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Suppose $v=(v_1,\dots ,v_r{)}^t$ belongs to the kernel of the displayed matrix $\Lambda$. Define the polynomial $q(z)={\sum }_{k=1}^r{v}_k{z}^k$. Then $q({\lambda }_j)=0$ for all $j$, and $q(0)=0$. These are $r+1$ distinct roots of a polynomial of degree at most $r$. Repeated application of the factor theorem therefore makes $q$ identically zero, so every $v_k=0$. Hence $\Lambda$ is injective and, by finite-dimensional rank-nullity theorem, invertible. Its transpose is invertible with inverse $({\Lambda }^{-1}{)}^t$, proving both kernels are trivial.

For the second matrix, membership of the all-ones vector in the kernel meansGroup the distinct nonzero values as ${\mu }_1,\dots ,{\mu }_r$, with multiplicities $m_1,\dots ,m_r>0$. Zero values contribute nothing to these positive powers. The first $r$ equations areor $\Lambda (\mu {)}^{t}m=0$. The proved transpose-kernel statement implies $m=0$, contradicting its positive entries if $r>0$. Therefore $r=0$ andConversely, all zero values plainly give a zero matrix. This proves the zero power sums force a finite complex multiset to vanish criterion, including repetitions and zero values rather than assuming all eigenvalues distinct.

Use the usual commutative unital-ring convention for Noetherian rings. A ring is Noetherian exactly when every ideal is finitely generated, equivalently when ascending chains of ideals stabilize. Indeed, the union of an ascending chain is an ideal; finitely many generators all lie in a single chain member. Conversely, if an ideal cannot be finitely generated, adjoining a new element at each stage constructs a strictly ascending chain.

First assume $R[X]$ is Noetherian. Evaluation at zero gives a surjective ring homomorphism $R[X]\to R$ with kernel $(X)$. A quotient of a Noetherian ring is Noetherian: lift an ideal to the source, take finitely many generators there, and map those generators down. Thus $R\cong R[X]/(X)$ is Noetherian.

For the converse, assume $R$ is Noetherian and take any ideal $I\subseteq R[X]$. For $d\ge 0$, let $J_d$ be the set of coefficients of $X^d$ in elements of $I$ of degree at most $d$. It is an ideal of $R$, since sums and scalar multiples of such polynomials stay in $I$. Multiplication by $X$ shows $J_d\subseteq J_{d+1}$. The ascending chain stabilizes, say at $N$.

For each $0\le d\le N$, choose finitely many generators $a_{d,j}$ of $J_d$ and corresponding $f_{d,j}\in I$ of degree at most $d$ having coefficient $a_{d,j}$ at $X^d$. Discard zero generators. We claim these finitely many polynomials generate $I$. If a nonzero $f\in I$ has degree $m\le N$, express its leading coefficient as ${\sum }_j{r}_j{a}_{m,j}$ and subtract ${\sum }_j{r}_j{f}_{m,j}$. The degree decreases. If $m>N$, use $J_m=J_N$, express the leading coefficient using the $a_{N,j}$, and subtractAgain the leading term cancels and the degree decreases. Induction on degree, including degree zero, expresses every $f$ in the claimed ideal. Therefore every ideal of $R[X]$ is finitely generated, proving the Hilbert basis theorem andThe argument is the bounded-degree coefficient proof of Hilbert basis theorem; it never divides by a leading coefficient and therefore does not require $R$ to be an integral domain. With the analogous left-ideal convention the same proof applies to left-Noetherian rings with a central indeterminate.

Let $p$ first be an arbitrary degree-$n$ polynomial with leading coefficient $a_n\ne 0$. Choose $R$ so large that on $|z|=R$,The homotopy $a_n{z}^n+s[p(z)-a_n{z}^n]$, $0\le s\le 1$, never meets zero on that circle. Hence the image curve $p(R{e}^{i\theta })$ has winding number $n$, the same as $a_n{R}^n{e}^{in\theta }$. The argument principle counts precisely $n$ zeros inside, with multiplicity, since there are no poles. Choose $R$ large enough that the same leading-term domination holds for every $|z|\ge R$; the lower-term ratios decrease with radius, so this also excludes all exterior zeros. This proves the fundamental theorem of algebra with its multiplicity count. A degree-zero nonzero polynomial has no zeros.

Now use the specified bound for the monic quartic. Compare $\Gamma$ with the closed curveIts first piece travels once counterclockwise around a circle, and its second stays at the endpoint, so its winding number is one. On the first piece, $|\Gamma -{\gamma }_0|<R^4$. On the second, both $p(R)$ and $p(iR)$ differ from $R^4$ by less than $R^4$; every convex combination of those errors also has modulus less than $R^4$. Thus the quoted winding number stability result gives

For real coefficients define $q(y)=y^4-b{y}^2+d$. It has no real zeros and positive leading coefficient, so $q(y)>0$ for every real $y$, including $d=q(0)>0$. Thereforelies strictly in the right half-plane for every real $y$. Also $p(x)>0$ for $x\ge 0$: there are no zeros there, and the polynomial is positive for sufficiently large positive $x$. These facts exclude roots on the positive real and imaginary axes.

