ASPECT OF NON-COMMUTATIVE ALGEBRA

HANGZHOU SUMMER SCHOOL

II. Localization of a Non-Commutative Ring

We will discuss the localization of a non-commutative ring. This localization method is useful for proof of the Structure Theorem set up by Coates, Schneider and Sujatha and it is also needed for Coates, Fukaya, Kato, Sujatha and Venjakob to formulate the GL_{2} main conjecture for elliptic curves without complex multiplication. Roughly speaking, localization is a systematic method of adding multiplicative inverse to a ring. In the commutative case, localization is a very powerful method.

We want to point out that localizing non-commutative rings is more difficult. The localization does not exist for every multiplicatively closed set S.

Most part of contents can be found in [S].

§1. Right Quotient Rings

Definition 1.1 Let A be a ring. A subset S of A is called a multiplicatively closed(m.c.) subset of A if s, t ∈ S ⟹ st ∈ S and 1 ∈ S.

Examples: (i) Let A be ring and x ∈ A. Then S = {1, x, x^{2}, ⋯, x^{n}, ⋯ } is a m.c. subset of A.

(ii) If P is a completely prime ideal of A, i.e., A/P is an integral domain, then A − P is a m.c. subset of A.

(iii) Let A be a ring. An element of A is called a regular element if it is a non-zero divisor of A. The set S_{reg} := {regular elements of A} is a m.c. subset of A. e.g., A = Λ(G), Iwasawa Algebra for some suitable p-adic group G. Then S_{reg} = Λ(G) − {0}, if we assume that A has no zero-divisors.

Remark: A ring A is an integral domain ⟺ x, y ∈ A, xy = 0 ⟹ x = 0 or y = 0(A not necessarily commutative.)

Definition 1.2 Let A be a ring and S a m.c. subset of A. Then a right quotient ring of A w.r.t. S is a ring A[S^{−1}] together with a ring homomorphism φ : A → A[S^{−1}] such that

(Q1.) φ(s) is invertible for all s ∈ S;

(Q2.) ∀q ∈ A[S^{−1}], q = φ(a)φ(s)^{−1} for some a ∈ A and s ∈ S;

(Q3.) φ(a) = 0 ⇔ as = 0 for some s ∈ S.

Similarly, we can define a left quotient ring of A.

2 HANGZHOU SUMMER SCHOOL

Lemma 1.1 (Universal property of a right quotient ring) If A[S^{−1}] exists, then for every homomorphism ψ : A → B such that ψ(s) is invertible in

B for every s ∈ S, there exists a unique homomorphism σ : A[S^{−1}] → B such that σφ = ψ. Proof Define σ : φ(a)φ(s)^{−1} → ψ(a)ψ(s)^{−1}.

(1) σ is well defined. If φ(a)φ(s)^{−1} = φ(b)φ(t)^{−1}, a, b ∈ A, s, t ∈ S, then ψ(a)ψ(s)^{−1} = ψ(b)ψ(t)^{−1} : φ(a)φ(s)^{−1} = φ(b)φ(t)^{−1} ⟹ φ(a) = φ(b)φ(t)^{−1}φ(s). φ(t)^{−1}φ(s) ∈ A[S^{−1}] ⟹ φ(t)^{−1}φ(s) = φ(c)φ(u)^{−1}, c ∈ A, u ∈ S ⟹ φ(s)φ(u) = φ(t)φ(c). So we have φ(a) = φ(b)φ(c)φ(u)^{−1} i.e., φ(a)φ(u) = φ(b)φ(c) ⟹ (au − bc)v = 0, for some v ∈ S (⟺ au − bc ∈ kerφ) and

(su − tc)v′ = 0, for some v′ ∈ S (⟺ su − tc ∈ kerφ) ⟹ ψ(a)ψ(u) = ψ(b)ψ(c), i.e., ψ(a) = ψ(b)ψ(c)ψ(u)^{−1}

ψ(s)ψ(u) = ψ(t)ψ(c), i.e., ψ(c)ψ(u)^{−1} = ψ(t)^{−1}ψ(s) Hence

ψ(a)ψ(s)^{−1} = ψ(b)ψ(c)ψ(u)^{−1}ψ(s)^{−1}
= ψ(b)ψ(t)^{−1}ψ(s)ψ(s)^{−1}
= ψ(b)ψ(t)^{−1}.

