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The mod-$p$ Iwasawa algebra is the completed group algebrawhere $U$ runs through open normal subgroups and the transition maps come from quotient maps. This is the mod-$p$ algebra, distinguished from the integral Iwasawa algebra ${\mathbb{Z}}_p[[K]]$.

Use the stated complete separated negative filtered ring structure, writing $F_{-n}={\mathfrak{m}}^n$ and $\nu (x)=\max \{n:x\in F_{-n}\}$ for a nonzero element. In the usual odd-prime congruence-group situation,The dimension-three assertion records the three independent entries of a trace-zero matrix; the argument below only needs the given graded polynomial-algebra property. Completeness and separatedness are with respect to this filtration and follow from the inverse-limit topology for the standard filtration. If $x,y\ne 0$, their leading symbols are nonzero, so their product in the graded domain is nonzero. Therefore $\nu (xy)=\nu (x)+\nu (y)$ and $xy\ne 0$. This proves the domain assertion by filtered-graded transfer for complete rings.

For a left ideal $I$, its leading symbols form a homogeneous left ideal $\mathrm{gr}I$ of the polynomial ring. The Hilbert basis theorem gives finitely many homogeneous generators; choose lifts $a_1,\dots ,a_s\in I$. Given $x\in I$, express its leading symbol using these generators, lift the homogeneous coefficients, and subtract the resulting $\sum r_j{a}_j$. The remainder is still in $I$ and has strictly larger valuation. Repeat. At a stage where the remainder has valuation $n$, the coefficient contribution for $a_j$ has valuation at least $n-\nu (a_j)$. Thus each accumulated coefficient is a Cauchy series, converging to an element of ${\Omega }_K$. The remainder tends to zero, and separatedness gives $x={\sum }_j{r}_j{a}_j$. Hence $I$ is algebraically generated by these finitely many lifts; no prior assumption that $I$ is closed was used. This proves $\boxed{{\Omega }_K\text{ is left Noetherian and a domain}}$ under the stated filtration hypothesis. The same proof for right ideals also gives right Noetherianity.

For the classical fractions, a left Noetherian domain satisfies the left Ore condition. Here is the common-multiple argument. If two nonzero principal left ideals $Ra,Rb$ had zero intersection, the ideals $Ra{b}^i$, $i\ge 0$, would form a direct sum: from a finite relation, its first term lies in $Ra\cap Rb$, and then cancellation of the nonzero $b$ proves the assertion inductively. Their partial sums give an infinite ascending chain, impossible in a left Noetherian ring. Thus $Ra\cap Rb\ne 0$, giving common left multiples for every two nonzero denominators. The right-handed argument also applies. Ore localization at all nonzero elements therefore produces the classical division ring of fractions, with expressions $s^{-1}a$ or $b{s}^{-1}$; every nonzero fraction is invertible.

There is a genuine prime restriction behind the auxiliary hypothesis. For $p=2$, the matrix $z=-I$ is a nonidentity element of $K$ of order two. In ${\mathbb{F}}_2[[K]]$, the nonzero element $z-1$ satisfies $(z-1{)}^2=z^2-2z+1=0$. It is nonzero already in a finite quotient separating $z$ from one. Thus the actual $p=2$ algebra is not a domain, and cannot have the supplied separated graded-domain filtration. The intended conclusion applies for odd $p$; at $p=2$ it instead applies to a suitable deeper torsion-free congruence subgroup, such as the kernel modulo four.

A uniform module is a nonzero module in which any two nonzero submodules intersect nontrivially. An essential extension $V\subseteq E$ means that $V$ is an essential submodule of $E$: every nonzero submodule of $E$ meets $V$. Since $R$ is a left Noetherian ring, the finitely generated $M$ is a Noetherian module.

Every nonzero Noetherian module contains a uniform submodule. Otherwise split a nonuniform submodule into two disjoint nonzero submodules, retain one for further splitting and set the other aside. If no uniform submodule is reached, the pieces set aside form an infinite direct sum, whose partial sums contradict the ascending chain condition. Now successively choose uniform submodules disjoint from the current direct sum. The process again must stop by Noetherianity, giving $V={⨁}_{i=1}^r{V}_i\subseteq M$. At termination no nonzero submodule is disjoint from $V$, soThis proves the finite uniform decomposition of a Noetherian module.

