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Let $K=k(z)$, where $z$ is an indeterminate, and form the right Artinian triangular ring that is not left NoetherianMultiplication uses the natural $k$–$K$ bimodule structure on the upper-right entry. With the two diagonal idempotents $e_1,e_2$, the right regular module is $e_{1}T\oplus e_{2}T$. The off-diagonal submodule of $e_{1}T$ is one-dimensional over the bottom-right field $K$, hence simple. Its quotient is one-dimensional over the top-left field $k$, hence simple. Also $e_{2}T$ is a simple right module over $K$. Thus $T_T$ has a composition series of length three, and in particular $T$ is right Artinian.

On the left, multiplication of an off-diagonal matrix givesEvery $k$-subspace $U\subseteq K$ therefore yields a left ideal of off-diagonal matrices with entries in $U$. The subspaces $U_m={\mathrm{span}}_k\{1,z,\dots ,z^m\}$ are strictly increasing because $z$ is transcendental. Their associated left ideals violate the ascending chain condition. Hence $T$ is not left Noetherian, even though it is right Artinian and right Noetherian.

In a commutative ring, if $x^m=0$, then the principal ideal $(x)$ satisfies $(x{)}^m=0$. Thus a semiprime ring has no nonzero nilpotent elements. Conversely, if $I^m=0$ for an ideal $I$, then every $x\in I$ satisfies $x^m=0$; absence of nonzero nilpotent elements forces $I=0$. Hence commutative semiprimeness is exactly reducedness.

For a prime ideal $P$, let $S=R\setminus P$. Every ideal $I$ of $R_P=S^{-1}R$ is the extension of its contraction $I^c$: if $r/s\in I$, multiply by the unit $s/1$ to get $r/1\in I$, so $r\in I^c$. If $R$ is Noetherian, write $I^c=(r_1,\dots ,r_m)$; their images generate $I$. This proves that $R_P$ is Noetherian.

The extended ideal $P{R}_P$ is proper. Indeed $1=p/s$ would imply $u(s-p)=0$ for some $u\notin P$, and then $us=up\in P$, contradicting primeness and $u,s\notin P$. A fraction with numerator outside $P$ is invertible, with inverse $s/r$. A fraction with numerator in $P$ lies in the proper ideal $P{R}_P$ and cannot be invertible. Thus the nonunits are precisely $P{R}_P$, making it the unique maximal ideal. ThereforeTo prove reducedness is detected by prime localizations, let $x$ be a nilpotent element of $R$. Its image in every $R_Q$ is zero because these localizations are semiprime, hence reduced. If $x\ne 0$, its annihilator $\mathrm{Ann}(x)$ is a proper ideal; choose a maximal ideal $Q$ containing it. The zero criterion for localization says $x/1=0$ in $R_Q$ only if $sx=0$ for some $s\notin Q$, which would put $s$ in $\mathrm{Ann}(x)\subseteq Q$, a contradiction. Hence $x=0$, and $R$ is semiprime. This argument does not need Noetherianity.

Primeness is not detected in the same way. Take $R=k\times k$. Its two prime ideals are $P_1=0\times k$ and $P_2=k\times 0$, and its corresponding localizations are both isomorphic to the field $k$. To see these are all the prime ideals, the orthogonal idempotents $e_1,e_2$ have product zero, so a prime ideal contains one of them and is then the kernel of the other projection. Both localizations are prime rings, but $R$ is not prime: $e_1,e_2\ne 0$ and $e_1{e}_2=0$.

The Bernstein inequality for Weyl algebra modules requires characteristic zero: for a nonzero finitely generated left or right module $M$ over $A_n(k)$,Equivalently, its Bernstein growth dimension of a Weyl algebra module is at least $n$. Without the characteristic-zero hypothesis this is false, as the module in part (c) has dimension zero in the growth sense. The proof below works over any characteristic-zero field, not just an algebraically closed one.

