frankl-Wilson theorem allows different sizes and different diagonal residues; in particular it applies when all sizes have one common residue outside $L$.

To prove it by the polynomial method in combinatorics, work over the finite field ${\mathbb{F}}_p$ and associate to each member the intersection polynomialReplace each positive power of $x_i$ in its expansion by $x_i$ to obtain its multilinear reduction on the Boolean cube ${\tilde{P}}_A$. This preserves evaluation on every characteristic vector, whose coordinates are zero or one, and leaves degree at most $s$. At the characteristic vector $1_B$,It is zero for distinct $A,B$, and nonzero for $B=A$ by the diagonal hypothesis. If ${\sum }_{A\in \mathcal{F}}{c}_A{\tilde{P}}_A=0$, evaluating at each $1_B$ forces $c_B=0$. Thus the multilinear polynomials are linearly independent. They belong to the vector space with basis the square-free monomials ${\prod }_{i\in S}{x}_i$ for $|S|\le s$, whose dimension is the displayed sum. This proves the nonuniform form, including $s=0$, where distinct members are impossible and the empty product is one.

For the sharp uniform form, we need the low-degree evaluation rank on a uniform layer. Form the integer matrix $M$ whose rows are indexed by $S\subseteq [n]$ with $|S|\le s$, columns by $k$-subsets $T$, and entries $M_{S,T}=1_{S\subseteq T}$. These are evaluations of the square-free monomials. Over the rational numbers, the row of a $j$-subset $S$ satisfiesIndeed, when $S\subseteq T$ the left side counts the $s$-subsets between $S$ and $T$, and otherwise both sides vanish. Since $s\le k$, the coefficient is a positive integer. Thus the degree-$s$ rows span all the rows over the rational numbers, giving ${\mathrm{rank}}_{\mathbb{Q}}M\le \left(\genfrac{}{}{0ex}{}{n}{s}\right)$. Every square submatrix of larger size has integer determinant zero, so its determinant remains zero modulo $p$. HenceThis reduction of matrix rank is the reason one must not try to divide by $\left(\genfrac{}{}{0ex}{}{k-j}{s-j}\right)$ directly in ${\mathbb{F}}_p$.

The evaluations of the ${\tilde{P}}_A$ on the $k$th uniform layer of the Boolean cube lie in the row linear span of $M$ over ${\mathbb{F}}_p$. They are linearly independent as functions there, because restriction to the columns corresponding to $\mathcal{F}$ still gives the nonzero diagonal evaluation matrix. Their number is consequently at most $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$. This completes the proof of the uniform Frankl-Wilson theorem as well.

For the odd-cardinality problem the sharper Oddtown theorem follows directly from linear algebra. Let $v_A=1_A\in {\mathbb{F}}_2^n$ be the characteristic vector. The parity conditions sayFor any linear combination ${\sum }_A{c}_A{v}_A=0$, take its dot product with $v_B$. All off-diagonal terms vanish, leaving $c_B=0$. Hence these vectors are linearly independent in an $n$-dimensional vector space, givingThe singleton sets show that the bound can be attained. Notice why the constant term in the general nonuniform Frankl-Wilson theorem is unnecessary here: these dot products already give a diagonal evaluation matrix using only the $n$ coordinate functions.

For even $n$, partition $[n]$ into the pairs $\{1,2\},\{3,4\},\dots ,\{n-1,n\}$ and take every union of a selection of these pairs. This gives exactly $2^{n/2}$ distinct sets, including the empty set. Each has even size; every pairwise intersection is again a union of entire pairs and therefore has even size. This attains the proposed bound.

For any set family satisfying the even-cardinality conditions, let $W$ be the linear span of its characteristic vectors over ${\mathbb{F}}_2$. Now the self-dot products vanish as well as the off-diagonal ones. By bilinearity, $u\cdot v=0$ for all $u,v\in W$, so $W\subseteq W^{\perp }$: it is a self-orthogonal binary subspace.