An injective hull $E(M)$ is an injective module containing $M$ essentially. It is equivalently a minimal injective extension; we use the general existence theorem for injective hulls. To prove uniqueness, let $E_1,E_2$ be two hulls of $M$. Injectivity of $E_2$ extends the identity on $M$ to $f:E_1\to E_2$. Its kernel is disjoint from $M$ and therefore zero by essentiality. Its image is injective, so its inclusion splits in $E_2$. The complementary summand is disjoint from $M$, hence zero. Thus $f$ is an isomorphism fixing $M$.

A finite direct sum of injective modules is injective, and a finite direct sum of essential inclusions is essential. For the latter assertion, start with a nonzero vector and process its coordinates in turn: if a coordinate lies outside its specified essential submodule, multiply by a ring element taking that coordinate nontrivially into that submodule. Already processed coordinates stay inside their submodules; the chosen coordinate keeps the vector nonzero. Essentiality is also transitive. Consequently both $E(M)$ and ${⨁}_{i}E(V_i)$ are injective hulls of $V$, so uniqueness givesThe essential extension of a uniform module is uniform, since intersecting any two nonzero submodules with the original uniform module already gives a nonzero common intersection.

Write $E=E(V_i)$ and $S={\mathrm{End}}_R(E)$. If $f\in S$ is injective, its image is isomorphic to the injective module $E$, hence is a direct summand. Uniformity leaves no nonzero complementary summand, so $f$ is an automorphism. Therefore the nonunits are exactly the noninjective endomorphisms. Let $J$ be their set. Two such maps have nonzero kernels, whose intersection is nonzero by uniformity; their sum kills this intersection. For a product, composition on the left preserves the original kernel. Composition on the right either has a noninjective right factor, or that factor is an automorphism and transports the original nonzero kernel. Thus $J$ is a proper two-sided ideal. Every element outside $J$ is a unit, so every nonzero residue class is invertible and every proper left or right ideal is contained in $J$. This proves the local endomorphism ring of a uniform injective module assertion:For the whole hull, let $E_i=E(V_i)$. Its endomorphism ring is the ring of finite block matriceswith multiplication given by summing compositions. The diagonal entries have the local rings just proved. Off-diagonal entries are homomorphism groups, not necessarily zero; in particular this ring is not generally the product of its diagonal rings. Repeated isomorphic summands give matrix blocks over their common local endomorphism ring.

For the left regular module of $A_1(\mathbb{C})$, the Bernstein filtration has graded ring $\mathbb{C}[x,\xi ]$, so finite leading-term cancellation proves that $A_1$ is left and right Noetherian and a domain. Let $Q$ be its classical division ring of fractions. It is an essential extension of $A_1$: a nonzero fraction $q=s^{-1}a$ satisfies $sq=a\ne 0$ in $A_1$.

To prove that $Q$ is injective, use the Baer criterion. For a homomorphism $f:I\to Q$ on a nonzero left ideal, fix $a\in I\setminus \{0\}$. For any nonzero $x\in I$, take common left multiples $sx=ta$ with $s,t\ne 0$. Then $sf(x)=tf(a)$, so in the division ringThus right multiplication by $a^{-1}f(a)$ extends $f$ from $I$ to $A_1$. The zero ideal is harmless. Hence $Q$ is injective and $E({}_{A_1}{A}_1)=Q$. For any $A_1$-linear endomorphism $F$ of $Q$, clearing the left denominator in $q$ gives $F(q)=qF(1)$. These are all right multiplications, and their composition reverses the multiplying elements. ThereforeThe opposite ring in the injective hull of an Ore domain computation is essential for left-module conventions.

Schur's lemma says that a nonzero intertwiner between irreducible representations is an isomorphism; over $\mathbb{C}$, every endomorphism of a finite-dimensional irreducible representation is scalar. To prove the Schur lemma, let $T:V\to W$ be a intertwining operator between irreducible representations. Its kernel and image of a linear map are invariant. If $T\ne 0$, irreducibility forces $\ker T=0$ and $\mathrm{im}T=W$. This proves the first assertion over any field. In particular the endomorphisms of an irreducible representation form a division ring.