For a left module, choose a nonzero finite-dimensional generating subspace $M_0$ and put $M_j=F_j{A}_n{M}_0$, where $F_j$ is the Bernstein filtration. Ordered monomials giveMultiplication defines the linear action mapWe prove this map injective by induction on $j$, which is the essential faithful finite-step action of a Weyl algebra argument. For $j=0$, an element is a scalar, and a scalar killing $M_0\ne 0$ is zero. Suppose $a\in F_j$ kills $M_j$. For any generator $z$ and $m\in M_{j-1}$, both $azm$ and $zam$ vanish because $zm\in M_j$ and $m\in M_j$. Thus $[a,z]\in F_{j-1}$ kills $M_{j-1}$, so induction gives $[a,z]=0$ for every generator. By the explicit commutator formulas in part (a), commuting with all $D_i$ forces all positive $x_i$ exponents of $a$ to vanish, and commuting with all $x_i$ forces all positive $D_i$ exponents to vanish. Characteristic zero is used to cancel the nonzero integer exponents. Hence $a$ is scalar and, because it kills $M_0$, is zero. This proves injectivity.

Taking dimensions yieldsThe left side is asymptotic to $j^{2n}/(2n)!$. Consequently ${\dim }_k{M}_{2j}\ge C{j}^n$ for all sufficiently large $j$ and some $C>0$. Taking logarithms and the limsup in the definition of Gelfand–Kirillov dimension of a module proves the claimed lower bound. For a good Bernstein filtration, the associated graded module is finite over $k[x_1,\dots ,x_n,{\xi }_1,\dots ,{\xi }_n]$; its cumulative Hilbert function is eventually polynomial, and the same comparison says its degree is at least $n$.

For right modules, use the opposite ring isomorphism $A_n(k)\cong A_n(k{)}^{\mathrm{op}}$ sending $x_i$ to $x_i$ and $D_i$ to $-D_i$. It respects the Bernstein filtration, so the identical estimate applies. The bound is sharp: the polynomial module $k[x_1,\dots ,x_n]$, generated by $1$, has $M_j$ equal to the polynomials of degree at most $j$, of dimension $\left(\genfrac{}{}{0ex}{}{j+n}{n}\right)$. Its Gelfand–Kirillov dimension of a module is exactly $n$.

In SI units, the source-free Maxwell equations in vacuum areUse complex amplitudes with phase $e^{i(\omega t-k\cdot x)}$ and take real parts at the end. The electric divergence equation gives $k\cdot E_0=0$. Faraday's equation gives $k\times E_0=\omega B_0$, soThe magnetic divergence equation follows because $k\cdot (k\times E_0)=0$. Ampere's equation becomes $k\times B_0=-{\mu }_0{ϵ}_0\omega E_0$. Since $k\times (k\times E_0)=-|k{|}^2{E}_0$, a nonzero wave satisfiesThus both fields are transverse, mutually perpendicular for linear polarization, and obey the vacuum plane-wave dispersion relation.

For the interface let $n=e_z$ point from the minus side to the plus side and write $[E]=E_{+}-E_{-}$, similarly for $B$. Applying the integral Maxwell equations to a thin pillbox and a thin loop gives the electromagnetic boundary conditionsHere $j$ is the tangential surface current. The jump notation fixes the signs, including the direction of the normal.

For the perfect conductor, take the time-varying interior fields to vanish. Choose positive $k$ with $\omega =ck$. The incident electric field has complex form $E_0{e}_x{e}^{i\omega t+ikz}$. To cancel its tangential value at the surface, use the reflected field $-E_0{e}_x{e}^{i\omega t-ikz}$, whose wavevector points upward. The total vacuum fields are thereforeThe sign of each magnetic contribution follows from $B_0=k\times E_0/\omega$. At $z=0$, the electric field is zero and the magnetic field is tangential. The jump conditions yieldFor a real incident amplitude this is $j=(2E_0/{\mu }_{0}c)\cos (\omega t)e_x$. It supplies the reflected wave while satisfying the normal-field and tangential-electric conditions. Any independently imposed static interior field is separate from this oscillatory response.

Choose the coordinate vectors $b=e_i$ and $c=e_j$ in the assumed bilinear identity. It gives $T_{ij}=0$ for each pair of indices, hence the tensor is zero.