(2) σ is a homomorphism:

(a) φ(a)φ(s)^{−1} ± φ(b)φ(t)^{−1} = φ(c)φ(u)^{−1}, s, t, u ∈ S ⟹ ψ(a)ψ(s)^{−1} ± ψ(b)ψ(t)^{−1} = ψ(c)ψ(u)^{−1}.

(b) φ(a)φ(s)^{−1} · φ(b)φ(t)^{−1} = φ(c)φ(u)^{−1} ⟹ ψ(a)ψ(s)^{−1} · ψ(b)ψ(t)^{−1} = ψ(c)ψ(u)^{−1}.

We only check (a). φ(a)φ(s)^{−1} + φ(b)φ(t)^{−1} = φ(c)φ(u)^{−1}

⟹ φ(a)φ(s)^{−1}φ(u) + φ(b)φ(t)^{−1}φ(u) = φ(c). Let φ(s)^{−1}φ(u) = φ(e)φ(v)^{−1} and φ(t)^{−1}φ(u) = φ(f)φ(w)^{−1}. We have

ψ(s)^{−1}ψ(u) = ψ(e)ψ(v)^{−1} ⟺ uv − se ∈ kerφ,

ψ(t)^{−1}ψ(u) = ψ(f)ψ(w)^{−1} ⟺ tf − uw ∈ kerφ.

ASPECT OF NON-COMMUTATIVE ALGEBRA 3

φ(a)φ(e)φ(v)^{−1} + φ(b)φ(f)φ(w)^{−1} = φ(c) ⟹ φ(a)φ(e) + φ(b)φ(f)φ(w)^{−1}φ(v) = φ(c)φ(v). φ(w)^{−1}φ(v) = φ(g)φ(x)^{−1} (⟺ vx − wg ∈ kerφ) ⟹ ψ(w)^{−1}ψ(v) = ψ(g)ψ(x)^{−1}. φ(a)φ(e) + φ(b)φ(f)φ(g)φ(x)^{−1} = φ(c)φ(v) ⟹ φ(a)φ(e)φ(x) + φ(b)φ(f)φ(g) = φ(c)φ(v)φ(x) ⟹ ψ(a)ψ(e)ψ(x) + ψ(b)ψ(f)ψ(g) = ψ(c)ψ(v)ψ(x) ⟹ψ(a)ψ(e) + ψ(b)ψ(f)ψ(g)ψ(x)^{−1} = ψ(c)ψ(v) ⟹ψ(a)ψ(e) + ψ(b)ψ(f)ψ(w)^{−1}ψ(v) = ψ(c)ψ(v) ⟹ ψ(a)ψ(e)ψ(v)^{−1} + ψ(b)ψ(t)^{−1}ψ(u) = ψ(c) ⟹ ψ(a)ψ(s)^{−1}ψ(u) + ψ(b)ψ(t)^{−1}ψ(u) = ψ(c) ⟹ ψ(a)ψ(s)^{−1} + ψ(b)ψ(t)^{−1} = ψ(c)ψ(u)^{−1}.

(3) σφ = ψ and σ is unique: σ : φ(a)φ(s)^{−1} → ψ(a)ψ(s)^{−1}. Take s = 1 ⟹ σφ(a) = ψ(a), ∀a ∈ A

⟹ σφ = ψ.