When $V$ is finite-dimensional over an algebraically closed field, an endomorphism $T$ has an eigenvalue $\lambda$. The endomorphism $T-\lambda I$ has nonzero kernel. By the first assertion it must be zero, so $T=\lambda I$. Consequently, for complex finite-dimensional irreducible representations, the space of intertwining operators has dimension zero for nonisomorphic representations and dimension one for isomorphic representations.

Every finite-dimensional representation of a complex semisimple Lie algebra is completely reducible. We prove the Weyl complete reducibility theorem using the allowed Casimir operator facts, without assuming a splitting in advance. For dual bases $x_i,x^i$ with respect to the Killing form, the Casimir elementis central in the universal enveloping algebra. Hence its Casimir operator $C_M={\sum }_i{\rho }_M(x_i){\rho }_M(x^i)$ commutes with the action on every Lie algebra representation $M$ and is compatible with subrepresentations, quotient representations, and intertwining operators. On the trivial Lie algebra representation it acts by zero. On every nontrivial finite-dimensional Irreducible Lie algebra representation it acts by a nonzero scalar.

For clarity, the last fact can be expressed by the Casimir eigenvalue formula: on an irreducible with dominant integral weight $\lambda$, the scalar is $(\lambda ,\lambda +2\rho )$, with the inner product induced by the Killing form and $\rho$ the half-sum of positive roots. On the real span of the weights this inner product is positive definite, and the scalar is positive for $\lambda \ne 0$. For a semisimple Lie algebra with several simple factors the scalars add, so a nontrivial representation still gives a nonzero scalar. These are properties of the Casimir operator being used here.

First establish Casimir splitting of a trivial quotient. Supposeis a short exact sequence of finite-dimensional Lie algebra representations, with trivial quotient. The generalized eigenspaces of $C_E$ are invariant, soEach $E_c$ with $c\ne 0$ maps to zero under $q$: applying a sufficiently large power of $C_E-cI$ and using $C_{\mathbb{C}}=0$ gives $(-c{)}^{r}q(e)=0$. Thus $q(E_0)=\mathbb{C}$.

Take a composition series of a module of $E_0$. The Casimir operator is nilpotent on $E_0$, so its scalar on every irreducible composition factor is zero. The stated Casimir operator property makes every such factor trivial. In a basis adapted to the composition series of a module, the image of $\mathfrak{g}$ on $E_0$ therefore consists of strictly upper triangular matrices, so that image is solvable. The allowed fact that a semisimple Lie algebra acts trivially on every one-dimensional representation implies that $\mathfrak{g}$ is a perfect Lie algebra: otherwise a nonzero linear functional on $\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]$ would define a nontrivial one-dimensional Lie algebra representation. Thus $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$, and its image is consequently a perfect Lie algebra too. A perfect Lie algebra that is also a solvable Lie algebra is zero, since its derived series of a Lie algebra is constant until it vanishes. Hence $\mathfrak{g}$ acts trivially on $E_0$. Choose $e\in E_0$ with $q(e)=1$. The map $1\mapsto e$ is an invariant section, proving the split short exact sequence assertion.

Now let $W\subseteq V$ be any nonzero subrepresentation. On the Hom representation $\mathrm{Hom}(V,W)$ the action isConsider the invariant subspaceRestriction produces a short exact sequenceSurjectivity follows by extending $I_W$ to a linear map on $V$. The quotient is trivial, because $I_W$ commutes with the action on $W$. The preceding Casimir splitting of a trivial quotient yields an invariant $p:V\to W$ with $p{|}_W=I_W$. Thusand $\ker p$ is invariant. The zero subrepresentation also has a complement. Repeatedly splitting off an irreducible subrepresentation now gives a direct sum of irreducibles, proving the Weyl complete reducibility theorem.

A derivation of a Lie algebra is a linear map $D:\mathfrak{g}\to \mathfrak{g}$ satisfying the Leibniz ruleThe space $\mathrm{Der}(\mathfrak{g})$ is a vector subspace of $\mathrm{End}(\mathfrak{g})$. Equip it with the commutator $[D,E]=DE-ED$. Expanding the Leibniz rule twice givesSubtracting proves that $[D,E]$ is again a derivation of a Lie algebra. Antisymmetry and the Jacobi identity hold for the commutator in every associative endomorphism algebra, so this defines the derivation Lie algebra.