Let $\Omega$ be the enclosed region. Apply the divergence theorem to the vector field $(b\cdot x)c$. Since $b,c$ are constant,ThereforeWriting the left side as $b_i{c}_j{\int }_S{x}_i{n}_{j}dS$ and using the initial tensor argument gives the boundary integral of position and normal componentsContract with the Kronecker delta to obtain ${\int }_{S}x\cdot ndS=3V$. Contract with the Levi-Civita symbol to obtainThus the requested integrals are $3V$ and the zero vector, respectively. These applications require the usual sufficiently regular closed boundary and outward-normal orientation of the divergence theorem.

Work in the classical setting of varieties over an algebraically closed field $k$. A prevariety is separated when its diagonal morphism ${\Delta }_X:X\to X{\times }_{k}X$ is a closed immersion; for prevarieties the diagonal is already an immersion, so this is equivalent to its image being closed. A complete variety has the following closed-projection property: for every variety $Y$, projection $X{\times }_{k}Y\to Y$ maps closed subsets to closed subsets. This is universal closedness; since a variety is separated and of finite type, it is equivalent to the structural map being proper.

First consider projective space. In two sets of homogeneous coordinates its diagonal is defined byThese equations hold exactly when the two nonzero coordinate vectors are proportional. They are homogeneous in each coordinate set, so they define a closed subset of ${\mathbb{P}}^n\times {\mathbb{P}}^n$. On a common affine chart, the diagonal map is the usual closed immersion defined by equality of the affine coordinates. Thus projective space is separated. If $X\hookrightarrow {\mathbb{P}}^n$ is a projective embedding, its diagonal is the intersection of that closed diagonal with $X\times X$. Consequently every projective variety is separated.

Here is an algebraic proof of completeness, rather than an appeal to projective compactness in an analytic topology. It suffices to prove closedness of projection from projective space over an affine chart $Y$, with coordinate ring $A=k[Y]$. A closed subset $Z\subseteq {\mathbb{P}}^n\times Y$ is cut out by a homogeneous ideal $I\subseteq A[T_0,\dots ,T_n]$. Put $B=A[T_0,\dots ,T_n]/I$ and write $B_d$ for its degree-$d$ part. Each $B_d$ is a finitely generated $A$-module, because it is a quotient of the free module on the degree-$d$ monomials.

Suppose the fibre over $y\in Y$ is empty. The specialized homogeneous equations then have no nonzero common zero. By the Hilbert Nullstellensatz, some power of each $T_i$ belongs to the specialized ideal. If those powers are $T_i^{r_i}$, then every monomial of degree $d=1+{\sum }_i(r_i-1)$ belongs to that ideal. HenceThe Nakayama lemma gives $(B_d{)}_{{\mathfrak{m}}_y}=0$. Since $B_d$ is finitely generated, there is $a\in A\setminus {\mathfrak{m}}_y$ with $(B_d{)}_a=0$: choose generators and multiply finitely many elements annihilating their localizations. On the principal open subset $D(a)$, all graded pieces of degree at least $d$ vanish too, since the quotient is generated in degree one. Every fibre over $D(a)$ is therefore empty: a projective point would have a nonzero coordinate whose degree-$d$ power could not vanish. Thus the complement of the projection of $Z$ is open, proving the required closed-projection property.

An affine open cover of an arbitrary $Y$ establishes the same result there. Finally, a closed subset of $X\times Y$ is closed in ${\mathbb{P}}^n\times Y$ when $X$ is a projective variety, so its image is closed by the preceding argument. ThereforeOver a non-algebraically-closed field, the corresponding statement concerns schemes or geometric points. Projection of rational points alone need not be closed, so it must not replace the completeness definition.

Half exponents in an odd-order root of unity group mean multiplication by the inverse of two modulo its order. This uniquely defines ${\zeta }_n^{a/2}$ inside $K_n$, and the definitions are compatible as $n$ varies. Put $d_a(z)=z^{-a/2}(1-z^a)$ on these roots. The symmetric cyclotomic unit is $c_n(a,b)=d_a({\zeta }_n)/d_b({\zeta }_n)$. Complex conjugation replaces each $d_a$ by $-d_a$, so the ratio is real. It is a global unit: choosing positive representatives for $a,b$ modulo $p^{n+1}$ gives geometric-series expressions for $(1-{\zeta }_n^a)/(1-{\zeta }_n)$ and its inverse, and similarly for $b$. The remaining factor is a root of unity.