For completeness the standard bilinear form on ${\mathbb{F}}_2^n$ is nondegenerate, because a vector orthogonal to every coordinate vector has every coordinate zero. Thus the map $v\mapsto (w\mapsto v\cdot w)$ from ${\mathbb{F}}_2^n$ onto the dual space $W^{\ast }$ is surjective: every linear functional on $W$ extends to the whole vector space by extending a basis, and every such functional on ${\mathbb{F}}_2^n$ is a dot product with a vector. Its kernel is $W^{\perp }$. The rank-nullity theorem consequently gives $\dim W^{\perp }=n-\dim W$. The inclusion $W\subseteq W^{\perp }$ implies $\dim W\le n/2$. Finally, distinct members have distinct characteristic vectors, all in $W$, and a $k$-dimensional vector space over ${\mathbb{F}}_2$ has exactly $2^k$ elements. ThereforeThis proves the Eventown theorem and also explains the suggested dimension contradiction if the set family had more than $2^{n/2}$ members.

The special linear group over a finite field $S{L}_n(q)$ consists of the determinant-one invertible matrices over ${\mathbb{F}}_q$. Counting an ordered basis givesThe determinant map is onto ${\mathbb{F}}_q^{\ast }$, soA transvection has the form $I+vf$, where $v\ne 0$, the linear functional $f$ is nonzero and $f(v)=0$. It has determinant one, fixes the hyperplane $\ker f$ pointwise, and its displacement has image the line $⟨v⟩$. In particular $E_{ij}(t)=I+t{e}_{ij}$ for $i\ne j$ is an elementary transvection when $t\ne 0$.

Row addition is multiplication by such a matrix. Gaussian elimination reduces a determinant-one matrix to diagonal form using row additions and determinant-one pivot exchanges. The latter, and all remaining diagonal factors, are themselves products of transvections: within two coordinates,A diagonal matrix with product of entries one is a product of these two-coordinate diagonal matrices. Thus transvections generate $S{L}_n(q)$. This generation statement actually holds for every $n\ge 2$ and every finite field; the exclusions in the question are needed for the subsequent simplicity assertion.

Commuting with all elementary matrices forces the group center to consist of scalar matrices $\lambda I$ with ${\lambda }^n=1$, so its order is $d=\gcd (n,q-1)$. The projective quotient acts faithfully on the one-dimensional subspaces. Its action is two-transitive: send an ordered pair of independent representative vectors to any other pair and adjust the determinant by rescaling a representative without changing its line. Therefore it is primitive.

The Iwasawa simplicity lemma says that in a faithful primitive action, if a point stabilizer has an abelian normal subgroup whose conjugates generate the group, every nontrivial normal subgroup contains the derived subgroup. In particular a nontrivial perfect group satisfying these conditions is simple. No proof of the lemma is required here.

For the point $\ell =⟨v⟩$, the subgroupis abelian, is normal in the line stabilizer, and its conjugates contain all transvections. Its image in the projective quotient has the same properties. Perfectness follows explicitly from elementary commutators. If $n\ge 3$, choose a third index $r$ to obtainIf $n=2,q>3$, choose $t$ with $t^2\ne 1$ and set $h(t)=\mathrm{diag}(t,t^{-1})$. ThenEvery upper transvection is a commutator, and conjugation gives the lower ones. The group and its projective quotient are perfect. Iwasawa's simplicity lemma now proves

The order of $PS{L}_2(4)$ is $4(4^2-1)=60$. Its faithful projective-line action has degree five, so its image is an index-two subgroup of $S_5$, necessarily $A_5$. Thus $PS{L}_2(4)\cong A_5$.

