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Work over $\mathbb{C}$, or another field of characteristic zero. The Weyl algebra $A_n$ is generated by $x_1,\dots ,x_n,{\partial }_1,\dots ,{\partial }_n$ with commuting $x$'s, commuting $\partial$'s and $[{\partial }_i,x_j]={\delta }_{ij}$. Its ordered monomial basis of a Weyl algebra consists of the monomials $x^{\alpha }{\partial }^{\beta }$. One way to verify independence is to apply a finite operator relation to the formal exponential $e^{\mathbf{t}\cdot \mathbf{x}}$: after canceling that exponential the relation becomes a polynomial identity $\sum c_{\alpha \beta }{\mathbf{x}}^{\alpha }{\mathbf{t}}^{\beta }=0$, so every coefficient vanishes. The relations already reorder every word into that basis.

Give both generators degree one. The Bernstein filtration isThus $\dim F_m{A}_n=\left(\genfrac{}{}{0ex}{}{m+2n}{2n}\right)$. Commutator terms have lower degree, accounting for the commutative symbol algebra. Since that algebra is Noetherian, leading-symbol generation followed by finite downward degree cancellation proves that $A_n$ is Noetherian on both sides.

A left finitely generated module $M$ has a good filtration, for example $M_m={\sum }_j{F}_m{A}_n{m}_j$ for a finite generating set of the module. More generally one allows fixed degree shifts of the generators. The associated graded module is finite over the symbol algebra. DefineThis is the characteristic variety of a Weyl algebra module for the Bernstein algebra filtration. It records where the module's leading symbols remain nonzero and makes noncommutative module growth accessible to commutative algebra.

For a fixed algebra filtration it is independent of the good filtration. Any two such filtrations differ by bounded shifts $M_i\subseteq {\tilde{M}}_{i+c}$ and conversely, since each finite set of generators occurs in bounded degree in the other filtration. If a degree-$d$ operator has symbol annihilating $\mathrm{gr}M$, its $r$th power maps $M_i$ into $M_{i+rd-r}$. For $r>2c$, bounded-shift comparison forces the power of that symbol to annihilate $\mathrm{gr}\tilde{M}$. Reversing the argument gives equal radicals of ideals of annihilators. This is independence of characteristic support from a good module filtration. Changing the algebra filtration itself, for example to the order filtration of a Weyl algebra with the $x_i$ in degree zero, need not preserve the same variety; the convention matters.

The Hilbert-Serre theorem gives an eventual cumulative polynomialHere $d$ is also the Gelfand–Kirillov dimension of a module, and $e(M)>0$ is an integer multiplicity of a filtered module for nonzero $M$. Finite generation bounds $d\le 2n$.

Here is a proof sketch of the essential lower bound, rather than an appeal to its name. For a nonzero $f\in F_m{A}_n$, choose a nonzero monomial of greatest total degree in its ordered expansion, say $x^{\alpha }{\partial }^{\beta }$. The operations $\mathrm{ad}{\partial }_i$ and $\mathrm{ad}(-x_i)$ differentiate this expansion with respect to $x_i$ and ${\partial }_i$. Apply them ${\alpha }_i$ and ${\beta }_i$ times. Lower-degree terms die, and every other monomial of the same degree also dies unless it has exactly those exponents. The result is a nonzero scalar multiple of the chosen coefficient. Characteristic zero is essential for the factorials. Expanding the repeated commutators therefore givesChoose $0\ne v\in M_0$. If $f$ acted as zero on all $M_m$, every $f{v}_{j}v$ would vanish, contrary to $v={\sum }_j{u}_{j}f{v}_{j}v$. Hence the linear action mapis injective. It follows that $\dim F_m{A}_n\le (\dim M_m)(\dim M_{2m})$. Comparing polynomial degrees gives $2n\le 2d$. ThusThis proves Bernstein inequality for Weyl algebra modules through the faithful finite-step action of a Weyl algebra.

The bound is sharp. The polynomial representation $\mathbb{C}[x_1,\dots ,x_n]\cong A_n/{\sum }_i{A}_n{\partial }_i$ has the zero section as its characteristic variety of a Weyl algebra module, namely $\xi =0$ and dimension $n$. It is a simple module: differentiate a nonzero polynomial sufficiently to obtain a nonzero constant, then multiply by the $x_i$ to generate the whole module. The module $A_n/{\sum }_i{A}_n{x}_i$ similarly has characteristic variety of a Weyl algebra module $x=0$, a fiber of the cotangent bundle. A nonzero free module has the whole cotangent bundle and dimension $2n$. No nonzero finite-dimensional module exists for $n\ge 1$, also seen by taking the trace of $[{\partial }_1,x_1]=I$.

Modules attaining dimension $n$ are holonomic Weyl algebra modules (with the zero module included conventionally). Every nonzero subquotient remains of dimension $n$ by the lower bound and monotonicity of growth. Compatible filtrations in a short exact sequence make the leading multiplicity of a filtered module additive. Since each nonzero factor has positive integer multiplicity of a filtered module, the number of factors in any strict submodule chain is bounded by $e(M)$. Together with Noetherianity, this proves finite composition length. It illustrates why characteristic variety of a Weyl algebra module and multiplicity of a filtered module provide useful structure beyond a list of presentations. In positive characteristic there are finite-dimensional Weyl modules, so the characteristic-zero inequality and this discussion must not be asserted without its field hypothesis.