The conjugates of ${\zeta }_n$ over $K_{n-1}$ are $\xi {\zeta }_n$ with $\xi \in {\mu }_p$. Because $p\nmid a$, multiplication by $a$ permutes ${\mu }_p$, andAlso ${\prod }_{\xi \in {\mu }_p}\xi =1$ for odd $p$, so $N_{n,n-1}({\zeta }_n)={\zeta }_{n-1}$. Applying the compatible half power proves $N_{n,n-1}(d_a({\zeta }_n))=d_a({\zeta }_{n-1})$. Dividing the analogous identities for $a,b$ proves

Embed the fields into the p-adic cyclotomic tower. Its interpolating Coleman power series isHere $(b-a)/2\in {\mathbb{Z}}_p$, and binomial expansion defines the half power. Both numerator and denominator vanish to order one at $T=0$, with coefficients $-a$ and $-b$, so cancellation gives an integral unit with $f_{a,b}(0)=a/b$. This applies also to negative integers $a,b$. Evaluation at ${\zeta }_n-1$ gives $c_n(a,b)$; the same root-product calculation shows $\mathcal{N}{f}_{a,b}=f_{a,b}$ for the Coleman norm operator.

For clarity, the normalization of the higher logarithmic derivative isIn the formal coordinate $T=e^z-1$, $D=d/dz$ andOnly the derivative of the logarithm is needed, so no choice of its constant term is involved. Differentiating gives the regular-at-zero expressionUsing the Bernoulli numbers defined by $z/(e^z-1)={\sum }_{r\ge 0}{B}_r{z}^r/r!$, with $B_1=-1/2$, givesThe two poles cancel. Taking $k-1$ further derivatives and evaluating at zero therefore yields all the requested values:In particular ${\delta }_1=0$, ${\delta }_2=(a^2-b^2)/12$, and ${\delta }_4=-(a^4-b^4)/120$. The odd case includes $k=1$; substituting $B_1$ into the even-index formula would incorrectly lose the symmetric normalization. All the displayed values lie in ${\mathbb{Z}}_p$, directly from the integral Coleman power series definition, even when the rational Bernoulli expression has an apparent p-adic denominator. For even $k\ge 2$ the same answer is $-(a^k-b^k)\zeta (1-k)$.

The relevant notion is a root-of-unity Euler system, which contains auxiliary-prime data as well as the p-power tower. Fix a finite set $S$ of primes containing two but not $p$, and let $W_S$ consist of all roots of unity whose orders are prime to every member of $S$. It is a map $\Phi :W_S\to {\overline{\mathbb{Q}}}^{\times }$ satisfying these three axioms:

The first axiom puts $\Phi (\zeta )$ in the maximal real subfield of $\mathbb{Q}(\zeta )$. For $\zeta ={\zeta }_n$, the second axiom is exactly norm compatibility from $K_n^{+}$ to $K_{n-1}^{+}$ for $n\ge 1$. Auxiliary roots with prime orders different from $p$ are essential: mere norm compatibility of a sequence in the tower is not a verification of all three axioms.

The real cyclotomic unit group at a p-power conductor is generated by $-1$ and the symmetric cyclotomic units $c_n(a,b)$ with $p\nmid ab$. Galois conjugates are included, since a conjugate of $c_n(a,b)$ is $c_n(ua,ub)$ for an integer $u$ prime to $p$. Thus a prescribed unit can be expressed aswhere $d_a(z)=z^{-a/2}-z^{a/2}$, with half exponents interpreted in the odd-order root group. Choose $S$ to contain two and every prime divisor of the $a_j$. For $z\ne 1$ in $W_S$ putEach factor at $z\ne 1$ is nonzero because raising $z$ to $a_j$ preserves its order. At $z=1$ each $d_{a_j}$ has a simple zero with first coefficient $-a_j$; the total exponent zero cancels the zeros and signs, giving the stated removable value. Consequently $\Phi ({\zeta }_n)=\epsilon$.