For $PS{L}_2(5)$ the order is $5(5^2-1)/2=60$. We construct a faithful degree-five action. In $S{L}_2(5)$ takeThey satisfy $i^2=j^2=-I$ and $ij=-ji$. Their quaternion subgroup projects to a Sylow 2-subgroup $V\cong C_2^2$. Its projective centralizer is $V$: a lift of a centralizing element conjugates each of $i,j$ to itself or its negative. In the matrix basis $I,i,j,ij$, those four sign choices each leave a one-dimensional space, and determinant one leaves just its two quaternion representatives. The matrixhas order three and cyclically permutes the three nonidentity elements of $V$. The normalizer quotient embeds in $S_3$ and has order dividing both $6$ and $60/4=15$, so it has order exactly three. Hence $|N(V)|=12$ and there are five Sylow 2-subgroups. Their conjugation action is nontrivial and transitive, and simplicity makes it faithful. Its image again has order $60$ in $S_5$, proving

Finally,In characteristic two an involutory matrix has form $I+N$ with $N^2=0$. Its Jordan blocks have size at most two. A projective involution in $PS{L}_3(4)$ has a unique involutory lift to $S{L}_3(4)$: its square is central, and squaring is an automorphism of the scalar center of order three, allowing a unique scalar adjustment. The nonidentity Jordan form in dimension three is only $(2,1)$. This GL-class remains a single SL-class, since the centralizer contains $\mathrm{diag}(I_2,c)$ of every determinant $c\in {\mathbb{F}}_4^{\ast }$. Thus $PS{L}_3(4)$ has one involution class.

In $PS{L}_4(2)=S{L}_4(2)=G{L}_4(2)$ there are two forms, $(2,1,1)$ and $(2,2)$, so there are two involution classes. More concretely, the first group's single class has $21\cdot 5\cdot 3=315$ elements, counting image line, kernel plane through it and nonzero map between the quotient and image. The rank-one class in dimension four over ${\mathbb{F}}_2$ has $15\cdot 7=105$ elements; the rank-two class has $35\cdot |G{L}_2(2)|=210$. Isomorphisms preserve conjugacy classes, so

First consider finite-dimensional Lie algebra representations. The operators representing an abelian Lie algebra commute. They have a common eigenvector: take an eigenspace for the first operator of a basis of the algebra, observe that every remaining operator preserves it, and repeat inside this nonzero space. The resulting line is an invariant subspace, so an irreducible representation must be one-dimensional. Its action has the form $xv=\lambda (x)v$ for a linear functional $\lambda$. Conversely, every such functional defines a Lie algebra representation, since both the Lie brackets in the algebra and the commutators of its scalar operators vanish. Distinct functionals give nonisomorphic representations. Thus the irreducibles are precisely the characters $\lambda \in {\mathfrak{g}}^{\ast }$.

Even without a finite-dimensional assumption on the module, this conclusion for a finite-dimensional abelian Lie algebra follows from the Weak Hilbert Nullstellensatz: its universal enveloping algebra is the polynomial algebra $\mathrm{Sym}(\mathfrak{g})$. A simple module over a commutative algebra is its quotient by a maximal ideal, and every maximal ideal of this polynomial algebra over $\mathbb{C}$ has residue field $\mathbb{C}$.

Indecomposable does not imply irreducible. For the one-dimensional abelian Lie algebra $\mathbb{C}x$, let $x$ act on ${\mathbb{C}}^2$ by the Nilpotent Jordan blockThe line spanned by the first coordinate vector is an invariant subspace, so the module is reducible. If it were a direct sum of two nonzero submodules, both would have dimension one; nilpotence would force $N$ to act as zero on both, contrary to $N\ne 0$. Thus it is an indecomposable representation.