Work near the identity $O$ of a generalized Weierstrass equation of an elliptic curve. On the projective chart where $Y\ne 0$, use the local parameter $t=-x/y$ and the auxiliary coordinate $w=-1/y$. Substitution givesThe coefficient of $w$ in the linearized equation at $(0,0)$ is one. Consequently, successive coefficient comparison determines a unique formal power series $w(t)$, without dividing by any coefficient of the equation. Its initial terms areRecover the affine coordinates as Laurent series $x=t/w(t)$ and $y=-1/w(t)$; in particular $x=t^{-2}+O(t^{-1})$ and $y=-t^{-3}+O(t^{-2})$.

The elliptic curve group law is a regular morphism near $(O,O)$. Applying the local parameter to the sum and passing to completed local rings producesOne can construct the series without any nonunit division. PutThe divided difference is an integral formal power series, including its tangent line specialization at $T_1=T_2$. The line $w=\lambda t+\nu$ meets the cubic in the two chosen points and a third point $R$. Substitution in the equation shows that its cubic and quadratic coefficients in $t$ are $-A,-B$, whereSince $\lambda$ has total degree at least two and $\nu$ at least three, $A$ is a unit. The sum of the roots therefore givesThe last expression reflects the third intersection according to the chord-and-tangent group law. Both denominators have constant term one. This proves integrality of the universal construction and also provides an explicit coefficient algorithm. Expanding it, the resulting formal group of an elliptic curve beginsThe integral coefficient recursion and the unit-denominator addition calculation work universally over $\mathbb{Z}[a_1,a_2,a_3,a_4,a_6]$; thus these coefficients specialize in every characteristic of a field. This is the initial coefficients of the elliptic formal group at infinity construction, rather than a definition requiring division by the elliptic-curve discriminant.

The identity, commutativity and associativity of the elliptic curve group law give the defining identities of a one-dimensional commutative formal group law:Reflection in the generalized Weierstrass equation of an elliptic curve gives its formal inverse explicitly:Successive addition gives the multiplication series $[n{]}_F(T)=nT+O(T^2)$.

In characteristic zero, an especially useful description comes from the invariant differential on an elliptic curve:Integrating with zero constant term constructs the formal logarithmTranslation invariance of the invariant differential on an elliptic curve implies $L(F(X,Y))=L(X)+L(Y)$. Thus the formal logarithm converts the formal group law into the formal additive group in characteristic zero. Expanding this identity to degree three also recovers the displayed coefficients of $F$; in positive characteristic the integral addition construction remains valid although the logarithmic denominators may not exist.

A Hilbertian field is a field over which generic irreducibility can be retained by specializing a parameter. More precisely, for polynomials $f_i(T,X)$ irreducible over $K(T)$ and separable in $X$, and a nonzero exclusion polynomial $h(T)$, the Hilbert subsetis infinite. Denominators, leading coefficients and discriminants may be incorporated into $h$. The infinite condition permits simultaneous avoidance of any finitely many bad places. Separability is part of the definition, particularly in positive characteristic.

The field-theoretic interpretation is useful in constructive Galois theory. If $L/K(T)$ is finite separable and generated by a root of $f$, specialization at a good $a$ is the residue extension at a place over $T-a$. Irreducibility of $f(a,X)$ means its residue degree is the entire degree $[L:K(T)]$. Thus a Hilbertian field has infinitely many unramified rational places which are completely inert in each prescribed finite separable extension. For a finite family, one can work in a primitive-element presentation of their separable compositum; preserving its full degree preserves the degrees of all the subextensions.

Hilbert's irreducibility theorem says that $\mathbb{Q}$ is Hilbertian, and gives the corresponding assertion for number fields, including infinitely many integral specializations in its classical form. Rational function fields $k(U)$ over arbitrary fields $k$ are also Hilbertian. In particular, global function fields are Hilbertian just as number fields are. These standard field classes and the specialization viewpoint are discussed in Haran and Jarden's introduction to Hilbertian fields. The theorem does not say every parameter is good: $X^2-T$ is generically irreducible, but a square parameter makes it split. For simultaneous irreducibility, one excludes more exceptional parameters, not the entire field.

Hilbertianity survives finite extensions. For the separable case, let $K$ be Hilbertian, $M/K$ finite separable and $f\in M(T)[X]$ irreducible and separable. A primitive element of $M(T)(\alpha )$ over $K(T)$ has degree $[M:K]\mathrm{deg}f$. Apply Hilbertianity to its minimal polynomial and exclude the presentation's denominators. The resulting residue field contains the constant copy of $M$ and has the same total degree over $K$, so its degree over $M$ is $\mathrm{deg}f$. Therefore $f(a,X)$ remains irreducible over $M$, even for infinitely many $a$ chosen in $K$. This argument also handles finitely many polynomials by a compositum. Purely inseparable finite extensions are treated by separating the separable part and transporting its rational places through the unique purely inseparable extension; the finite-extension stability statement remains valid.