σ is unique: If σ′ is a homomorphism such that σ′φ = ψ, then

σ′ : φ(a) → ψ(a), φ(s)^{−1} → ψ(s)^{−1} ⟹ σ′ : φ(a)φ(s)^{−1} → ψ(a)ψ(s)^{−1} ⟹ σ = σ′. Corollary 1.2 A[S^{−1}] is unique up to isomorphism. Corollary 1.3 If both A[S^{−1}] and [S^{−1}]A exist, then they are isomorphic. Now we are going to study when A[S^{−1}] exists. Theorem 1.4 let A be a ring and S a m.c. subset of A. Then A[S^{−1}] exists if and only

if S satisfies: S1. For s ∈ S and a ∈ A, there exist t ∈ S and b ∈ A such that sb = at; S2. If sa = 0 with s ∈ S, then at = 0 for some t ∈ S. When A[S^{−1}] exists, it has the form

A[S^{−1}] = A × S/ ∼

4 HANGZHOU SUMMER SCHOOL

where ∼ is the equivalence relation:

(a, s) ∼ (b, t) ⟹ there exist c, d ∈ A such that ac = bd and sc = td ∈ S. Remark A m.c. subset of A is said to satisfy the right Ore condition if S1 is satisfied. Proof ” ⟹:” Q2 ⟹ ∀a ∈ A, s ∈ S, φ(s)^{−1}φ(a) = φ(b′)φ(t′)^{−1} for some b′ ∈ A and t′ ∈ S. So φ(at′) = φ(sb′). Q3 ⟹ ∃u ∈ S such that at′u = sb′u. Put b = b′u, t = t′u ⟹ S1. If sa = 0 then φ(sa) = 0 ⟹ φ(s)φ(a) = 0. φ(s) is invertible ⟹ φ(a) = 0. Q3 ⟹ ∃t ∈ S such that at = 0 ⟹ S2. ”⟸:” First, it is not difficult to prove that the relation ”∼” is an equivalent relation. In A[S^{−1}] = A × S/ ∼, we define ”+”: (a, s) + (b, t) = (ac + bd, u), where u = sc = td ∈ S. ”·”: (a, s) · (b, t) = (ac, tu), where sc = bu, u ∈ S. We need to show

(1) ”+” and ”·” are independent of the choice of representing couples;

(2) A[S^{−1}] = A × S/ ∼ is a ring;

(3) φ : A → A[S^{−1}] = A × S/ ∼, a → (a, 1) is a right quotient ring, i.e., φ is a ring homomorphism and Q1, Q2 and Q3 are satisfied.

It is easy to see that (a, s) + (b, t) = (b, t) + (a, s).

(1) ”+” is independent of the choice of representing couples. Assume (a, s) = (a′, s′) and (b, t) = (b′, t′). Need to prove : (a, s) + (b, t) = (a′, s′) + (b′, t′). It is suffice to show (a, s) + (b, t) = (a′, s′) + (b, t). In fact, if it is true, then (a, s) + (b, t) = (a′, s′) + (b, t) = (a′, s′) + (b′, t′). By definition, (a, s) + (b, t) = (ac + bd, u), where u = sc = td ∈ S and

ASPECT OF NON-COMMUTATIVE ALGEBRA 5

(a′, s′) + (b, t) = (a′c′ + bd′, u′), where u′ = s′c′ = td′ ∈ S. We have (a, s) = (ac, sc) = (ac, u), (sc ∈ S) and (a′, s′) = (a′c′, s′c′) = (a′c′, u′), (s′c′ ∈ S). (a, s) = (a′, s′)⟹ ∃x, y ∈ A, s.t. acx = a′c′y, ux = u′y ∈ S. But u = td, u′ = td′, so tdx = td′y ⟹ t(dx − d′y) = 0, t ∈ S. S2 ⟹ ∃z ∈ S, s.t.

(dx − d′y)z = 0, i.e., bdxz = bd′yz

From acx = a′c′y, we get acxz = a′c′yz. ⟹ (ac + bd)xz = (a′c′ + bd′)yz and ux = u′y ⟹ uxz = u′yz. ux, u′y ∈ S, z ∈ S ⟹ uxz = u′yz ∈ S. Therefore, (ac + bd, u) = (a′c′ + bd′, u′), i.e., (a, s) + (b, t) = (a′, s′) + (b, t). Exercise Check that if (a, s) = (a′, s′) and (b, t) = (b′, t′), then (a, s) · (b, t) =

(a′, s′) · (b′, t′).