For a finite-dimensional module $V$, put $d=\dim V$ and define its generalized weight spaces intrinsically byTo obtain the decomposition, choose a basis $x_1,\dots ,x_r$ of $\mathfrak{g}$, split into generalized eigenspaces of $\rho (x_1)$, and successively split each piece by $\rho (x_2),\dots ,\rho (x_r)$. All the pieces are submodules because the acting operators commute. On one resulting piece, $\rho (x_j)$ has a unique eigenvalue ${\lambda }_j$. The commuting operators can be simultaneously upper triangularized by repeating the common-eigenvector argument on quotients. Hence on this piece every $\rho (x_j)-{\lambda }_{j}I$ is strictly upper triangular. Any linear combination is also strictly upper triangular, and its $d$th power vanishes. Extending ${\lambda }_j$ linearly therefore identifies the piece with $V^{\lambda }$. Different tuples give disjoint pieces; if a component has a different tuple from $\lambda$, at least one $\rho (x_j)-\lambda (x_j)I$ is invertible there. Consequentlywith only finitely many nonzero summands. Since the definition uses every $x$ rather than a chosen basis, this generalized-weight decomposition for a nilpotent Lie algebra is canonical. Its summands need not be irreducible, as the Jordan block example already shows.

Use the printed bilinear pairing ${\varphi }_u(v)=\int uv$, which is complex linear in both arguments. For the conjugate exponents $p,q$, Holder inequality gives $|{\varphi }_u(v)|\le \Vert u{\Vert }_q\Vert v{\Vert }_p$, proving well-definedness, continuity, and $\Vert {\varphi }_u\Vert \le \Vert u{\Vert }_q$. Linearity of $u\mapsto {\varphi }_u$ follows from linearity of integration.

For $u\ne 0$, takewith the numerator defined as zero where $u=0$. Since $(q-1)p=q$, this has $\Vert v{\Vert }_p=1$ and ${\varphi }_u(v)=\Vert u{\Vert }_q$. Hence $\boxed{\Vert {\varphi }_u\Vert =\Vert u{\Vert }_q}$; the same identity holds for $u=0$. This also proves injectivity.

For surjectivity, let $\Lambda$ be a nonzero bounded linear functional on $L^p$. Choose $f$ with $\Lambda (f)=1$ and let $K=\ker \Lambda$. The closest point theorem for a closed convex subset of Lp supplies the nearest point $h\in K$; write $w=f-h\ne 0$. For every $z\in K$, the whole line $h+tz$ lies in $K$, so the norm derivative at the minimum vanishes for real $t$. Applying this also to $iz$ shows that the complex linear functionalvanishes on $K$. Since $\Lambda (w)=1$, each $v-\Lambda (v)w$ is in $K$, giving $\Psi (v)=\Lambda (v)\Psi (w)$ and $\Psi (w)=\Vert w{\Vert }_p^p$. ThereforeMembership follows from $(p-1)q=p$. The zero functional is represented by $u=0$. This proves the asserted Lp duality isometric isomorphism, including surjectivity, without a separate representation theorem.

An idempotent $e$ gives a projective module $Ae$, since $A=Ae\oplus A(1-e)$. It is primitive when it cannot be written as a sum of two nonzero orthogonal idempotents. This is equivalent to indecomposability of $Ae$: a projection onto a direct summand is an idempotent of ${\mathrm{End}}_A(Ae)\cong (eAe{)}^{\mathrm{op}}$, and conversely such an idempotent splits $Ae$. Starting with $1$, split any nonprimitive summand repeatedly. The process terminates because each splitting increases the number of nonzero vector-space summands in a finite-dimensional algebra. Thuswith orthogonal primitive idempotents and indecomposable projective modules. Each summand is generated by its idempotent.

Let $D={\mathrm{Hom}}_k(-,k)$. A Frobenius algebra is a finite-dimensional algebra with $A\cong D(A)$ as left modules over $A$, where $(a\varphi )(x)=\varphi (xa)$. Equivalently there is a linear functional $\lambda :A\to k$ for which $(x,y)\mapsto \lambda (xy)$ is nondegenerate. A symmetric algebra satisfies the stronger bimodule isomorphism $A\cong D(A)$, with $(a\varphi b)(x)=\varphi (bxa)$; equivalently the functional can be chosen with $\lambda (xy)=\lambda (yx)$. This is the associative-algebra meaning of symmetric, rather than the symmetric algebra of a vector space.