Consider the positively oriented quarter-disk boundary: first the arc from $R$ to $iR$, then the imaginary axis to zero, then the real axis back to $R$. Under $p$, its last two pieces form a path from $p(iR)$ to $p(R)$ entirely within the right half-plane. The chord closing $\Gamma$ also lies there. Since this half-plane is convex and excludes zero, deforming one closing path into the other changes no winding number. Thus the full image of the quarter-disk boundary has winding number one. By the argument principle there is exactly one root inside that quarter-disk, counted with multiplicity.

Finally, all polynomial roots lie in $|z|<R$ under the given strict coefficient bound: for $|z|\ge R$, the sum of lower terms divided by $|z{|}^4$ is smaller than one. HenceIts multiplicity is one. This is the quarter-sector winding test for a real quartic; it accounts for the axis exclusions and does not replace the sector boundary by a chord without justifying the deformation.

The holomorphic convex hull is usually defined analytically byWith that convention the requested equivalence is the definition. Its substantive planar content is the geometric descriptionOne can instead define the hull by the right side and prove the analytic equivalence; we do so after proving Runge theorem.

The prescribed-pole form of Runge theorem is as follows. If $K\subset \mathbb{C}$ is compact and $A\subset \mathbb{C}\setminus K$ meets each bounded complementary component, every holomorphic function on a neighborhood of $K$ is uniformly approximable on $K$ by rational functions with finite poles in $A$. Polynomial terms are allowed, accounting for the unbounded component. In particular, if every complementary component meets the complement of a domain $\Omega$, the approximants can be chosen as holomorphic functions on $\Omega$. Connected complement permits polynomial approximants.

For a proof, surround $K$ by the interior of a finite union of small grid squares whose closure lies in the neighborhood where $f$ is a holomorphic function. Orient their boundaries positively and cancel shared edges, obtaining finitely many piecewise linear boundary curves $\Gamma$ a positive distance from $K$. The Cauchy integral formula, applied to the union, givesUniform continuity of this integrand for $z\in K$ makes its Riemann sums converge uniformly on $K$. Each Riemann sum is a finite linear combination of reciprocals $(z-b{)}^{-1}$ with $b\in \Gamma \subset \mathbb{C}\setminus K$.

It remains to move these poles into the allowed components. Let $\mathcal{A}$ be the uniform closure on $K$ of rational functions with finite poles in $A$ and polynomial terms. This is a closed algebra. DefineFor $b\in E$ and $|c-b|<\mathrm{dist}(b,K)$, the uniformly convergent geometric serieslies in $\mathcal{A}$, so $E$ is relatively open. If $b_n\in E$ tends to $b\notin K$, the corresponding reciprocals converge uniformly on $K$, so $E$ is also relatively closed. Thus containing one point of a complementary component means containing the entire component. Each bounded component has its prescribed point in $A\subset E$. The unbounded component meets $E$ because, for $|b|>{\sup }_K|z|$,is uniformly approximated by polynomials. Therefore $E=\mathbb{C}\setminus K$. Approximate the finitely many reciprocal terms in a Riemann sum closely enough to control their combined error. This completes the proof of Runge theorem without assuming its conclusion during the pole-moving argument.

Now let $H$ denote the geometric filled set displayed above. For $w\in K$ the required inequality is automatic. If $w$ lies in a filled component $U$, then $f$ is a holomorphic function near its compact closure and $\partial U\subset K$. The maximum modulus principle gives $|f(w)|\le {\sup }_K|f|$. Hence $H\subseteq {\hat{K}}_{\Omega }$.

Conversely, suppose $w\in \Omega \setminus H$. Its component $V$ of $\mathbb{C}\setminus K$ meets $\mathbb{C}\setminus \Omega$, or is unbounded. To see the alternative, a bounded component contained in $\Omega$ has boundary in $K$, so its closure is compact in $\Omega$ and would have been filled. Apply the same open-and-closed reciprocal argument to the algebra generated by polynomials and reciprocals with poles outside $\Omega$. The component $V$ has a starting reciprocal either at a point outside $\Omega$ or near infinity. Thus there is $h\in \mathcal{O}(\Omega )$ approximating $(z-w{)}^{-1}$ as closely as desired on $K$. Let $M={\sup }_K|z-w|>0$ and choose the error less than $1/(2M)$. ThenThis holomorphic function separates $w$ from the analytic hull. Thus $w\notin {\hat{K}}_{\Omega }$, proving the equivalence with the geometric definition as well. For empty $K$, the filled hull is empty and the constant function one excludes every point when the supremum of the nonnegative modulus over the empty set is taken as zero; the nonempty case is the one used above.

We prove both inequalities, including the BK inequality, for the inhomogeneous product measure in the question. Identify a configuration with its set of coordinates equal to one. An up-set is then an increasing event. For an increasing event $A$, a set $S$ of open coordinates is a witness if the configuration with precisely $S$ open belongs to $A$: upward closure ensures that those coordinates alone force $A$, whatever the remaining coordinates do. The event $A\square B$ means that there are disjoint open witnesses $S,T$ for $A,B$ respectively.