(2) A × S/ ∼ is a ring: In A × S/ ∼, 1 = (1, 1), 0 = (0, 1).

(3) φ : A → A × S/ ∼ is a right quotient ring. φ : a → (a, 1), (1 ∈ S) φ is a homomorphism(easy to check). ∀s ∈ S, φ(s) = (s, 1) and

(s, 1)(1, s) = (1, 1),
(1, s)(s, 1) = (1, 1).

So, φ(s) is invertible ⟹ Q1.

∀(a, s) ∈ A × S/ ∼, (a, s) = (a, 1)(1, s) = φ(a)φ(s)^{−1} ⟹ Q2. Kerφ = {a ∈ A|(a, 1) = (0, 1) ⟺ ∃p, q ∈ A, ap = 0 · q, 1 · p = 1 · q ∈ S}

= {a ∈ A|as = 0 for some s ∈ S} ⟹ Q3.

6 HANGZHOU SUMMER SCHOOL

§2 Applications Definition 2.1 (1) S is called a right denominator set if S is a m.c. subset of A and

satisfies S1 and S2.

(2) S is called a right Ore set if S is a m.c. subset of A and satisfies S1. If S_{reg} is a right Ore set, we say that A satisfies the right Ore condition.

(3) A right ideal of the form r(B) = {a ∈ A|Ba = 0} is called a right annihilator, where B is a subset of A.

Remarks (1) 0 ∈ S ⟹ A[S^{−1}] does not exist.

(2) If S ⊆ S_{reg} is a m.c. subset of A, then A[S^{−1}] exists ⟺ S is a right Ore set. In this case, φ : A → A[S^{−1}] is injective.

The following proposition shows that under circumstances S1 ⟹ S2. Proposition 2.1 Assume A satisfies ACC(ascending chain condition) on right anni-

hilator. If S is a right Ore set of A, then S is a right denominator set(i.e., S also satisfies S2).

Proof If sa = 0 for some s ∈ S, consider the following ascending chain r(s) ⊆ r(s^{2}) ⊆ ⋯ ⊆ r(s^{n}) ⊆ · · ·. ACC⟹ ∃n, r(s^{n}) = r(s^{n+1}). S1⟹ ∃b ∈ A, t ∈ S, s.t. s^{n}b = at. Then s^{n+1}b = sat = 0 ⟹ b ∈ r(s^{n+1}) = r(s^{n}) ⟹ at = s^{n}b = 0 ⟹ S2. Corollary 2.2 (Ore) A[S_{reg}^{−1}] exists if and only if A satisfies the right Ore condition. If A has no zero-divisors, then A satisfies the right Ore condition ⟺ aA ∩ sA ≠ 0 for

all a, s ∈ A ⟺ 𝔞 ∩ 𝔟 ≠ 0 for all non-zero right ideals 𝔞 and 𝔟.

Corollary 2.3 Let A be a ring without zero-divisors, so S_{reg} = A − {0}. If A satisfies the right Ore condition(A is called a right Ore domain). Then A[S_{reg}^{−1}] is a skew-field.

Theorem 2.4 Every right Noetherian ring without zero divisors is a right Ore domain. Proof Let a, b ∈ A with a, b ≠ 0. Let 𝔞_{n} = bA + abA + ⋯ + a^{n}bA be a right ideal of A.

Then

⋯ ⊆ 𝔞_{n} ⊆ 𝔞_{n+1} ⊆ · · · .

A is Noetherian ⟹ ∃n, s.t. 𝔞_{n} = 𝔞_{n+1}.

Suppose n is the smallest integer. Then a^{n+1}b ∈ 𝔞_{n+1} = 𝔞_{n} ⟹ a^{n+1}b = bc_{0} + abc_{1} + ⋯ + a^{n}bc_{n}. If bc_{0} = 0, then a(a^{n}b) = a(bc_{1} + ⋯ + a^{n−1}bc_{n}). a ≠ 0 ⟹ a^{n}b = bc_{1} + ⋯ + a^{n−1}bc_{n} ⟹ 𝔞_{n} = 𝔞_{n−1}.