The property is not inherited by arbitrary algebraic extensions. An algebraic closure is not Hilbertian: in characteristic zero, $X^2-T$ is irreducible over $K(T)$ but every specialization has a root in $K$. In characteristic $p$, use $X^p-X-T$ instead; a hypothetical rational-function root would give its right side a pole of order divisible by $p$, contradicting its simple pole at infinity. All constant specializations split over the algebraically closed field.

Finite fields also fail. For $K={\mathbb{F}}_q$ of characteristic $p$, considerIts right-hand additive datum has a pole of order $q+1$, not divisible by $p$, and cannot be $b^p-b$ for a rational function $b$. The polynomial is therefore irreducible over $K(T)$: its roots are $\alpha +c$ for $c\in {\mathbb{F}}_p$, so its splitting-field group has prime order unless a root already belongs to the base field. At every $a\in K$, however, $a^q-a=0$ and the specialization is $X^p-X$, which splits. This gives a concrete failure of the required specialization property.

The main constructive application is Galois-preserving Hilbert specialization. Suppose $E/K(T)$ is finite Galois with group $G$, and use a primitive element of the entire splitting field, not merely a generator of one root field. A good specialization retaining degree $|G|$ has residue group embedded in $G$ by unramified reduction. Since its order is still $|G|$, it is exactly $G$. Preserving only the irreducibility of the original degree-$n$ polynomial would prove transitivity, which is weaker than preserving its full group. Resolvents or the primitive generator eliminate that ambiguity.

Consequently a regular realization of $G$ over $K(T)$, with constant field $K$, yields realizations over the Hilbertian field $K$. This is the bridge from branch-cycle constructions to the Inverse Galois problem over number fields. A realization over $\mathbb{C}(T)$ alone cannot be specialized to $\mathbb{Q}$; descent of its coefficients and constants is an additional step. Rational rigidity can supply that descent under its appropriate hypotheses, while ordinary complex rigidity answers only the geometric uniqueness question.

For example, the family $X^n-X-T$ has full symmetric geometric monodromy in characteristic zero. It is irreducible because it is linear in $T$. Its derivative has $n-1$ distinct critical roots satisfying $n{X}^{n-1}=1$, and their distinct critical values are $-(n-1)X/n$. Each simple critical point gives a transposition. These finite branch cycles generate the monodromy, since the infinity loop is the inverse product of the finite loops. Their transitive connected-edge action gives $S_n$. The arithmetic group is also $S_n$, so the constant field is $\mathbb{Q}$. Hilbert specialization then produces symmetric Galois groups over $\mathbb{Q}$. For $n=2$ the same conclusion follows directly from the square-root family.

Finally, the good parameters form a large arithmetic set without being every point or generally a Zariski-open set. Reducibility can occur at infinitely many exceptional points, as the square example shows. The language of thin sets packages those exceptions as images of generically finite separable covers of degree greater than one, together with finitely many points. Hilbertianity asserts that such finitely many exceptional constructions cannot exhaust the rational line. It is an existence principle with powerful constructive consequences, not by itself an algorithm that certifies a particular chosen parameter.

For a finite point $p$, use the valuation $v_p$ with local parameter $u=t-p$; at infinity use $u=1/t$. Its extensions $w$ to $E$ are permuted transitively by $G$. The completed residue field is $F$, because $F$ is algebraically closed, so the local decomposition group equals its inertia group.

Write $e=e(w/v_p)$. Characteristic zero makes all local ramification tame. One can choose a uniformizer $\pi$ in the completion for whichIndeed, initially $u$ is ${\pi }^e$ times a unit; that unit has an $e$th root in $F[[\pi ]]$, by its algebraically closed residue field and the invertibility of $e$. Absorb it into $\pi$. The local automorphisms are then $\pi \mapsto \zeta \pi$, $\zeta \in {\mu }_e(F)$, and form a cyclic group of order $e$. Define $e_p=e$.

To specify an element rather than a subgroup, fix a compatible choice of primitive roots of unity ${\zeta }_e\in F$. Define ${\tau }_w$ by its uniformizer characterand let $C_p$ be its conjugacy class in $G$. Over $\mathbb{C}$, take ${\zeta }_e=e^{2\pi i/e}$, corresponding to positive local continuation around $p$. At an unramified point, $e_p=1$ and $C_p=\{1\}$.

Changing the uniformizer does not affect this character. If ${\pi }^{\prime }=a\pi$ with $a$ a unit, inertia fixes its residue, soThe character is injective: a nonidentity finite-order automorphism with derivative one would have $\tau (\pi )=\pi +b{\pi }^r+\cdots$ for $r\ge 2$, and its $N$th iterate would have coefficient $Nb$ at that first nonzero displacement. Characteristic zero contradicts finite order. Thus the chosen primitive character identifies a unique generator.

For another place $w^{\prime }=\sigma w$, conjugation gives $D_{w^{\prime }}=\sigma D_w{\sigma }^{-1}$, preserves its order, and carries ${\tau }_w$ to ${\tau }_{w^{\prime }}$, since $\sigma$ fixes the constants and hence ${\zeta }_e$. This proves that $e_p$ and the branch-cycle conjugacy class are independent of the place and uniformizer. Without the stated roots-of-unity convention, only the cyclic inertia-subgroup conjugacy class is intrinsic: replacing ${\zeta }_e$ by another primitive root replaces the generator by a coprime power, which need not lie in the same conjugacy class.