We verify the axioms individually. First, $d_a(z^{-1})=-d_a(z)$; the product's sign is $(-1{)}^{\sum e_j}=1$, so $\Phi$ is even. The unique half-root operation commutes with every field automorphism, all exponents are integers and $\eta$ is rational. Hence $\Phi (\sigma z)=\sigma \Phi (z)$, including at $z=1$.

Second, $q\notin S$ is odd and does not divide any $a_j$. For a variable $x$, choose half powers consistently and useHere $\rho \mapsto {\rho }^a$ permutes ${\mu }_q$, and the product of its half powers is one. Raising these identities to $e_j$ and multiplying gives the distribution identity, since ${\eta }^q=\eta$. If $x=1$ or $x^q=1$, take the removable values after cancellation; the rational-function identity remains valid there. Thus the distribution axiom holds also at the special roots where the uncancelled factors vanish.

Third, fix $q\notin S$, a root $z$ of order prime to $q$, and a prime $\mathfrak{q}$ above $q$. Every $\rho \in {\mu }_q$ reduces to one modulo $\mathfrak{q}$. If $z\ne 1$, its reduction retains its exact order: prime-to-q roots have distinct reductions because their order polynomial is separable in characteristic $q$. Therefore $1-z^{a_j}$ is a local unit, and all the factors and inverse factors are regular at that reduction. Reduction of the expression for $\Phi (\rho z)$ then gives $\Phi (z)$, since the half roots reduce compatibly. This proves the congruence and local integrality for nontrivial $z$.

At $z=1$, write $\rho =1+T$. The cancelled expressionis a unit formal power series over ${\mathbb{Z}}_q$, because every $a_j$ and two are q-adic units. Its constant term is $\eta {\prod }_j{a}_j^{e_j}$. Evaluating at $T=\rho -1$ proves the remaining congruence modulo every prime over $q$. This also covers $\rho =1$. Hence every real cyclotomic unit at any layer is a value of an Cyclotomic Euler system satisfying all three axioms.

For completeness, at a nontrivial root $z$ every ratio $d_a(z)/d_b(z)$ is a global unit. After removing a root-of-unity factor, both it and its inverse are geometric sums because $a,b$ are invertible modulo the order of $z$. Thus the values used at nontrivial roots really are units, not just nonzero algebraic numbers. Values at one may be rational S-units, which is compatible with the definition.

The auxiliary norm factor can also be read off explicitly. If $z$ has order $m>2$ and $q\nmid m$, then for $\rho \ne 1$ of order $q$,Here the arithmetic Frobenius sends $z$ to $z^q$. This convention makes the sign of the Euler factor explicit, while the case of a repeated prime $p$ gives the exact tower norm relation already proved in Question 2.

Write $F_n=K_n^{+}$, $F_{\infty }={\bigcup }_n{F}_n$, $G=\mathrm{Gal}(F_{\infty }/\mathbb{Q})={\Delta }^{+}\times \Gamma$, where $|{\Delta }^{+}|=(p-1)/2$ and $\Gamma \simeq {\mathbb{Z}}_p$. Set $A={\mathbb{Z}}_p[[G]]$. Choosing a topological generator $\gamma$ of $\Gamma$ identifies its Iwasawa algebra with $\Lambda ={\mathbb{Z}}_p[[S]]$, $S=\gamma -1$; this group variable is distinct from the local-unit interpolation variable $T$. The Teichmüller character decomposes $A$ into a product of copies of $\Lambda$, one for each character of ${\Delta }^{+}$, equivalently each even Teichmüller character of $\mathrm{Gal}(K_0/\mathbb{Q})$. All characteristic-ideal statements below are componentwise in this product.

Let $U_n^1$ be the local principal units at the unique prime above $p$. Let $E_n^1$ and $C_n^1$ be respectively the p-adic closures of global units and cyclotomic units inside $U_n^1$, and take their inverse limits under the field norms. Define $X_{\infty }$ as the Galois group of the maximal abelian pro-p extension of $F_{\infty }$ unramified away from $p$, and $Y_{\infty }$ as that of the maximal everywhere unramified extension. This is the real tower: its prime-ramified module is torsion, by the Leopoldt theorem for abelian number fields. It must not be confused with the positive-rank prime-ramified module of the full complex cyclotomic tower. The unramified Iwasawa torsion theorem gives torsion for $Y_{\infty }$.