First prove the Harris' inequality by induction on the number of coordinates. For the last coordinate, of parameter $p$, write $A_0\subseteq A_1$ and $B_0\subseteq B_1$ for the two sections in the remaining coordinates, and put $a_i=\mathbb{P}(A_i)$, $b_i=\mathbb{P}(B_i)$. Induction applied to each section givesSubtract the product of the marginal probabilities from the right-hand side:The base case of zero coordinates is immediate. Thus

For disjoint occurrence, first assume every coordinate parameter is at most $1/2$, and again use induction. Put $c_{ij}=\mathbb{P}(A_i\square B_j)$ on the remaining coordinates. The zero-section of $A\square B$ is exactly $A_0\square B_0$. The one-section is exactlyIndeed, the last open coordinate can belong to at most one of two disjoint witnesses. Assigning it to the $A$ witness leaves a witness for $A_1$ and a witness for $B_0$; assigning it to the $B$ witness gives the other event. If neither witness uses it, both descriptions still hold. Conversely, witnesses for either displayed event lift to disjoint witnesses by adding the last coordinate only to the appropriate one.

The intersection of the two displayed events contains $A_0\square B_0$, because the sections are increasing. ThereforeThe coefficient $1-2p$ is nonnegative. Applying the induction hypothesis to each term givesThe product of the marginal probabilities is exactly this last expression plusThis proves the desired inequality whenever every parameter is at most $1/2$.

To remove that restriction, suppose first $p_i<1$ for every coordinate. Replace coordinate $i$ by the logical disjunction of $m_i$ independent Bernoulli bits, each with parameterChoose $m_i$ large enough that $q_i\le 1/2$. Different blocks are independent, and their OR values have exactly the original distribution. Let $\tilde{A},\tilde{B}$ be the inverse images of $A,B$ under this block-OR map; they are increasing events. If the OR configuration has disjoint witnesses $S,T$, choose one open bit from each block indexed by $S$ or $T$. The bits chosen for $S$ force $\tilde{A}$, those for $T$ force $\tilde{B}$, and the two bit sets are disjoint. Thus the inverse image of $A\square B$ is contained in $\tilde{A}\square \tilde{B}$. The small-parameter result provesParameters equal to one follow by a limit: probabilities of any fixed event on a finite cube are polynomials in the coordinate parameters, hence continuous. Zero parameters already cause no difficulty. We have obtained

For the graph application, take coordinates to be the edges of the complete graph. Let $A$ be the increasing event that $x$ is connected by a graph path to $U$, and $B$ the corresponding event for $W$. Their probabilities are $\alpha ,\beta$. A simple graph path from $U$ to $W$ through $x$ splits at $x$ into two graph paths with disjoint edge sets. Those sets witness $A$ and $B$, so this graph path event is contained in $A\square B$. The BK inequality givesOn the other hand, if both $A$ and $B$ occur, concatenate an $x$-to-$U$ graph path and an $x$-to-$W$ graph path to obtain a walk from $U$ to $W$. Removing loops yields a simple graph path, although it need not pass through $x$. Thus $A\cap B$ is contained in the unrestricted connection event, and the Harris' inequality givesThe distinction between a simple graph path through $x$ and an unrestricted connection is essential: a walk may revisit an edge, so a through-$x$ walk would not in general provide disjoint witnesses.

For a sheaf of abelian groups $\mathcal{F}$ on a topological space $X$, putThese arbitrary families of germs form a sheaf, and restrictions are projections. Extending a family by zero proves that $C(\mathcal{F})$ is flasque. Sending a section to all its germs gives an injective sheaf map $\mathcal{F}\to C(\mathcal{F})$, since a section with every germ zero vanishes locally and hence globally.

Set $Q^{-1}=\mathcal{F}$, $I^j=C(Q^{j-1})$ and $Q^j=\mathrm{coker}(Q^{j-1}\to I^j)$, using the sheaf cokernel. Compose $I^j\to Q^j\to I^{j+1}$ to obtain the Godement resolutionIt is exact by construction, with flasque terms. Define sheaf cohomology by the cohomology of its global-section complex:The resolution principle for sheaf cohomology identifies this with the computation using any flasque resolution.

The elementary input for acyclicity is the flasque-kernel section-lifting lemma: in a short exact sequence with flasque kernel, sections of the quotient lift over every open subset. Consequently a quotient of two flasque sheaves is flasque, because a quotient section lifts and that lift extends. If $\mathcal{F}$ itself is flasque, induction through $0\to Q^{j-1}\to I^j\to Q^j\to 0$ makes every $Q^j$ flasque. The same lifting lemma makes these sequences exact on global sections. Hence the global-section complex has no positive-degree cohomology, provingThis deduction uses the lifting lemma, not the desired cohomological vanishing as an assumption.

For the rational-section construction, add two representative sections after restricting to the intersection of their dense domains. Multiply by a representative rational function in the same way. Finite intersections of dense open sets are dense, and further restricting both representatives does not change their resulting class. The module axioms hold on their common domain. Thus rational sections form a module over $\mathrm{Rat}(X)$; irreducibility is not needed for this module statement. When $X$ is irreducible, $\mathrm{Rat}(X)$ is the function field $K=k(X)$.