The derived series of a Lie algebra is defined recursively by ${\mathfrak{g}}^{(0)}=\mathfrak{g}$ and ${\mathfrak{g}}^{(r+1)}=[{\mathfrak{g}}^{(r)},{\mathfrak{g}}^{(r)}]$, where the bracket denotes the linear span of all brackets of pairs of elements. A solvable Lie algebra is one whose derived series of a Lie algebra terminates at zero:The solvable radical $\mathrm{rad}\mathfrak{g}$ of a finite-dimensional Lie algebra is its largest solvable ideal of a Lie algebra. A semisimple Lie algebra has no nonzero solvable ideals of a Lie algebra, equivalentlyThese are definitions; identifying semisimplicity with a direct sum of simple ideals additionally uses the structure theorem in characteristic zero. The calculations and representation theorems below are over $\mathbb{C}$, as in the complex representation setting of this paper.

The proof of the linearity of a uniform pro-p group has three distinct steps: pass to a finite-dimensional Lie algebra, integrate a faithful representation on a sufficiently small open subgroup, and extend it to the entire group. Working first on a small subgroup is essential because a faithful Lie algebra representation need not send the original integral lattice into the convergence domain of the matrix exponential.

Let $G$ be a uniform pro-p group and let $L_G$ be its intrinsic powerful Lie lattice. Thenis a finite-dimensional Lie algebra over a field of characteristic zero. The Ado theorem, the classical Lie-theoretic ingredient in this proof, supplies a faithful finite-dimensional Lie algebra representationOne must use a faithful representation rather than just the Adjoint representation, which annihilates the center of a Lie algebra.

Choose a p-adic lattice $\Lambda \subset V$. Since $L_G$ is finitely generated as a module, the entries of the representing matrices of a module basis have bounded denominators. Hence for sufficiently large $k$,For odd $p$, the matrix exponential and p-adic logarithm converge and are inverse between $p\mathrm{End}(\Lambda )$ and $1+p\mathrm{End}(\Lambda )$. Put $H=G^{p^k}$; under the uniform pro-p group and powerful Lie lattice correspondence it is the BCH group on $p^k{L}_G$. DefineHere ${\exp }_G$ denotes the inverse of the intrinsic logarithm identification, not an assumed matrix representation of $G$. The Lie algebra homomorphism $\rho$ respects the convergent Baker--Campbell--Hausdorff formula, soThus $\tau$ is a continuous group homomorphism. If $\tau ({\exp }_{G}X)=1$, taking the matrix logarithm gives $\rho (X)=0$, whence $X=0$. Therefore $\tau$ is faithful and all its matrices preserve $\Lambda$. This proves linearity of the open normal subgroup $H$.

To finish, use the extension of a faithful representation from an open normal subgroup. Choose representatives $t_1=1,t_2,\dots ,t_r$ of the finite left cosets $G/H$, where $r=[G:H]$. On the finite free module $W={⨁}_{j=1}^r\Lambda$, define the action of $g\in G$ by writing uniquelyand mapping the $j$-th block to the $i$-th block by $\tau (h_{ij})$. If $g^{\prime }{t}_j=t_{\ell }{h}^{\prime }$ and $g{t}_{\ell }=t_{i}h$, then $g{g}^{\prime }{t}_j=t_{i}h{h}^{\prime }$; this proves the representation identity because $\tau (h{h}^{\prime })=\tau (h)\tau (h^{\prime })$. Each block matrix is invertible over ${\mathbb{Z}}_p$, and the map is continuous because the finite coset permutation is locally constant and the remaining matrix entries are obtained continuously from $\tau$.

If $g\notin H$, its permutation of $G/H$ moves the identity coset, so its block representation is not the identity. If $g\in H$ and its block representation is the identity, the block corresponding to $t_1=1$ is $\tau (g)=1$, and faithfulness of $\tau$ gives $g=1$. Thus this induced representation is faithful. Choosing a basis of $W$ provesThe continuous injection is a topological embedding, since $G$ is compact and the target is Hausdorff. The trivial group is immediate. For uniform groups at $p=2$, the same proof uses the usual stronger powerfulness condition and the convergence ball $4\mathrm{End}(\Lambda )$ instead; the preceding questions concern odd $p$.

The homotopy fibre is the homotopy pullback of the path fibration over the target sphere along $f$. The Sullivan fibre-model theorem computes it by taking a relative acyclic model of that path fibration and tensoring it over the target model with the source model. Here the construction can be written out and checked directly.

Adjoin $a$ of degree $2n-1$ and $b$ of degree $4n-2$ to the sphere model, withThis is a differential, because $D(v-au)=u^2-u^2=0$. It is an acyclic relative Sullivan algebra: putting $v^{\prime }=v-au$ makes its two pairs $Da=u$ and $Db=v^{\prime }$. The first is a polynomial generator killed by an odd generator; the second is an odd generator killed by an even polynomial generator. Each pair has cohomology only $\mathbb{Q}$, the second by the ordinary polynomial antiderivative calculation in characteristic zero. Hence this models the contractible path space, and its augmentation models the chosen basepoint.