ASPECT OF NON-COMMUTATIVE ALGEBRA 7

This contradicts the minimality of n. bc_{0} = a(a^{n}b − bc_{1} − ⋯ − a^{n−1}bc_{n}) ≠ 0 ⟹ bA ∩ aA ≠ 0. Example(Coates, Fukaya, Kato, Sujatha and Venjakob) Let G be a compact p-

adic Lie group and Λ(G) the Iwasawa Algebra. Assume there is a closed normal subgroup H of G such that G/H ≅ Z_{p}. Let S be the set of all f in Λ(G) such that Λ(G)/Λ(G)f is a finitely generated Λ(H)-module.

Theorem(CFKSV) (1) S is a m.c. subset of Λ(G) which is a left and right Ore set. The element of S are non-zero divisor in Λ(G) ⟹ Λ(G)_{S} := Λ(G)[S^{−1}] exists.

(2) Let S^{∗} = ⋃_{n≥0} p^{n}S. Then S^{∗} is a m.c. subset of Λ(G) which is a left and right Ore set and it has no zero divisors. Λ(G)_{S^{∗}} := Λ(G)[S^{∗}^{−1}] exists.

Definition 2.2 A ring A is called a ring of quotients if every non-zero divisor of A is invertible.(So, for any m.c. S ⊆ S_{reg}, A = A[S^{−1}] )

Proposition 2.5 If A satisfies DCC on principal right ideals, then A is a ring of quotients.

This is a corollary of the following proposition. Proposition 2.6 Suppose that A satisfies DCC on principal right ideals. The following

properties of an element are equivalent:

(i) a is invertible; (ii) a is regular; (iii) r(a) = 0. Proof (i)⟹ (ii)⟹ (iii) are clear. (iii)⟹ (i). Let a ∈ A, r(a) = 0. Consider

aA ⊇ a^{2}A ⊇ · · · .

By hypothesis, ∃n, a^{n}A = a^{n+1}A. So a^{n} = a^{n+1}b ⟹ a^{n}(1 − ab) = 0.

r(a) = 0⟹ 1 − ab = 0 ⟹ ab = 1. On the other hand, aba = a i.e. a(ba − 1) = 0, by r(a) = 0 again, we get ba = 1.

§3 Goldie’s Theorem Definition 3.1 Let Q be a ring of quotients. A subring A of Q is called a right order in

Q if Q = A[S_{reg}^{−1}], where S_{reg} is the set of all regular elements in A.

Definition 3.2 A ring A is said to have finite right rank if A doesn’t contain any right ideal of the form 𝔞 = ⊕_{i=1}^{∞}𝔞_{i} with 𝔞_{i} ≠ 0 for any i.

Clearly, every Noetherian ring has of course finite right rank. Lemma 3.1 Let A be a ring. Then the following are equivalent:

8 HANGZHOU SUMMER SCHOOL

(a) A is semi-simple.

(b) All right A-modules are semi-simple.

(c) All right A-modules are projective.

(d) All right A-modules are injective.

(e) Every right ideal of A is direct summand of A.

(f) A ≅ M_{n_{1}}(D_{1}) × ⋯ × M_{n_{k}}(D_{k}) for some skew-fields D_{1}, ⋯, D_{k}.

(g) A has no proper essential right ideals.

(h) A is right Artinian and J(A) = 0. Lemma 3.2 If A is semi-prime and satisfies ACC on right annihilator ideals, then A

has no right or left nil ideals ≠ 0.

Proof Clearly, Aa is nil ⟺ aA is nil. On the other hand, we only need to consider the left(or right) ideals of the form Aa(aA). So assume Aa is nil. A satisfies ACC ⟹ there exists b ∈ Aa with maximal annihilator. ∀c ∈ A, let k be the smallest integer such that (cb)^{k} = 0, (k ≥ 1). We have r(b) ⊆ r(cb)^{k−1} if k ≥ 2. r(b) maximal ⟹ r(b) = r((cb)^{k−1} So we always have cb ∈ r(b), (k = 1, cb = 0). ⟹ bAb = 0 ⟹ (AbA)^{2} = 0. A is semi-prime ⟹ AbA = 0 ⟹ b = 0. This implies Aa = 0. Lemma 3.3 If A satisfies ACC on right annihilators, then the right singular ideal of A

is nilpotent.