A convenient formulation of the Iwasawa main conjecture uses the p-adic zeta pseudomeasure ${\zeta }_p$ on $G$. With the sign conventionits product with the augmentation ideal $I(G)$ is integral and principal. The pole on the trivial-character component is cancelled by that ideal; simply writing an integral quotient by ${\zeta }_p$ would be incorrect. The main assertion in this convention isWe explain the local calculation, the remaining global obstruction, and how the cyclotomic Euler system removes it.

The local calculation is the Iwasawa theorem on local cyclotomic units:One can see why this theorem has exactly this analytic term. For a norm-fixed Coleman power series $f$, its corrected logarithm isThe ratio lies in $1+pR$, so the expression is integral. The norm identity puts its Amice transform in the kernel of the Coleman trace operator, hence gives a measure on ${\mathbb{Z}}_p^{\times }$. Its kth moment is $(1-p^{k-1}){\delta }_k(f)$. The full local exact sequence has p-power roots of unity at its kernel and cokernel. Passing to the real, or even, part removes both, since complex conjugation acts as minus one on their Tate module and $p$ is odd. The corrected logarithm therefore identifies $U_{\infty }^1$ with $A$.

Question 2 now shows that the image of $c(a,b)$ is $([b]-[a]){\zeta }_p$: its even moments are $(1-p^{k-1})(a^k-b^k)B_k/k$. A choice of integer $e$ generating ${\mathbb{Z}}_p^{\times }$ topologically gives a norm-compatible generator $c(e,1)$, after multiplication by its constant Teichmuller lift to make it principal. That constant has zero logarithmic derivatives. Its conjugates generate the closed cyclotomic-unit module, and $[e]-[1]$ generates $I(G)$. This yields the displayed local quotient and its characteristic ideal. The generation statement is an important ingredient; the mere calculation of a few moments would not prove the theorem.

The class-field unit sequence gives the global comparisonMultiplicativity of characteristic ideals in exact sequences givesThus the local theorem proves the main conjecture precisely when one proves $\mathrm{char}(B)=\mathrm{char}(Y_{\infty })$. These terms measure the global-unit and ideal-class obstructions. Discarding them would silently impose a much stronger arithmetic hypothesis.

The first global step is the Euler-system divisibility for real cyclotomic class modules: a generator $f_Y$ of $\mathrm{char}(Y_{\infty })$ divides a generator $f_B$ of $\mathrm{char}(B)$ on every character component. Here is the mechanism. Choose auxiliary primes $q$ splitting completely in a finite layer and satisfying $q\equiv 1(\mathrm{mod}{p}^m)$. For a generator ${\sigma }_q$ of its auxiliary cyclic group, putApply products of these operators to the unit Cyclotomic Euler system of Question 3. The norm axiom and this group-ring identity make the derived elements invariant modulo $p^m$-th powers, so Kummer descent gives classes in $F_n^{\times }/(F_n^{\times }{)}^{p^m}$. The congruence axiom identifies the valuation at each newly introduced prime with a residue symbol of the preceding class; valuations away from the chosen auxiliary primes vanish modulo $p^m$. The Chebotarev density theorem supplies primes in prescribed ideal classes with the required residue-symbol behavior. Adding these primes successively forces the elementary divisors of the class group to divide the available cyclotomic-unit index. Taking the norm limit yields $f_Y\mid f_B$. Finite control defects do not survive localization at height-one primes; using two coprime annihilators of such a finite defect also shows that no hidden p-power factor is left. This is the substantive global argument furnished by the Cyclotomic Euler system, not a consequence of Iwasawa's local theorem alone.

For the reverse comparison use the cyclotomic unit index formula and norm descent, keeping invariants as well as coinvariants. At the base field put $A_0$ for its p-primary ideal class group, and let $N_{\infty }(E_0^1)$ be the universal norm of units in a Zp-extension. Class-field and unit descent, using total ramification and the principal prime over $p$, giveThese are the required norm-defect control statements. The cyclotomic unit index formula, with the p-adic closure justified by Leopoldt, gives $\#A_0=\#(E_0^1/C_0^1)$. Also $B^{\Gamma }=0$: its coinvariants are finite by the displayed control, so its invariants are finite by the Iwasawa module structure theorem, whereas $B$ is a submodule of the cyclic local quotient $H$, which has no nonzero finite submodule. One must retain $Y_{\infty }^{\Gamma }$; dropping it would incorrectly identify all global units with universal norms.