Tensoring with the projective-space model via the map in part (i) givesThe last differential is necessary: using only $Db=f^{\ast }v$ would give $D^{2}b=x^{2n}\ne 0$. Make the invertible generator change $z=y-xa$, of degree $2n+1$. Since $x$ is even,For $n\ge 2$, all nonzero differentials are decomposable and the generator ordering $x,a,z,b$ is Sullivan. Thus no further minimization is needed:For example at $n=2$ this is $(\Lambda (x_2,a_3,z_5,b_6),Da=x^2,Db=xz)$.

The indecomposable generators of a simply connected finite-type minimal model are dual to its rational homotopy groups. Therefore, for $n\ge 2$,These four degrees are distinct. The exact homotopy sequence provides an independent check: ${\mathbb{CP}}^n$ has rational homotopy only in degrees $2,2n+1$, while $S^{2n}$ has it in degrees $2n,4n-1$. For $n\ge 2$ these source and target degrees do not overlap, so the source classes survive in the fibre and the target classes contribute their one-degree-lower connecting classes, giving the same four groups.

At $n=1$, the displayed relative fibre algebra is not minimal: $Da=x$ and $Db=z$ are linear pairs. Cancelling both acyclic pairs leaves $\mathbb{Q}$. AccordinglyThis agrees with ${\mathbb{CP}}^1\cong S^2$ and the degree-one map being a homotopy equivalence. The case distinction is required before reading homotopy groups from the generators.

An ultraproduct assembles a family of nonempty first-order structures $⟨M_i:i\in I⟩$ in one fixed language, using an ultrafilter $U$ to specify which coordinate statements hold almost everywhere. Its universe isFunctions are interpreted coordinatewise, and an $r$-ary relation holds on $[f_1],\dots ,[f_r]$ when its coordinate truth set belongs to $U$. Changing representatives changes such a truth set only off a $U$-large set, so these definitions are well defined. A principal ultrafilter simply recovers its distinguished factor; a nonprincipal ultrafilter can produce new elements and structures.

The fundamental result is the Łoś theorem:Prove it by structural induction on first-order formulas. Atomic formulas are the definition. Intersections handle conjunction, while exactly one of a set and its complement belongs to an ultrafilter, handling negation. For the existential step, if the coordinate existential truth set is in $U$, use choice to select witnesses there and arbitrary values elsewhere; their product function is an ultraproduct witness. Conversely a product witness makes the coordinate existential truth set contain a $U$-large set. Other connectives and universal quantification follow.

In an ultrapower, all factors are one structure $M$. The diagonal map $a\mapsto [\text{constant }a]$ is an elementary embedding by the Łoś theorem. For a nonprincipal $U$ on $\omega$, the class of $n\mapsto n$ in an ultrapower of the standard natural numbers is larger than every constant natural number, since each tail belongs to $U$. This gives a nonstandard model of Peano arithmetic. Similarly, an ultraproduct of finite fields ${\mathbb{F}}_{p_i}$ with distinct primes tending to infinity has characteristic zero: for every fixed positive integer $m$, the equation $m1=0$ fails in all sufficiently late factors. An infinite structure can therefore arise from finite factors.

Ultraproducts prove the compactness theorem. If each finite subset $\Delta$ of a theory $T$ has a model $M_{\Delta }$, index these models by $[T{]}^{<\omega }$. The cones $\{\Delta :\phi \in \Delta \}$ have the finite intersection property. Extend their filter to an ultrafilter using the ultrafilter lemma. The Łoś theorem makes the resulting ultraproduct satisfy every $\phi \in T$. This construction also gives elementary extensions realizing prescribed finitely satisfiable types.

A useful stronger property is countable saturation of an ultraproduct over omega. Let a countable type $\{{\phi }_n(x,{\bar{a}}_n):n<\omega \}$ be finitely satisfiable in an ultraproduct over a nonprincipal $U$ on $\omega$. Choose coordinate representatives of its parameters. Let $E_n$ be the coordinates where the first $n$ formulas have a simultaneous witness; then $E_n\in U$, the $E_n$ decrease, and we can replace $E_n$ by $E_n\cap [n,\infty )$. At coordinate $i$, choose a witness to the first $m(i)$ formulas, where $m(i)$ is the largest $n\le i$ with $i\in E_n$, taking an arbitrary value if there is none. For each fixed $n$, the set where $m(i)\ge n$ is $U$-large. Its class realizes the entire type. This proof uses countable incompleteness of $U$, not arbitrary completeness or automatic saturation at larger cardinals.

Finally, complete ultrafilters connect ultraproducts to large cardinals. A countably complete ultrafilter prevents an externally infinite descending membership chain in an ultrapower of a well-founded universe: intersect the countably many coordinate truth sets and obtain an impossible descending chain at one coordinate. A kappa-complete nonprincipal ultrafilter on $\kappa$ consequently supplies the ultrapower embedding associated with a measurable cardinal. Thus the same construction supports both elementary compactness arguments and large-cardinal embeddings, with the completeness hypothesis deciding which well-foundedness and closure properties survive.