For a finite-dimensional module,The socle is its largest semisimple submodule, and the radical of a module gives its largest semisimple quotient. With $J=J(A)$ one has $\mathrm{Rad}M=JM$ and $\mathrm{Soc}M=\{m:Jm=0\}$: $A/J$ is semisimple, and $J$ annihilates every simple module.

Now let $P=Ae$ for a primitive idempotent, allowing the field $k$ to be nonsplit. Symmetry identifies $Ae$ with $D(eA)$ through the pairingIt is nondegenerate on these corners. If $v=ve\ne 0$, choose $a$ with $\lambda (av)\ne 0$; then $\lambda (eav)=\lambda (ave)=\lambda (av)\ne 0$. The other variable is treated similarly. The pairing respects the left $A$-action, soThe right-module quotient $eA/eAJ$ is simple. A primitive idempotent remains primitive modulo the nilpotent radical: idempotents lift through that radical, so a decomposition in the quotient would lift to one in $eAe$. In the semisimple algebra $A/J$, the left simple $Ae/JAe$ and the dual of the right simple $eA/eAJ$ belong to the same matrix block. There is exactly one left simple isomorphism class in that block, even over a nonalgebraically closed field. Hence the head-socle identity for a symmetric algebra isThis is an isomorphism of left modules, and need not be canonical.

For the modular examples put $r=(123)$, $s=(12)$, so $r^3=s^2=1$ and $srs=r^{-1}$. The coefficient-of-identity functional on the group algebra is symmetric and nondegenerate, so the identity just proved applies.

In characteristic two, the central idempotents are $b_0=1+r+r^2$ and $b_1=1-b_0$. The first block has basis $b_0,b_{0}s$ and is $k{C}_2\cong k[u]/(u^2)$. For the second, choose a primitive cube root $\omega$ and useThe two $r$-eigenlines are interchanged by $s$, proving irreducibility. Polynomial interpolation in $r$ gives diagonal matrix units, and multiplying by $s$ gives the off-diagonal units. Thus the four-dimensional block maps onto $M_2(k)$ and is isomorphic to it:There are two simple modules, the trivial one-dimensional module $T$ and the two-dimensional module $V$; the sign module equals $T$. Their projective covers are $P_T=k[u]/(u^2)$, with layers $T,T$, and $P_V=V$. The regular module is $P_T\oplus V^{\oplus 2}$, and the Cartan matrix of a group algebra is $\mathrm{diag}(2,1)$. Every module is a sum of $T,P_T,V$: the first block is classified by Jordan chains for a nilpotent linear map of lengths one and two, while the matrix block is semisimple. Thus there are exactly three indecomposable isomorphism classes. This describes the 2-modular blocks of S3.

In characteristic three put $u=r-1$. The ideal $I=uA$ satisfies $I^3=0$, and $A/I\cong k{C}_2\cong k\oplus k$, so $J(A)=I$. The only simples are the trivial and sign modules $S_{+},S_{-}$. Put $e_{\pm }=(1\pm s)/2$. The projectives $P_{\pm }=A{e}_{\pm }$ have dimension three. A convenient radical generator is $v=r-r^{-1}$:The basis $e_{\pm },v{e}_{\pm },v^2{e}_{\pm }$ gives the radical layersThus $A=P_{+}\oplus P_{-}$ and its Cartan matrix of a group algebra isThe elements $e_{-}v{e}_{+}$ and $e_{+}v{e}_{-}$ give one arrow in each direction between the two simples. All paths of length three vanish, and the six paths of lengths zero, one and two are linearly independent. Hence $A$ is the path algebra of that two-vertex oriented cycle modulo the ideal of paths of length three. Its connected quiver also proves there is a single block of S3 in characteristic three.