By definition, the right singular ideal of a ring A is

Z = {a ∈ A|r(a) is essential right ideal of A}.

One can check that Z is a two-sided ideal.

Proof If Z^{n} ≠ 0, choose a ∈ Z such that Z^{n−1}a ≠ 0 and r(a) is maximal. ∀b ∈ Z, r(b) ∩ aA ≠ 0 (r(b) is essential since b ∈ Z) ⟹ ∃c ∈ A such that ac ≠ 0, bac = 0 ⟹ r(ba) ⊃ r(a), but r(ba) ≠ r(a).

ASPECT OF NON-COMMUTATIVE ALGEBRA 9

We have ba ∈ Z, so Z^{n−1}ba = 0. b is arbitrary ⟹ Z^{n}a = 0. Z^{n−1}a ≠ 0, Z^{n}a = 0 ⟹ ∃n, such that Z^{n} = 0. Theorem 3.4(Goldie) Let A be a ring. Then the following are equivalent:

(a) A is a right order in a semi-simple ring.

(b) A has finite right rank, satisfies ACC on right annihilators and is a semi-prime ring. Corollary 3.5 If A is a Noetherian semi-prime ring, then S_{reg} is a right denominator

set and A[S_{reg}^{−1}] is a semi-simple ring.

Proof In fact, we will prove that (a)⟹ (b) ⟹ (c) ⟹ (a), where (c) is the following:

(c) A right ideal of A is essential if and only if it contains a regular element. (a)⟹ (b). Let Q be the semi-simple right ring of quotients of A. (b1) A has finite rank. We will show that a direct sum of right ideals in A ⟹ a corresponding direct sum of

right ideals in Q.

If 𝔞 ∩ 𝔟 = 0 for right ideals 𝔞 and 𝔟 of A, then 𝔞Q ∩ 𝔟Q = 0. In fact, if (a, s) = (b, t) ∈ 𝔞Q ∩ 𝔟Q with a ∈ 𝔞, b ∈ 𝔟 and s, t regular, then (a, s) = (b, t) ⟺ ∃ p, q ∈ A, s.t. ap = bq, sp = tq ∈ S ⟹ ap = bq ∈ 𝔞 ∩ 𝔟 = 0 ⟹ (a, s) = (ap, sp) = 0,

(b, t) = (bq, tq) = 0. So Q has finite rank implies that A has finite rank too. (b2) A satisfies ACC on right annihilators. Q is Noetherian ⟹ Q satisfies ACC on right annihilators. So r_{A}(B) = r_{Q}(B) ∩ A (B ⊆ A) implies (b2). (b3) A is semi-prime (𝔞^{m} = 0 ⟹ 𝔞 = 0). We show that l(𝔞) is an essential right ideal of A hence l(𝔞)Q is essential in Q. If 𝔟 ≠ 0 is a right ideal of A, then ∃ a largest integer n ≥ 0 such that 𝔟𝔞^{n} ≠ 0. But

𝔟 ∩ l(𝔞) ⊇ 𝔟𝔞^{n} ≠ 0, so l(𝔞) is essential in A. On the other hand, for any non-zero right ideal T of Q, we have

0 ≠ (T ∩ A) ∩ l(𝔞) ⊂ T ∩ l(𝔞)Q

since l(𝔞) is essential. So l(𝔞)Q is essential in Q.

10 HANGZHOU SUMMER SCHOOL

Q is semi-simple ⟹ l(𝔞)Q = Q ⟹ ∃ a_{i} ∈ l(𝔞), q_{i} ∈ Q, s.t. 1 = ∑ a_{i}q_{i}. One can check that ∃ s ∈ A such that q_{i}s ∈ A for all i. Then

s𝔞 = ∑ a_{i}q_{i}s𝔞 ⊂ ∑ a_{i}𝔞 = 0.