Let $\lambda =({\lambda }_1\ge {\lambda }_2\ge \cdots )$ be a partition of an integer $n$. Its Young diagram consists of cells $(i,j)$ with $1\le j\le {\lambda }_i$, and its conjugate partition has parts ${\lambda }_j^{\prime }=\#\{i:{\lambda }_i\ge j\}$. A numbered Young tableau fills these cells bijectively with $1,\dots ,n$. Its row and column stabilizers $R_t,C_t$ permute entries within each row and each column respectively. A tabloid $\{t\}$ remembers the sets in the ordered rows, forgetting their internal order. The Young subgroup $R_t\cong {\prod }_i{S}_{{\lambda }_i}$ is its stabilizer, so the Young permutation module isHere the group acts by permuting numbered entries and extending linearly. A submodule is an invariant linear subspace. Give the tabloid basis its orthonormal symmetric bilinear form, denoted $⟨,⟩$. This form is nondegenerate and invariant under the group. Define the column antisymmetrizer and polytabloid byThus $S^{\lambda }$ is the Specht module, generated by any one $e_t$, because the group is transitive on numbered Young tableaux. The orthogonal complement $(S^{\lambda }{)}^{\perp }$ consists of vectors pairing to zero with every polytabloid.

The key calculation isTo prove it, first take $v=\{u\}$. If two entries of one column of $t$ occur in one row of $u$, their transposition $s$ fixes $\{u\}$; pairing the terms indexed by $g$ and $gs$ cancels $b_t\{u\}$. This also works in characteristic two, where the paired equal terms have sum zero. Otherwise each row of $u$ contains at most one entry from each column of $t$. The first row has ${\lambda }_1$ entries and there are ${\lambda }_1$ columns, so it uses every column once. Removing those entries, the second row must use exactly the columns of remaining positive height, and induction does the same for all rows. Consequently there is a unique $h\in C_t$ with $\{u\}=\{ht\}$. Uniqueness follows from $R_t\cap C_t=1$. In that caseLinearity proves the identity.

Now let $U\le M^{\lambda }$ be a submodule. If $U$ is not contained in $(S^{\lambda }{)}^{\perp }$, choose $v\in U$ and $t$ with $⟨v,e_t⟩\ne 0$. The displayed identity puts $e_t$ in $U$, so its group translates put all of $S^{\lambda }$ in $U$. This proves the James submodule theorem without a semisimplicity assumption.

Put $R=S^{\lambda }\cap (S^{\lambda }{)}^{\perp }$. If $R\ne S^{\lambda }$, every proper submodule of $S^{\lambda }$ lies in $R$, by the same theorem. Hence $R$ is its unique maximal submodule, and $D^{\lambda }=S^{\lambda }/R$ is irreducible. If $R=S^{\lambda }$, the quotient is zero. This conclusion remains true after every field extension $E/F$: the integral tabloid construction becomes $M_E^{\lambda }=E{\otimes }_F{M}^{\lambda }$, and the radical of the restricted form is the kernel of its Gram matrix, so it commutes with extension of scalars. A nonzero Gram entry stays nonzero in $E$. Applying the theorem over $E$ makes $E{\otimes }_F{D}^{\lambda }$ irreducible. Thus a nonzero quotient is absolutely irreducible.

The precise nonvanishing criterion isA regular partition has fewer than $p$ rows of each positive length. Here is the integral pairing argument behind this condition. Write $m_j$ for the number of rows of length $j$. Expanding two polytabloids into tabloids, simultaneous permutations of equal-length rows partition of an integer all contributions to their pairing into free orbits of size ${\prod }_j{m}_j!$; the signs agree within each orbit because the same row permutation is applied in every relevant column on both sides. Therefore every pairing is divisible by ${\prod }_j{m}_j!$. If some $m_j\ge p$, the entire restricted form is zero modulo $p$. Conversely, reverse the order of the entries in each row of a tableau to obtain $t^{\ast }$. Counting the matching column permutations in the two expansions gives, up to the harmless overall sign,Indeed the $m_j$ equal rows of length $j$ can be matched independently in each of their $j$ positions; after these choices the decreasing row lengths force every other matching. For a regular partition this integer is nonzero modulo $p$, so the form and its radical quotient are nonzero. In characteristic zero it is nonzero for every shape. This supplies the requested justification of the criterion.

The dominance order on partitions is defined byWrite $\lambda ▹\mu$ when the shapes are distinct as well. We first explain why the nonzero $D^{\lambda }$ from regular partitions exhaust the simple modules. The argument also provides the required constraint on composition factors.

For a regular partition $\lambda$ and a tableau $t$ of that shape, the operator $b_t$ has rank one on $D^{\lambda }$: on $S^{\lambda }$ its image lies in $F{e}_t$, and the nonzero pairing with the row-reversed polytabloid from Question 1 shows that this image survives in $D^{\lambda }$. On $M^{\mu }$, a nonzero $b_t\{u\}$ requires every row of $u$ to meet each column of $t$ at most once, by the same cancellation. Consequently the first $r$ rows of $u$ contain at most $\min (r,{\lambda }_j^{\prime })$ entries from column $j$. Summing givesThus nonzero column action forces $\lambda ⊵\mu$. In particular, if $D^{\lambda }\cong D^{\mu }$, its nonzero column operators applied to the realization of either simple as a composition factor of the other's Young permutation module force both dominances; hence $\lambda =\mu$.