Suppose now that $\mathcal{F}$ is locally free and $X$ irreducible. A germ at $P$ is represented on a nonempty open neighbourhood, which is dense, so it defines a rational section. Two representatives of the germ agree on a neighbourhood of $P$ and therefore give the same class. If the class is zero, trivialize $\mathcal{F}$ on a neighbourhood of $P$. Its coefficient functions vanish on a dense open subset; regular functions on a reduced irreducible variety with this property vanish identically. The germ is therefore zero. This proves the stalk inclusion into rational sections of a locally free sheaf:Local freeness matters: a nonzero torsion germ can vanish on a dense open set.

On an irreducible algebraic curve, every proper closed subset is finite, and every open subset is quasi-compact because the Zariski space is Noetherian. Let $R=\mathrm{Rat}(\mathcal{F})$. The constant sheaf $\mathcal{R}(\mathcal{F})$ has sections $R$ on every nonempty open set: such opens are irreducible and hence connected. DefineRestriction drops the components outside the smaller open set. For a compatible family over an open cover of $U$, take a finite subcover by quasi-compactness. The union of the finite supports of these finitely many sections is finite. Their agreeing components therefore glue to a unique finite-support tuple on $U$. This verifies the sheaf gluing axiom, including arbitrary covers, rather than merely asserting that a presheaf direct sum is always a sheaf.

A rational section is regular on some dense open set, whose complement on a curve is finite. Its classes $[s{]}_P\in R/{\mathcal{F}}_P$ therefore have finite support, defining a sheaf map $\mathcal{R}(\mathcal{F})\to \mathcal{P}(\mathcal{F})$. At $P$, the first sheaf has stalk $R$. The second has stalk $R/{\mathcal{F}}_P$: any finite tuple can be restricted to a neighbourhood omitting all its other support points, and its $P$-component is unaffected by such restriction. The resulting sequence on stalks isExactness of sheaves can be tested on stalks, so this proves the rational principal-parts resolution on an algebraic curveBoth right-hand sheaves are flasque: restrictions for the constant rational-section sheaf are identities between nonempty opens, and restrictions for the rational principal-parts sheaf are projections, with extension by zero. ConsequentlySurjectivity on stalks has not been confused with surjectivity on global sections.

For an explicit failure of global surjectivity take $X={\mathbb{P}}^1$ and $\mathcal{F}=\mathcal{O}(-2)$. Use coordinate $z=X_1/X_0$ and frame $e_0=X_0^{-2}$ on the finite chart. At infinity use $w=1/z$ and $e_{\infty }=X_1^{-2}=z^{-2}{e}_0$. Prescribe the rational principal part of $z^{-1}{e}_0$ at zero and zero at every other point. If a rational section $r(z)e_0$ realized this tuple, then $r-z^{-1}$ would be regular at zero and $r$ would be regular at every other finite point. Thus $r=z^{-1}+p(z)$ with $p\in k[z]$. Regularity at infinity would requireto be regular at $w=0$. Its $w^{-1}$ term cannot cancel with any term of the polynomial contribution, whose powers are at most $-2$. This is impossible. Hence the global rational-principal-part map need not be surjective.

For an affine curve it is always surjective, and here is a direct construction. Write $X=\mathrm{Spec}A$, $K=\mathrm{Frac}A$ and $\mathcal{F}=\tilde{M}$. By the affine module sheaf construction, $R=M{\otimes }_{A}K=:M_K$ and ${\mathcal{F}}_P=M_{{\mathfrak{m}}_P}$. Choose representatives $v_P\in M_K$ for a prescribed finite tuple, and choose one nonzero $a\in A$ with $a{v}_P\in M$ for all of them. The zero set $T=V(a)$ is finite. The ring $A/(a)$ is Artinian, since it is Noetherian of dimension zero, and its Artinian decomposition into local factors givesPrescribe the class of $a{v}_P$ in the indicated component for each requested point in $T$, and zero in the other components. A requested point outside $T$ already has zero principal part because $a$ is a unit there. Lift the tuple to $w\in M$. Then $s=w/a\in M_K$ satisfies $s-v_P\in M_{{\mathfrak{m}}_P}$ at every requested point in $T$, and is regular at every other point of $T$. Outside $T$ it is regular because $a$ is invertible. It therefore realizes exactly the prescribed tuple. This proves affine principal-parts interpolation for a locally free sheaf and givesNo nonsingularity assumption was used; the construction applies to singular irreducible affine curves as well.

On the standard affine charts $U_i=\{X_i\ne 0\}$, use the formal frame $e_i=X_i^m$ and glue rank-one free sheaves byThe transition functions are regular units on overlaps and satisfy the cocycle identity. This constructs the twisting sheaf on projective space $\mathcal{O}(m)$ as an invertible sheaf for every integer $m$, including negative $m$.

On a nonempty $U$, a section has local expressions $h_i{e}_i$, where $h_i\in \mathcal{O}(U\cap U_i)$. Pulling to the punctured affine cone gives $h_i(X/X_i)X_i^m$. Their agreement on overlaps makes them one homogeneous rational function $r$ of degree $m$, regular on ${\pi }^{-1}U$. To make its polynomial representation explicit, write one nonzero local coefficient as $p(y)/q(y)$ in the affine coordinates. Homogenize $p,q$ to degrees $a,b$; the resulting rational expression isIf the exponent of $X_i$ is negative, move its power to the denominator. Cancel common factors; the greatest common divisor of homogeneous polynomials can be chosen homogeneous. Thus $r=F/G$ with $F,G$ coprime homogeneous polynomials, $G\ne 0$, and $\mathrm{deg}F-\mathrm{deg}G=m$.