Define the finite Euler characteristic of a one-variable Iwasawa module by ${\chi }_{\Gamma }(M)=\#M_{\Gamma }/\#M^{\Gamma }$. The preceding identities now give an exact cancellation:For a torsion $\Lambda$-module with finite invariant and coinvariant groups, the structure theorem gives ${\chi }_{\Gamma }(M)=|f_M(0){|}_p^{-1}$. Write $f_{B,\chi }=f_{Y,\chi }{h}_{\chi }$ using the Euler-system divisibility. After forgetting ${\Delta }^{+}$, the characteristic series are the products of their character-component series. Equality of these Euler characteristic of a one-variable Iwasawa module values givesEvery summand is nonnegative, since $h_{\chi }\in {\mathbb{Z}}_p[[S]]$. Each is therefore zero. A formal power series with p-adic-unit constant term is a unit, so every $h_{\chi }$ is a unit. Hence $\mathrm{char}(B)=\mathrm{char}(Y_{\infty })$, and cancellation in the class-field exact sequence completes the boxed main conjecture.

The familiar reflected odd class-group formulation follows from Kummer reflection in Iwasawa theory, with the inversion of the group variable and the Tate twist both included. For example, for nontrivial even $\chi$, if $f_{\chi }((1+p{)}^s-1)=L_p(\chi ,s)$, then the p-ramified series is $g_{\chi }(S)=f_{\chi }((1+p)(1+S{)}^{-1}-1)$, and $g_{\chi }((1+p{)}^{1-s}-1)=L_p(\chi ,s)$. The proof above includes the exceptional trivial component through augmentation regularization. It assumes neither a vanishing real class group nor Vandiver's conjecture: the Euler-system divisibility and the norm-defect cancellation replace that extra hypothesis.

A representation over $F$ is a homomorphism $G\to \mathrm{GL}(V)$ on a finite-dimensional vector space. It is irreducible if it has no nonzero proper invariant subspace, and has absolute irreducibility of a group representation if it remains irreducible after every field extension. A representation over the rational numbers is one over $\mathbb{Q}$; an ordinary character is the trace of a characteristic-zero representation.

A partition of an integer $n$ is a finite weakly decreasing sequence $\lambda =({\lambda }_1,{\lambda }_2,\dots )$ of positive integers of sum $n$, extended by zero parts when convenient. Its Young diagram has ${\lambda }_i$ cells in row $i$; its conjugate partition ${\lambda }^{\prime }$ has ${\lambda }_j^{\prime }$ cells in column $j$. Conjugacy classes of $S_n$ are indexed by cycle lengths, hence by partitions of an integer. The number of ordinary irreducible characters equals the number of conjugacy classes; we now construct that many mutually distinct representations over the rational numbers.

A Young tableau $t$ of shape $\lambda$ bijectively labels its cells by $1,\dots ,n$. Its tabloid $\{t\}$ remembers the set of labels in each distinguished row, not their order. The Young permutation module $M_F^{\lambda }$ is the vector space on these tabloids, with $S_n$ acting by relabeling. Its tabloid bilinear form makes the tabloid basis orthonormal. Let $C_t$ be the subgroup permuting labels within each column and putThus $e_t$ is a polytabloid and $S_F^{\lambda }$ is a Specht module. Relabeling takes $e_t$ to $e_{gt}$, so any polytabloid generates the module. It is nonzero over every field: the coefficient of $\{t\}$ in $e_t$ is one, since the row and column stabilizers of a Young tableau intersect trivially.