Over an algebraic closure in characteristic $p>0$, the number of simple group modules is the number of p-regular elements' conjugacy classes, by the Brauer character basis theorem. Such classes in $S_n$ have cycle lengths not divisible by $p$. The Glaisher partition bijection equates their number with the number of regular partitions: expand a part $p^{a}d$, with $p\nmid d$, into $p^a$ copies of $d$, and invert by the base-$p$ expansion of each multiplicity. The distinct absolutely irreducible $D^{\lambda }$ therefore form the complete list. Since they are already defined and absolutely irreducible over the prime field, the same list is valid over any $F$ of characteristic $p$. In characteristic zero, the analogous count uses all partitions of an integer and all conjugacy classes.

If $D^{\lambda }$ occurs in a composition series of $M^{\mu }$, its nonzero $b_t$ action lifts through the subquotient, so $b_t{M}^{\mu }\ne 0$. The inequality above proves $\lambda ⊵\mu$. When $\lambda =\mu$ is regular, $b_t$ has rank one on $M^{\mu }$ and rank one on $D^{\mu }$. An operator preserving a finite filtration has rank at least the sum of its ranks on the successive quotients: relative to an adapted basis its matrix is block triangular, and nonzero pivot minors of the diagonal blocks combine to a nonzero minor. Therefore $D^{\mu }$ can occur at most once. It occurs at least once as the quotient of $S^{\mu }\subset M^{\mu }$. We have provedFor a nonregular $\mu$, equality is impossible because $D^{\mu }=0$.

Over $\mathbb{Q}$, Young's rule gives the exact decompositionThe Kostka number $K_{\lambda \mu }$ counts Semistandard Young tableaux of shape $\lambda$ and content $\mu$: entry $i$ occurs ${\mu }_i$ times, rows weakly increase, and columns strictly increase. In particular $K_{\mu \mu }=1$ and nonzero multiplicity requires dominance.

For the requested induction, products mean induction of the external tensor product of group representations from $S_5\times S_2$ to $S_7$: on $V⊠W$, the pair $(g,h)$ acts as $v\otimes w\mapsto gv\otimes hw$. One can compute using Young's rule alone. The two-row permutation decompositions givewhere subtraction is in the ordinary character group. Reordering row blocks gives isomorphic permutation modules. Inducing after multiplication by the trivial $S^{(2)}$ gives $[M^{(3,2,2)}]-[M^{(4,2,1)}]$. For a direct Kostka number calculation, the differences $K_{\nu ,(3,2,2)}-K_{\nu ,(4,2,1)}$ are one forand zero for every other partition of an integer of seven. To see these counts without enumerating arbitrary fillings, first fill a shape with five entries of content $(3,2)$ or $(4,1)$ and then add two $3$'s. The added cells must be a horizontal strip. Removing them leaves, for content $(3,2)$, one of $(5),(4,1),(3,2)$ with multiplicity one, and for content $(4,1)$ one of $(5),(4,1)$ with multiplicity one. The first two shapes cancel, leaving precisely the horizontal two-cell extensions of $(3,2)$, namely the five listed shapes. ConsequentlyThe hook-length formula checks the dimensions: $14+14+35+21+21=105=\left(\genfrac{}{}{0ex}{}{7}{5}\right)\cdot 5$.

Since $S^{\mu }\subset M^{\mu }$, Question 2 shows that every composition factor of $S^{\mu }$ is $D^{\lambda }$ for a regular partition $\lambda ⊵\mu$. If $\mu$ is regular, its radical quotient $D^{\mu }$ occurs once: it occurs as a quotient, and its multiplicity is bounded above by the multiplicity one in $M^{\mu }$. All remaining factors have strictly dominating labels. A nonregular row has only strictly dominating regular labels.

Define $d_{\mu \lambda }=[S^{\mu }:D^{\lambda }]$. Order the row labels, all partitions of an integer of $n$, and the column labels, the regular partitions of an integer, by the same total order refining decreasing dominance. ThenThe matrix is generally rectangular. Its submatrix on regular rows is lower unitriangular in this decreasing convention; placing least dominant shapes first would transpose the visual triangular convention, not the indices.

This is the decomposition matrix, not just an abstract composition factor table. The standard polytabloid basis makes $S_{\mathbb{Z}}^{\mu }$ a free invariant integral lattice whose extension to characteristic zero is the ordinary irreducible $S^{\mu }$. Reducing this same lattice modulo $p$ gives exactly the modular $S^{\mu }$. On p-regular elements, reduction of its ordinary character is the sum of the Brauer characters of its composition factors, with coefficients $d_{\mu \lambda }$. These are the defining decomposition numbers; the resulting class in the modular representation ring is independent of the chosen invariant lattice.