Any module is a $\mathbb{Z}/2$-graded vector space given by the $s$-eigenspaces, with $v$ interchanging the grades and $v^3=0$. Constructing Jordan chains with homogeneous bases in the kernel and image filtrations and homogeneous lifts decomposes it into graded chains. Each chain has length $1,2$ or $3$ and is determined by its top sign. It is a uniserial module and hence indecomposable. Consequently there are exactly six indecomposables, the length-$\ell$ quotients $P_{\pm }/J^{\ell }{P}_{\pm }$, $\ell =1,2,3$.

For a finite group, define the Frattini subgroup bythe intersection is $G$ when there are no maximal subgroups, which for a finite group happens only for $G=1$. A second definition is the set of non-generators: an element $x$ such that $⟨S,x⟩=G$ always implies $⟨S⟩=G$.

If $x\in \Phi (G)$ and $H=⟨S⟩<G$, put $H$ in a maximal subgroup $M$. Then $x\in M$, so $⟨S,x⟩\le M<G$. Conversely, if $x\notin \Phi (G)$, some maximal subgroup $M$ omits it; maximality gives $⟨M,x⟩=G$, although $M\ne G$. This proves equivalence.

Every group automorphism permutes maximal subgroups, so $\Phi (G)$ is a characteristic subgroup, in particular a normal subgroup. Alsoa proper $H$ would lie in a maximal subgroup containing both factors.

Set $F=\Phi (G)$ and take a Sylow subgroup $P$ of $F$. For $g\in G$, normality of $F$ makes $P^g$ a Sylow subgroup of $F$. Conjugating inside $F$ therefore gives the Frattini argument factorization $G=F{N}_G(P)$. The preceding generating property forces $N_G(P)=G$. Thus every Sylow subgroup of $F$ is normal, and part (i) shows that $F$ is a nilpotent group.

Now suppose $G$ is a finite p-group. The normalizer condition makes every maximal subgroup $M$ normal: $M<N_G(M)$ and maximality force $N_G(M)=G$. The quotient $G/M$ has no proper nontrivial subgroup, hence is cyclic of prime order; its p-power order makes that prime $p$. Consequently every such $M$ containswhere the two factors are the commutator subgroup and the subgroup generated by pth powers. Thus $D\le \Phi (G)$.

The quotient $G/D$ is an elementary abelian group, hence a vector space over ${\mathbb{F}}_p$. If $x\notin D$, extend the nonzero vector $xD$ to a basis and choose a linear functional nonzero on it. Its kernel pulls back to a subgroup of index $p$, a maximal subgroup omitting $x$. Therefore $x\notin \Phi (G)$, provingThe two factors in the last formula are multiplied; the superscripted expression in the TeX aid is not the formula printed in the PDF.

We first prove the algebraic extension step needed for Hahn-Banach separation theorem. Suppose $p$ is a finite sublinear functional on a real vector space and $g$ is a linear functional on a vector subspace $W$, with $g\le p$ there. For $v\notin W$, choose a real number $c$ betweenThis interval is nonempty: for any $w,z\in W$,Taking $w=0$ or $z=0$ also bounds the endpoints by finite numbers. Define $\tilde{g}(w+tv)=g(w)+tc$. For $t>0$, the upper bound for $c$, applied to $w/t$, proves $\tilde{g}(w+tv)\le p(w+tv)$. For $t<0$, writing $t=-s$ and using the lower bound at $w/s$ proves the same inequality. The case $t=0$ is the original domination. Thus a dominated linear functional extends over one further direction. Order all dominated extensions by extension of their domains. A chain has its union as an upper bound, so Zorn's lemma supplies a maximal extension. The one-direction argument shows its domain is the whole vector space. This proves the required algebraic Hahn-Banach theorem, rather than assuming the separation result.