Conversely, for such a homogeneous rational function regular on ${\pi }^{-1}U$, its coefficient $r/X_i^m$ is regular on $U\cap U_i$: restrict to the slice $X_i=1$. These coefficients obey the displayed transition rule and hence define a section. This proves the homogeneous rational sections of a twisting sheaf description in both directions, for arbitrary open $U$ rather than only a standard chart.

Now lift a regular degree-zero rational function $f=P/Q$ to the punctured cone. The quotient rule givesThis is a rational function homogeneous of degree $-1$. It is independent of the chosen representation because differentiation is a derivation of the rational function field. It is regular wherever $f$ is regular: a derivation of a polynomial ring extends to each localization by the quotient rule, and regular functions are locally such fractions. The preceding homogeneous-section description therefore provesZeros of a particular displayed denominator do not invalidate this argument; regularity is a property of the rational function, which can have another local representation.

The map $f\mapsto ({\partial }_{0}f,\dots ,{\partial }_{n}f)$ is a $k$-linear sheaf derivation with values in $\mathcal{O}(-1{)}^{\oplus (n+1)}$. The universal property of Kähler differentials consequently definesMultiplication by the coordinate section $X_i\in \Gamma ({\mathbb{P}}^n,\mathcal{O}(1))$ defines the other mapFor a degree-$d$ homogeneous polynomial, differentiating each monomial gives ${\sum }_i{X}_i{\partial }_{i}P=dP$. Applying this to the equal-degree numerator and denominator gives ${\sum }_i{X}_i{\partial }_{i}f=0$ for degree-zero $f$, so $\sigma D=0$. This identity is valid in every characteristic.

To prove all the exactness assertions, fix $U_i$ and put $y_j=X_j/X_i$ for $j\ne i$. The Kähler differential sheaf is free there on $d{y}_j$, because a derivation of the polynomial ring is determined freely by its values on the coordinates. Trivialize each $\mathcal{O}(-1)$ with frame $X_i^{-1}$. In this frame,where $e_j$ now denotes the $j$th coordinate vector of the direct sum. The map $\sigma$ is surjective, since its $i$th coefficient is one. Its kernel consists exactly of tuples with $b_i=-{\sum }_{j\ne i}{y}_j{b}_j$, so the $n$ displayed vectors $e_j-y_j{e}_i$ form a free basis of that kernel. The map $D$ sends the differential basis bijectively to this kernel basis and is therefore injective. Exactness on these charts proves the cotangent Euler sequence in homogeneous coordinatesThe proof uses no division by an integer and hence no characteristic-zero assumption.

For $n\ge 1$, the cohomology of twisting sheaves on projective space givesIn particular, $H^0(\mathcal{O})=k$, $H^{q>0}(\mathcal{O})=0$, and $H^q(\mathcal{O}(-1))=0$ for every $q$. The long exact sequence in sheaf cohomology of the displayed Euler sequence beginsand gives zero in the remaining degrees. ThusThe nonzero group is generated by the connecting image of the section $1$ of $\mathcal{O}$. If $n=0$ is admitted, ${\mathbb{P}}^0$ is a point and its cotangent sheaf is zero, so the sole requested group $H^0$ is zero.

Fix a finite-dimensional nonnegative filtration of a ring giving $R$ its almost commutative algebra structure, and choose a good filtration of a module on $M$. When $\mathrm{gr}R$ is a standard graded algebra, the Hilbert-Serre theorem makes the cumulative Hilbert function $h_M(j)={\dim }_k{F}_{j}M$ eventually a polynomial. For $M\ne 0$, define the dimension of a filtered module and multiplicity of a filtered module byUse $d(0)=-\infty$ and $m(0)=0$. With positive weighted generators instead of a standard grading, the cumulative Hilbert function is eventually a quasipolynomial with a common positive leading coefficient across residue classes. Equivalently, $h_M(j)=c{j}^d+O(j^{d-1})$, and the definitions are $d(M)=d$ and $m(M)=d!c$. The common leading coefficient follows from monotonicity: interlacing consecutive residue classes forces their leading coefficients to agree. For a standard grading, multiplicity is a positive integer, since an integer-valued polynomial has an integral expansion in the binomial polynomials; weighted normalization can instead give a positive rational number.

Two good filtrations of a module for this fixed algebra filtration bound each other after fixed index shifts. The corresponding Hilbert functions therefore satisfy $h_F(j-a)\le h_G(j)\le h_F(j+b)$ for fixed $a,b$. Comparing their leading growth shows that the degree and leading coefficient agree. Thus $d(M)$ and $m(M)$ are independent of the module filtration; the fixed algebra filtration is part of the multiplicity convention.