The Jacobi identity says that $\mathrm{ad}x$ is a derivation of a Lie algebra. Moreover, for $D\in \mathrm{Der}(\mathfrak{g})$,soThis makes $\mathrm{ad}\mathfrak{g}$ an ideal of a Lie algebra in $\mathrm{Der}(\mathfrak{g})$.

Finally suppose $\mathfrak{g}$ is semisimple. Let $\mathfrak{g}$ act on $\mathrm{Der}(\mathfrak{g})$ by $x\cdot D=[\mathrm{ad}x,D]$. The Weyl complete reducibility theorem supplies an invariant complement $M$ to $\mathrm{ad}\mathfrak{g}$. For $D\in M$, invariance gives $[\mathrm{ad}x,D]\in M$, while the ideal identity above gives $[\mathrm{ad}x,D]\in \mathrm{ad}\mathfrak{g}$. Their intersection is zero, so $\mathrm{ad}(Dx)=0$ for every $x$. The center of a Lie algebra of a semisimple Lie algebra is zero, hence $Dx=0$ for every $x$, and $M=0$. ThereforeEvery derivation of a Lie algebra is inner, and the element giving it is unique because the center of a Lie algebra vanishes.

Use $q\in k^{\times }$ and $YX=qXY$ for the quantum plane. Its monomials $X^i{Y}^j$, $i,j\ge 0$, form a basis, with multiplicationOne way to verify the basis assertion without assuming it is to define this multiplication on the vector space with the displayed formal basis. This defines an associative algebra: for a third monomial $X^w{Y}^z$, the two products of three monomials have the same exponent $ju+jw+vw$ of $q$, and its generators satisfy the required relation. Conversely, the relation puts every word into this form, establishing the presentation. Order exponent pairs by the lexicographic order. The largest monomials of two nonzero finite sums give the uniquely largest monomial of their product, with coefficient $cd{q}^{ju}\ne 0$. Therefore the quantum plane is a domain. If one allows $q=0$, the assertion fails because $YX=0$ with both factors nonzero.

A uniform module is a nonzero module in which any two nonzero submodules have nonzero intersection. Suppose the right regular module of a right Noetherian domain $A$ were not a uniform module. Choose nonzero $a,b$ from two right ideals with zero intersection. Then $aA\cap bA=0$. The right idealsform a direct sum. For if ${\sum }_{i=0}^m{b}^{i}a{c}_i=0$, then $a{c}_0\in aA\cap bA$ is zero, so $c_0=0$ by the noncommutative domain property. Cancel the nonzero factor $b$ on the left and repeat to obtain every $c_i=0$. Each summand is nonzero, so their finite partial sums form a strictly ascending chain of right ideals. This contradicts the ascending chain condition of a right Noetherian ring. Hence $A$ is a uniform right module.

It follows that $aA\cap sA\ne 0$ whenever $a,s\ne 0$: there are nonzero $u,v$ with $au=sv$. This is the right Ore condition for the multiplicative set $S=A\setminus \{0\}$; zero numerators cause no difficulty. The Ore localization theorem therefore constructs the ring of right fractionsThe map is injective because an element mapping to zero is annihilated on the right by some nonzero denominator, impossible in a noncommutative domain. To make the denominator convention concrete, if $su=tv\ne 0$ thenFor multiplication, choose $bu=sc$ with $u\ne 0$; thenCommon right multiples make these operations independent of the chosen representatives. Every nonzero $a{s}^{-1}$ has inverse $s{a}^{-1}$, so $Q$ is a division ring. The original field $k$ is central in $A$ and therefore in the inverses as well, making $Q$ a division algebra over $k$. No commutative fraction field construction is being assumed.

If $P$ is a prime ideal of the commutative Artinian ring $R$, then $R/P$ is an Artinian domain. For $0\ne x\in R/P$, the descending chain $(x)\supseteq (x^2)\supseteq \cdots$ stabilizes, so $x^n=x^{n+1}a$ for some $n$; cancellation gives $xa=1$. Thus $R/P$ is a field and $P$ is maximal.