For $S_3$, take rows $(3),(2,1),(1,1,1)$ in that order. In characteristic zero, or $p>3$, all three Specht modules are irreducible by Maschke's theorem, andFor $p=2$ the regular labels are $(3),(2,1)$ andIndeed the sign representation equals the trivial representation. The two-dimensional $S^{(2,1)}$ is the sum-zero subspace of the three-point permutation module. Over an algebraic closure, a $3$-cycle has two distinct nontrivial third-root eigenvalues there, and a transposition exchanges their lines. Neither line is invariant under $S_3$, proving absolute irreducibility.

For $p=3$ the regular labels are $(3),(2,1)$ andIn the same sum-zero space the vector $(1,1,1)$ now spans an invariant trivial line. Its one-dimensional quotient is sign: the determinant of the sum-zero representation is sign, and the determinant of the invariant line is trivial. Therefore $S^{(2,1)}$ has one trivial and one sign factor, with sign its simple head $D^{(2,1)}$. The bottom Specht module is itself sign. This proves every entry of the three possible matrices.

Use $[\lambda ]$ for the ordinary irreducible character of $S_{|\lambda |}$. Multiplication of different degrees is the product in the induction ring of symmetric-group characters: it induces an external tensor product of group representations. For $\alpha \subseteq \nu$, the skew Young diagram $\nu /\alpha$ is the set of cells of $\nu$ outside $\alpha$. A filling of it is a semistandard Young tableau when entries weakly increase along rows and strictly increase down columns. Its content $\beta$ means that $i$ occurs ${\beta }_i$ times. Its reading word takes each row right to left and the rows top to bottom. A lattice word has, in every initial segment, at least as many $i$'s as $(i+1)$'s for every $i$.

The Littlewood–Richardson rule sayswhere the Littlewood–Richardson coefficient $c_{\alpha \beta }^{\nu }$ counts such skew semistandard fillings with content $\beta$ and lattice reading word. If $\alpha$ is not contained in $\nu$, the coefficient is zero. Over characteristic zero this is an actual direct-sum multiplicity statement; over other fields the analogous statement describes a Specht filtration, not necessarily a decomposition into simple modules.

For the required coefficient, the five skew cells areTheir columns are all distinct. Content $(3,2)$ uses three $1$'s and two $2$'s. The first reading letter must be $1$ by the lattice word condition, and row monotonicity forces both top cells to be $1$. The remaining three cells, in reading order, can then be filled by $122$, $212$, or $221$. The resulting words $11122$, $11212$, $11221$ all satisfy the lattice inequalities, and exhaust the possibilities. Thus

For the determinantal form of a symmetric-group character, set $[0]=1$ in degree zero and $[r]=0$ for $r<0$. The one-row character $[r]$ is trivial. The one-row specialization of the Littlewood–Richardson rule is the Pieri rule: multiplication by $[r]$ adds a horizontal strip of $r$ cells, since all entries are $1$ and no column can contain two of them.

Here is an explicit determinant argument, so that the desired form is genuinely deduced. Under the Frobenius characteristic map, $[r]$ corresponds to the complete homogeneous symmetric polynomial $h_r$, whose monomials enumerate weakly increasing sequences of $r$ entries. Fix $N$ variables and $l$ rows. Let a directed lattice path run right or up from $A_i=(-i,1)$ to $B_j=({\lambda }_j-j,N)$, and assign weight $x_k$ to a horizontal step at height $k$ and weight one to a vertical step. The sum of its weights is $h_{{\lambda }_j-j+i}$. Expanding the determinant of this path matrix sums signed families connecting the sources to a permutation of the destinations.

Cancel intersecting families as follows. Choose the first intersection in a fixed order of vertices, and then the first pair of paths meeting there. Exchange their tails after that vertex. The total weight is unchanged, while the endpoint permutation changes by a transposition, reversing the sign. Choosing the intersection in this fixed order makes the operation an involution. The surviving families are nonintersecting; the ordered sources and ordered destinations force the identity endpoint pairing. Record the heights of the horizontal steps of path $i$ as row $i$ of a tableau. Rows weakly increase, and nonintersection of successive paths is precisely strict increase down the shared columns. Conversely every semistandard Young tableau gives one surviving path family. Thus the determinant has the semistandard-tableau generating function $s_{\lambda }$.

Young's rule identifies this generating function with the characteristic of $[\lambda ]$: its coefficient at a monomial of weight $\mu$ is $K_{\lambda \mu }$, the multiplicity obtained by pairing with the Young permutation module of type $\mu$. The Frobenius characteristic map is injective and respects induction products. Transposing the path matrix does not change its determinant, soThis is the character version of the Jacobi–Trudi identity.

For the example the matrix isExpanding gives the concise answerAt the identity the six induction-product dimensions are $210,140,105,35,42,21$, respectively; their alternating sum is $21$, agreeing with the hook-length formula for $(3,2,2)$.

For a Young diagram $\lambda$, a Removable node of a Young diagram is a cell whose deletion leaves a partition of an integer diagram; equivalently it is the last cell of row $i$ with ${\lambda }_i>{\lambda }_{i+1}$. Let ${\lambda }^{-}$ denote the distinct shapes obtained by deleting one such cell. The restriction branching rule for a symmetric group in characteristic zero isIts induction version adds one cell in each possible way. Over an arbitrary field, restriction instead has a Specht filtration with these successive quotients, each once; it need not split.