Here is the elementary identity behind irreducibility. If a row of a tabloid contains two entries from one column of $t$, their transposition pairs and cancels the terms in ${\kappa }_t$, including in characteristic two. Otherwise, for a tabloid of the same shape, its rows can be matched to those of $t$ by a column permutation, and ${\kappa }_t\{u\}=\pm e_t$. Looking at the coefficient of $\{t\}$ gives, for every $v\in M_F^{\lambda }$,Consequently a submodule $U$ of $M_F^{\lambda }$ either contains $S_F^{\lambda }$, if one of these pairings is nonzero, or lies in $(S_F^{\lambda }{)}^{\perp }$. This proves the James submodule theorem used below.

Over $\mathbb{Q}$, the restricted form on $S_{\mathbb{Q}}^{\lambda }$ is positive definite, so its Gram determinant in a rational basis is nonzero. Extending scalars to any characteristic-zero field preserves that nonzero determinant and hence keeps the restricted form a nondegenerate bilinear form. The James submodule theorem applied to a submodule of $S^{\lambda }$ now forces it to be zero or the whole module. Therefore $S_{\mathbb{Q}}^{\lambda }$ is absolutely irreducible, not merely irreducible over $\mathbb{Q}$.

To distinguish shapes, the cancellation argument above works for a $\mu$-tabloid too. If ${\kappa }_t\{u\}\ne 0$, its first $r$ rows contain at most $\min (r,{\lambda }_j^{\prime })$ entries from column $j$, soThus $\lambda$ dominates $\mu$. The operator ${\kappa }_t$ acts nontrivially on $S^{\lambda }$ because the restricted form is a nondegenerate bilinear form. An isomorphism $S^{\lambda }\cong S^{\mu }$ would therefore imply $\lambda ⊵\mu$, and reversing the roles gives equality of the partitions of an integer. Counting conjugacy classes proves exhaustion. All matrices in a rational polytabloid basis have rational entries, giving the requested representations over the rational numbers.

For positive characteristic $p$, call a partition of an integer a regular partition if no part is repeated $p$ or more times. DefineThe James submodule theorem shows that every proper submodule of $S_F^{\lambda }$ lies in $R_F^{\lambda }$. Whenever the restricted form is nonzero, $R_F^{\lambda }$ is proper and its quotient is simple.

We verify exactly when this happens. Let $m_j$ be the number of rows of length $j$. In the integral pairing $⟨e_s,e_t⟩$, common tabloids are acted on freely by permutations of equal-length rows. Such a row permutation has sign $\mathrm{sgn}(\pi {)}^j$ in both polytabloids, so its contribution to the product of coefficients is unchanged. Each orbit has size ${\prod }_j{m}_j!$. Thus that integer divides every pairing. On the other hand, reverse each row of $t$ to obtain $t^{\ast }$. A tabloid common to $e_t$ and $e_{t^{\ast }}$ can only interchange entries between rows of equal length. There are $m_j!$ independent choices in each of their $j$ columns, and the two signs agree. HenceThese two products have exactly the same prime divisors. All pairings vanish modulo $p$ precisely when some $m_j\ge p$. Therefore $D_F^{\lambda }\ne 0$ exactly for $p$-regular $\lambda$.

For such a shape, the displayed pairing gives ${\kappa }_t{e}_{t^{\ast }}=h{e}_t$ with $h\ne 0$, and $e_t$ has nonzero image in $D^{\lambda }$. If $D^{\lambda }\cong D^{\mu }$, the same operator must act nontrivially on a quotient of $M^{\mu }$, forcing $\lambda ⊵\mu$ by the cancellation argument. Interchanging the shapes proves $\lambda =\mu$.

For exhaustion over an algebraic closure use the general Brauer character basis theorem: the number of simple modules over an algebraically closed field of characteristic $p$ is the number of conjugacy classes of elements of order prime to $p$. Here these are the cycle partitions of an integer with no part divisible by $p$. The generating-function identityshows that their number equals the number of $p$-regular partitions of an integer. We have already constructed that many distinct simple modules, so they exhaust all simples. The form, its radical and its quotient commute with field extension, and the same simplicity proof holds after every extension. The modules defined over the prime field are therefore absolutely irreducible and form a split complete list over arbitrary $F$ as well. In summary,For characteristic zero the list is instead all $S_F^{\lambda }$. In positive characteristic a Specht module itself need not be irreducible; the radical quotient is essential.