Choose $a\in A$, and put $C=A-a$. This is a radially open convex set containing zero. It is absorbing: along each line through zero, sufficiently small multiples of any vector lie in $C$. Its Minkowski functionalis finite and nonnegative. Positive homogeneity follows by rescaling $t$. If $x\in sC$ and $y\in tC$, convexity gives $x+y\in (s+t)C$, proving subadditivity by taking infima. Moreover,Indeed, $p(x)<1$ gives $x\in sC$ for some $s<1$, hence $x\in C$ by convexity and $0\in C$. Conversely, radial openness at $x\in C$ permits $(1+ϵ)x\in C$ for some $ϵ>0$, giving $p(x)<1$.

Since $A\cap U=\varnothing$, $a\notin U$ and $u-a\notin C$ for every $u\in U$. Define a linear functional on $U+\mathbb{R}a$ byIt is well defined because $a\notin U$. For $t<0$, write $s=-t$; then $p(u+ta)=sp(u/s-a)\ge s=g(u+ta)$. For $t\ge 0$, domination follows from $g(u+ta)\le 0\le p(u+ta)$. The extension step therefore gives $L:V\to \mathbb{R}$ with $L\le p$, $L{|}_U=0$ and $L(a)=-1$. If $x\in A$, thenso $L(x)<0$. ConsequentlyBecause $L(a)\ne 0$, this kernel of a linear map has codimension one and is a hyperplane. This establishes the geometric Hahn-Banach theorem for radially open sets without assuming a norm or a topology on $V$.

For the convex function, use its strict epigraph in $V\times \mathbb{R}$. It is nonempty and convex, and does not contain $(0,0)$. It is also radially open: on every affine line, a finite convex function is continuous. To justify this last fact, convexity orders secant slopes; on a smaller interval the slopes are bounded above and below by slopes to two fixed exterior endpoints, giving a local Lipschitz bound. Thus $t\mapsto f(x+ty)-\beta -ts$ is continuous, and a strict negative value persists near $t=0$.

Apply the separation theorem with the zero subspace. A nonzero separating linear functional has form $L(x,\beta )=g(x)+c\beta$. Its values on the convex epigraph cannot have both signs, since a line segment would then meet its kernel. Choose the sign so that $L>0$ on the epigraph. At $x=0$ and $\beta >0$ this gives $c>0$. Letting $\beta$ decrease to $f(x)$ yields $g(x)+cf(x)\ge 0$. ThereforeThe resulting linear functional need not be nonzero: for example, the zero functional is the only linear minorant of $f(x)=x^2$ on $\mathbb{R}$.

Now suppose the norm and continuity hypotheses hold. Domination at $x$ and $-x$ givesIf $x_n\in \ker l$ and $x_n\to x$ in norm, apply this inequality to $x-x_n$. Since $l(x-x_n)=l(x)$ and both bounding values tend to $f(0)=0$, we obtain $l(x)=0$. The kernel of a linear map is therefore closed.

Finally, a linear functional with closed kernel on a normed vector space is continuous. If $l=0$, this is immediate. Otherwise choose $v$ with $l(v)=1$ and put $M=\ker l$. Closedness gives $d=\mathrm{dist}(v,M)>0$. For $l(x)\ne 0$, $x/l(x)-v\in M$, so $d\le \Vert x\Vert /|l(x)|$. The same resulting bound holds when $l(x)=0$:This proves continuity and explains exactly how closedness of the kernel is used.

False. Take the separable Banach space $E={\ell }^1$, whose continuous dual space is ${\ell }^{\infty }$. We prove that first countability of the weak unit ball would force this dual to be norm separable, producing a contradiction.