For a submodule $N$, use its subspace filtration and the quotient filtration on $M/N$. These are good because $\mathrm{gr}R$ is Noetherian. The short exact sequence of filtered pieces givesNonzero leading coefficients are positive, soIf $d(N)=d(M/N)$, those leading coefficients add andThis is dimension and multiplicity in a filtered exact sequence.

For the Weyl algebra $A_n(k)$ in characteristic zero, Bernstein inequality for Weyl algebra modules asserts $d(M)\ge n$ for every nonzero finitely generated module $M$. Write its generators as $x_i,D_i$, with $[D_i,x_j]={\delta }_{ij}$ and all other generator commutators zero. The ordered monomials $x^{\alpha }{D}^{\beta }$ give the Bernstein filtrationTake a finite-dimensional generating subspace $V\ne 0$ of $M$ and put $M_j=A_{j}V$.

To prove faithful finite-step action of a Weyl algebra, consider $0\ne a\in A_j$ and choose a nonzero coefficient $c_{\alpha \beta }$ of a monomial of maximal total degree $d\le j$. Applying ${\prod }_i(\mathrm{ad}{D}_i{)}^{{\alpha }_i}(\mathrm{ad}{x}_i{)}^{{\beta }_i}$ to $a$ givesIndeed, these commutators differentiate the ordered coefficients: lower-degree monomials vanish, and another degree-$d$ monomial can survive only if all its exponents dominate $(\alpha ,\beta )$, which forces equality. This is scalar extraction by iterated Weyl commutators.

Expanding these iterated commutators writes $\lambda ={\sum }_{\ell }{u}_{\ell }a{v}_{\ell }$ with $u_{\ell },v_{\ell }\in A_j$. If $a$ annihilated $M_j$, every summand would annihilate $V$, contrary to $\lambda V\ne 0$. Therefore the action map is injective:Taking dimensions and comparing polynomial degrees giveswhich proves Bernstein inequality for Weyl algebra modules. The polynomial representation $k[x_1,\dots ,x_n]$ has dimension $n$, so the bound is sharp.

The associated graded ring of the Weyl algebra is a polynomial ring, so ascending filtered-graded transfer of Noetherianity makes $A_n(k)$ a left Noetherian ring. Thus every submodule of a finitely generated module is finitely generated. If $d(M)=n$, every nonzero subquotient has dimension at most $n$ by the exact-sequence formula, and at least $n$ by Bernstein inequality for Weyl algebra modules. Every such subquotient therefore has dimension $n$ and positive integral multiplicity.

In a strictly descending chain $M=M_0⊋M_1⊋\cdots$, each nonzero quotient $M_i/M_{i+1}$ consumes at least one unit of multiplicity, by dimension and multiplicity in a filtered exact sequence. Hence there are at most $m(M)$ strict steps. This proves the descending chain condition, so $M$ is Artinian. In fact this holonomic Weyl algebra module has module length at most $m(M)$, as in multiplicity bounds the length of a holonomic Weyl module.

For a cell $(i,j)$ in the Young diagram of $\lambda$, its hook of a Young diagram consists of that cell, the cells to its right in row $i$ and the cells below it in column $j$. Its length isA removable rim hook is a connected border strip $\lambda /\mu$ containing no $2\times 2$ square, whose removal leaves a Young diagram. Its leg length is the number of rows it meets minus one. The hook graph is the Young diagram labeled at each cell by $h_{ij}$; here “graph” means this labeled array, not a graph of possible hook-walk moves.

Choose $l\ge \ell (\lambda )$, pad with zeros and form the beta numbersThey are distinct nonnegative integers in decreasing order and recover ${\lambda }_i={\beta }_i-l+i$. For $l=\ell (\lambda )$ they equal the first-column hook lengths, since $h_{i1}={\lambda }_i+l-i$. If $l$ is increased, old beta numbers shift upward by one and a new bead at zero is inserted.

For a prime-modulus partition abacus arrange the nonnegative positions in $p$ runners: position $pq+r$ lies at level $q$ on runner $r$, $0\le r<p$. Put beads at the beta numbers and leave the other positions empty. Replacing a bead at $b$ by a bead at a vacant $b-k\ge 0$ removes a rim hook of length $k$; its leg length is the number of beads strictly between these positions. This follows by tracing the boundary of the diagram: bead positions record vertical steps and gaps record horizontal steps, and exchanging one bead and gap removes the intervening border strip. In particular a removable $p$-hook is exactly a bead that can slide up one place on its own runner.

The partition core at modulus $p$ is the result of sliding beads up until there are no gaps above any bead on a runner. If runner $r$ initially has $b_r$ beads, its final occupied levels must be $0,1,\dots ,b_r-1$, independently of slide order. Thus its packed beta set, and hence its core partition of an integer, is unique. Increasing the number of beta numbers merely re-expresses the same boundary by shifting positions and adding the new initial bead; the packed configuration represents the same partition of an integer. This proves well-definedness both with respect to removal order and to the chosen padding. The number of slides isThe partition quotient components record how far each runner's beads lie above their packed positions, giving the hook-removal interpretation used in Question 5.