There are only finitely many maximal ideals. Otherwise, distinct maximal ideals $P_i$ are pairwise comaximal and their finite products give the strictly descending chaincontradicting the Artinian condition. This proves the statement about the prime ideals of a commutative Artinian ring.

The nilradical $N$ is the ideal of all nilpotent elements, equivalently the intersection of all prime ideals. Its powers stabilize, say $N^r=N^{r+1}=I$. If $I\ne 0$, choose an ideal $J$ minimal subject to $IJ\ne 0$, then choose $x\in J$ with $Ix\ne 0$. Since $I^2=I$, minimality gives $Rx=Ix$, so $x=ax$ for some $a\in I$. But $a$ is nilpotent, hence $1-a$ is a unit, contradicting $(1-a)x=0$. Therefore the nilradical of a commutative Artinian ring is nilpotent.

If the maximal ideals are $P_1,\dots ,P_s$, the Chinese remainder theorem givesa finite product of fields. Each layer $N^j/N^{j+1}$ is an Artinian module over this product and therefore finite-dimensional. Since $N$ is nilpotent, these finitely many layers show that every ideal of $R$ is finitely generated. This proves the Artinian commutative ring is Noetherian theorem.

Now assume $R$ is local with maximal ideal $P$ and ${\dim }_{R/P}P/P^2=1$. The Nakayama lemma gives $P=(x)$. Since $P$ is nilpotent, its powers form a finite chain ending in zero. For a nonzero ideal $J$, choose the largest $r$ with $J\subseteq (x^r)$ and take $y\in J\setminus (x^{r+1})$. Write $y=x^{r}u$; then $u\notin P$, so $u$ is a unit. Henceand every ideal of $R$ is principal.

Finally, the Artin–Wedderburn theorem says that a finite-dimensional semisimple $k$-algebra has the formwhere each $D_i$ is a finite-dimensional division ring over $k$; the factors are unique up to permutation and isomorphism. By the assumed complete reducibility, write the right regular module aswith pairwise nonisomorphic simple right modules $S_i$. The Schur lemma says that $D_i={\mathrm{End}}_T(S_i)$ is a division ring and that homomorphisms between distinct $S_i$ vanish. Left multiplication and the endomorphism ring of the displayed direct sum therefore giveConversely, the regular module of $M_n(D)$ is a direct sum of $n$ simple column modules, proving that every algebra on the right is semisimple. Uniqueness follows from uniqueness of the simple summands and their multiplicities.

Because $M$ is a finite-dimensional semisimple module, it has an isotypic decompositionwhere the $S_i$ are pairwise nonisomorphic simple right $R$-modules. By Schur lemma,is a division ring, while ${\mathrm{Hom}}_R(S_i,S_j)=0$ for $i\ne j$. Consequently every endomorphism preserves the isotypic summands and is a matrix of entries from $D_i$ on each one. ThereforeIf the ground field is algebraically closed and the $S_i$ are finite-dimensional over it, Schur lemma gives $D_i=k$.

A prime ring is a nonzero ring whose zero ideal is a prime ideal of a noncommutative ring. Equivalently, the product of any two nonzero two-sided ideals is nonzero. This condition is weaker than absence of zero divisors: a full matrix ring over a division ring is prime even when its size is greater than one. For a commutative ring, primeness is exactly being an integral domain.

A noncommutative domain whose right ideals satisfy the ascending chain condition is a right Noetherian domain. Its right regular module is a uniform module, so any two nonzero principal right ideals intersect. Thus its nonzero elements satisfy the right Ore condition and it embeds in a division ring by Ore localization.

If two nonzero principal right ideals in a right Noetherian domain had zero intersection, the sums $bR\oplus abR\oplus \cdots \oplus a^{m}bR$ would be strictly increasing. Directness follows by splitting the lowest term into $bR$ and the other terms into $aR$, then cancelling $a$ and repeating. This contradicts the ascending chain condition. A nonzero common element $au=bv$ gives the right Ore condition. The left version follows from the opposite ring. Consequently a noncommutative domain Noetherian on both sides has a single division ring of fractions with both left and right fraction presentations.