Suppose that $B$ had a countable weak neighbourhood base $(U_n)$ at zero. Inside each $U_n$, choose a basic relative weak neighbourhood specified by finitely many continuous linear functionals $F_n$. The union $F={\bigcup }_n{F}_n$ is countable. For any $f\in E^{\prime }$ and $ϵ>0$, some $U_n$ lies in $\{x\in B:|f(x)|<ϵ\}$. On the vector subspacethis implies $\Vert f{|}_{M_n}\Vert \le ϵ$, by scaling vectors into $B$. The Hahn-Banach theorem extends $f{|}_{M_n}$ to $h\in E^{\prime }$ with $\Vert h\Vert \le ϵ$. The functional $f-h$ vanishes on $M_n$ and is therefore a linear combination of $F_n$: it factors through the finite-dimensional coordinate map $x\mapsto (g(x){)}_{g\in F_n}$, and a linear functional on that map's image extends to the finite-dimensional coordinate space. HenceEvery $f\in E^{\prime }$ thus belongs to the norm closure of $\mathrm{span}F$. Taking rational coefficients, or rational real and imaginary parts, gives a countable norm-dense subset of $E^{\prime }$. This proves that weak-ball metrizability requires a norm-separable dual.

For ${\ell }^1$, every bounded sequence $a=(a_j)$ defines $f_a(x)={\sum }_j{a}_j{x}_j$ with $\Vert f_a\Vert =\Vert a{\Vert }_{\infty }$. Conversely, the values $f(e_j)$ of any bounded linear functional determine such a sequence and recover $f$ by density of finitely supported vectors. Thus $E^{\prime }={\ell }^{\infty }$. The uncountable family $\{0,1{\}}^{\mathbb{N}}$ in ${\ell }^{\infty }$ has distance one between distinct members, so this dual is not norm separable: disjoint balls of radius less than $1/2$ would require distinct members of any countable dense set. Since every metrizable space is first countable, the weak unit ball of ${\ell }^1$ is not metrizable.

A uniform algebra on a compact Hausdorff space $K$ is a closed complex subalgebra $A\subseteq C(K)$ containing the constants and separating points of $K$, with the supremum norm. It is therefore a commutative unital Banach algebra. Its character space of an algebra ${\Phi }_A$, also called its carrier space, consists of nonzero multiplicative linear functionals, equipped with the Gelfand topology.

For $x\in K$, evaluation gives the character of an algebra ${\delta }_x(f)=f(x)$. The map $\delta :K\to {\Phi }_A$ is continuous: composing it with the coordinate $\phi \mapsto \phi (f)$ gives the continuous function $f$ for every $f\in A$. It is injective because $A$ separates points. The Gelfand topology is Hausdorff because distinct algebra characters differ on some element of $A$. A continuous injection from compact $K$ into a Hausdorff space is a homeomorphism onto its image: closed subsets of $K$ have compact, hence closed, images. Its image is also compact and closed. Thusis the asserted homeomorphism. A natural uniform algebra is one for which this image is all of ${\Phi }_A$.

Here is the matrix form of Rado's theorem. A matrix $A$ with entries in the rational numbers and columns $a_1,\dots ,a_n$ is a partition regular matrix if every finite coloring of the positive integers admits a vector $x\in {\mathbb{N}}^n$ whose coordinates are monochromatic and for which $Ax=0$. Its columns condition asks for an ordered partition into nonempty blocks $B_1,\dots ,B_s$, withRado's theorem states that these two conditions are equivalent. Coordinates need not be distinct; positivity is required for every coordinate.

First prove necessity using the finite separating-functional proof of the columns condition. Clearing denominators does not change either the kernel or the columns condition, so assume that $A$ has integer entries. For each pair of disjoint index sets $I,B\subseteq \{1,\dots ,n\}$, with $B$ nonempty, for which $b_B={\sum }_{i\in B}{a}_i$ is outside the rational linear span of the $I$-columns, choose an integer linear functional ${\ell }_{I,B}$ such thatSuch a linear functional exists: extend a basis over the rational numbers of the indicated linear span by $b_B$, prescribe values zero on the former basis and one on $b_B$, extend to a basis of the ambient vector space over the rational numbers, and then clear the functional's denominators. There are only finitely many pairs $(I,B)$. Choose a prime number $p$ larger than the absolute values of all their nonzero evaluations; if there are none, choose any prime number.