A block of a group algebra in characteristic $p$ for $F{S}_n$ is a two-sided ideal $B=F{S}_{n}e$ for a primitive central idempotent $e$, meaning a nonzero central idempotent that is not a sum of two nonzero orthogonal central idempotents. The identity is the sum of these block idempotents and the group algebra is their direct product. An indecomposable module $V$ belongs to $B$ when $eV=V$, equivalently every other block idempotent acts as zero. Indeed $V={⨁}_{e}eV$; indecomposability makes exactly one summand nonzero.

The Nakayama block theorem, historically called the Nakayama Conjecture, statesEquivalently the ordinary characters assigned to one modular block have a common partition core at modulus $p$. Its simple modules are the $D^{\lambda }$ with regular $\lambda$ having that core. Only cores $\kappa$ for which $n-|\kappa |$ is a nonnegative multiple of $p$ occur. The block weight is $(n-|\kappa |)/p$; a core alone determines the block at this fixed degree.

One route to the proof separates the combinatorial core invariant from the algebraic center invariant. Let $c_r(\lambda )$ be the number of cells of residue $r=j-i(\mathrm{mod}p)$. A length-$p$ rim hook has one cell of every residue: along its border strip the contents are consecutive integers. Therefore removal subtracts one from every $c_r$. More precisely, choose the number $l$ of beta numbers divisible by $p$ and write the runner charges as $q_r=b_r-l/p$. Reading the boundary, or adding a cell and checking its two changed beta positions, givesThe charges determine the packed runners and hence the core. For fixed $n$, two equal cores have the same weight, so their full residue multisets are equal. Conversely equal residue multisets give equal charges and the same core. This establishesfor partitions of an integer of the same integer. The fixed-degree condition matters.

For the center invariant, use the commuting Jucys–Murphy elements $J_k={\sum }_{i<k}(ik)$, with $J_1=0$. In the Young seminormal form basis, $J_k$ has eigenvalue equal to the content of the cell containing $k$. A symmetric polynomial in them therefore acts on $S^{\lambda }$ as the same scalar for every Young tableau: its value on the multiset of all contents of $\lambda$. The integral Specht lattice makes this a scalar identity before modular reduction, so modulo $p$ these central scalars depend only on the residue multiset, and every composition factor inherits them.

The substantive center theorem used in this proof is that the center of the symmetric-group algebra, integrally and over a field, consists of symmetric polynomials in the Jucys–Murphy elements. Centrality of elementary symmetric polynomials follows fromthe coefficients are central class-sum combinations. The reverse inclusion requires the integral triangular class-sum argument: expand symmetric monomials in the $J_k$ as products of transpositions, group the leading terms by reduced cycle type, and eliminate these leading class sums in the resulting unitriangular order. The integral coefficients, with no division by multiples of $p$, are crucial; knowing the center only over $\mathbb{Q}$ would not prove the modular assertion. This is the central algebraic ingredient of this proof route.

Thus equal residue multisets yield equal characters of the entire modular center. Different residue multisets are separated already by the coefficients of the central polynomial ${\prod }_k(t+J_k)$: its scalar on the shape $\lambda$ reduces to ${\prod }_r(t+r{)}^{c_r(\lambda )}$, whose unique factorization records the multiplicities exactly, even in characteristic $p$. In a finite-dimensional split algebra, two simple modules are in the same block exactly when their central characters agree: the center of each block is local, while different block idempotents separate different blocks. It follows that equal cores give one block and different cores give distinct blocks. This proves both directions rather than just observing that core is constant on known blocks.

Finally, each ordinary Specht lattice has a single reduced central character, so all constituents of its reduction lie in that one block. Question 1 supplies every simple module and its label; the regular labels therefore identify all the simples in each core block. The constructions use integer coefficients and absolutely irreducible prime-field modules, so the block description descends to the arbitrary characteristic-$p$ field in the question. This completes the main steps of the block proof without presupposing semisimplicity in positive characteristic.

Write $K={\sigma }_A(x)$ and $R_x(\zeta )=(\zeta 1-x{)}^{-1}$. Suppose first that a unital complex-algebra homomorphism sends $Z$ to $x$. If $\lambda \notin U$, the holomorphic function $1/(Z-\lambda )$ belongs to $\mathcal{O}(U)$ and is the multiplicative inverse of $Z-\lambda$. Its image is an inverse for $x-\lambda 1$. Thus existence forces $K\subset U$, even before continuity is used.

For the converse, assume $K\subset U$. Choose a bounded finite union of polygonal regions $V$ with $K\subset V$, $\overline{V}\subset U$, and boundary avoiding $K$. Orient its boundary cycle $\Gamma$ positively around $V$, including negative orientations on holes. Its winding number is one near $K$ and zero outside $U$. DefineThe integral exists in $A$ because the integrand is continuous on finitely many compact contour pieces. The Cauchy theorem shows it is unchanged by replacing $\Gamma$ with another such spectral cycle: the difference has zero winding around the possible singularities in $K$. Banach-valued Cauchy statements follow by applying bounded linear functionals. We use the scalar Cauchy integral formula in its cycle form: a cycle with index one at a point and surrounding only a holomorphy region integrates $f(\zeta )/(\zeta -z)$ to $2\pi if(z)$.

Linearity is immediate and, for a fixed cycle,This is a bound by one compact-uniform seminorm, proving continuity for local uniform convergence. To obtain ${\Theta }_x(1)=1$, deform the resolvent integral, which is holomorphic on $\mathbb{C}\setminus K$, to a large circle and integrate its Neumann series. Similarly $\zeta R_x(\zeta )=1+x{R}_x(\zeta )$ gives ${\Theta }_x(Z)=x$.

For multiplicativity choose an outer spectral cycle for $f$ and an inner one for $g$, nested in $U$ so the inner cycle lies inside the outer region. The resolvent identity isSubstitute it into the double integral for ${\Theta }_x(f){\Theta }_x(g)$. In the term containing $R_x(\eta )$, first integrate $f(\zeta )/(\zeta -\eta )$ around the outer cycle, obtaining $f(\eta )$. In the term containing $R_x(\zeta )$, first integrate $g(\eta )/(\zeta -\eta )$ around the inner cycle, obtaining zero because $\zeta$ lies outside it. Interchange is valid for the continuous integrands on disjoint compact cycles. HenceThis proves the required continuous unital homomorphism, the holomorphic functional calculus.

For uniqueness use the following version of the allowed Runge theorem: on any open $U\subseteq \mathbb{C}$, rational functions with all finite poles outside $U$ approximate each holomorphic function locally uniformly; poles at infinity are allowed, giving polynomials. A unital homomorphism sending $Z$ to $x$ is forced on polynomials and on every inverse $(Z-\lambda {)}^{-1}$, $\lambda \notin U$, and hence on these rational functions. Continuity then forces its value on their limits. Thus ${\Theta }_x$ is unique. This argument works for disconnected $U$ as well and does not assume polynomials alone are dense. It proves continuity and uniqueness of holomorphic functional calculus.

We will use the character description of spectra, with a short justification. A character of an algebra $\phi$ is a nonzero complex multiplicative linear functional; it satisfies $\phi (1)=1$. Applying it to an inverse shows $\phi (a)\in {\sigma }_A(a)$, so $|\phi (a)|\le \Vert a\Vert$ and the character is automatically continuous. Conversely, for noninvertible $a-\lambda 1$, its principal ideal is proper because $A$ is commutative. Put it in a maximal ideal $M$. The closure of $M$ is proper: an ideal containing an element sufficiently close to one contains an invertible element and thus the identity. Maximality makes $M$ closed. The quotient is a complex Banach division algebra. By the nonempty-spectrum result of Question 3, each quotient element differs from some scalar by a noninvertible element, which in a division algebra must be zero. Thus the quotient is $\mathbb{C}$ and its quotient map is a character taking $a$ to $\lambda$. This provesIt is the spectrum equals character values in a commutative Banach algebra statement and does not assume semisimplicity.

For any character, continuity permits passing it inside the contour integral. Since $\phi (R_x(\zeta ))=(\zeta -\phi (x){)}^{-1}$ and $\phi (x)\in K$, the scalar Cauchy formula givesApply the character description first to ${\Theta }_x(f)$ and then to $x$ to obtain the holomorphic spectral mapping theorem

On ${\Pi }_{+}$, use the analytic logarithm with argument in $(-\pi /2,\pi /2)$ and set $s(z)=\exp (\frac{1}{2}\mathrm{Log}z)$. It satisfies $s(z{)}^2=z$ and takes its values in ${\Pi }_{+}$. The principal square root in a commutative Banach algebra isFor uniqueness let $v$ be another root with spectrum in ${\Pi }_{+}$. For each character, $\phi (y)$ and $\phi (v)$ are scalar roots of $\phi (x)$ with positive real part, so both equal $s(\phi (x))$. Hence every character takes the nonzero value $2s(\phi (x))$ at $y+v$. The character description implies $y+v$ is invertible. Commutativity gives $(y-v)(y+v)=y^2-v^2=0$, and multiplying by the inverse yields $y=v$. Equal character values alone would not imply equality in a possibly nonsemisimple algebra; the invertible sum is the essential extra step.

For the unheaded representation request, retain the question's standing unital hypothesis. Let $K$ be the character space of an algebra of $A$, with its weak-star or Gelfand topology. The standard commutative Banach-algebra results allowed here give a nonempty compact Hausdorff character space and ${\sigma }_A(a)=\{\phi (a):\phi \in K\}$. In particular, character compactness follows from the Banach-Alaoglu theorem: the equations $\phi (1)=1$ and $\phi (ab)=\phi (a)\phi (b)$ define a weak-star closed subset of the dual unit ball.

The Gelfand transform is the unital homomorphismEvery element of the commutative C-star algebra is normal. Part i and the character spectrum formula therefore giveThus $\Gamma$ is an isometry, hence injective, and its range is closed because $A$ is complete. Part ii gives $\Gamma (a^{\ast })=\overline{\Gamma (a)}$. The range contains constants and separates points of $K$, since two distinct characters differ on some element of $A$. The complex Stone-Weierstrass theorem states that a unital, conjugation-closed subalgebra of $C(K)$ separating points of a compact Hausdorff space is uniformly dense. Its application here, together with closedness of the range, yieldsThis proves the Commutative Gelfand--Naimark theorem from the preceding parts and the stated approximation theorem.