So 𝔞 = 0.

(b) ⟹ (c). One direction is easy. In fact, the following is true: If A has finite right rank and r(s) = 0, then sA is a essential right ideal. To show this, consider any right ideal 𝔞 ≠ 0. If sA ∩ 𝔞 = 0, then

𝔞 ⊕ s𝔞 ⊕ ⋯ s^{n}𝔞 ⊕ ⋯ ⊆ A.

But s^{n}𝔞 ≠ 0(r(s) = 0). This contradicts to A has finite right rank. So sA ∩ 𝔞 ≠ 0. We prove now that if 𝔞 is an essential right ideal of A, then 𝔞 contains a regular element. By lemma 3.3, 𝔞 is not nil ⟹ ∃ a_{1} ∈ 𝔞, a_{1} ≠ 0, r(a_{1}) = r(a_{1}^{2}). If 𝔞 ∩ r(a_{1}) = 0, then we let c = a_{1}. If 𝔞 ∩ r(a_{1}) ≠ 0, then we choose a_{2} ∈ 𝔞 ∩ r(a_{1}) such that r(a_{2}) = r(a_{2}^{2}). If 𝔞 ∩ r(a_{1}) ∩ r(a_{2}) ≠ 0. Choose a_{3} ∈ 𝔞 ∩ r(a_{1}) ∩ r(a_{2}) and continue the process. We claim that the process must stop at some stage, where 𝔞 ∩ r(a_{1}) ∩ ⋯ ∩ r(a_{k}) = 0. In fact, at each step, we have a direct sum

a_{1}A ⊕ ⋯ ⊕ a_{k}A.

We show this by induction.

Suppose a_{1}A ⊕ ⋯ ⊕ a_{k−1}A is direct and

a_{k}b = a_{1}b_{1} + ⋯ + a_{k−1}b_{k−1}.

But a_{1}a_{k} = 0 and a_{1}a_{2} = ⋯ = a_{1}b_{k−1} = 0 ⟹ a_{1}^{2}b_{1} = 0, i.e., b_{1} ∈ r(a_{1}^{2}) = r(a_{1}) ⟹ a_{1}b_{1} = 0, i.e., a_{k}b = a_{2}b_{2} + ⋯ + a_{k−1}b_{k−1} a_{2}a_{k} = 0, a_{2}a_{3} = ⋯ = a_{2}a_{k−1} = 0(k ≥ 3) ⟹ a_{2}b_{2} = 0,

· · · · · ·

So we conclude that a_{i}a_{k} = 0 if i < k. ⟹ b_{i} ∈ r(a_{i}^{2}) = r(a_{i}) ⟹ ∑_{i=1}^{k−1} a_{i}b_{i} = 0 ⟹ direct sum.

ASPECT OF NON-COMMUTATIVE ALGEBRA 11

𝔞 essential ⟹ r(a_{1}) ∩ ⋯ ∩ r(a_{k}) = 0. Write c = a_{1} + ⋯ + a_{k} ∈ 𝔞. Then r(c) = 0. In fact, for any t ∈ r(c), 0 = ct = a_{1}t + ⋯ + a_{k}t ⟹ a_{i}t = 0, i.e., ∀i, t ∈ r(a_{i}). But ∩r(a_{i}) = 0 ⟹ t = 0. So c will be a regular element in 𝔞, if we can show that l(c) = 0. We see that x ∈

l(c), i.e., xc = 0 implies c ∈ r(x), i.e., r(x) ⊃ cA, cA is essential ⟹ r(x) is essential ⟹ l(c) ⊂ Z.

A is semi-prime ⟹ Z = 0 ⟹ l(c) = 0.

(c) ⟹ (a)

(1) Verify the right Ore condition: Let s ∈ S_{reg} and a ∈ A. So sA is essential. (cA : a) = {c|ac ∈ sA} is essential: ∀𝔟 ≠ 0, if a𝔟 ≠ 0, then a𝔟 ∩ sA ≠ 0. So ∃b ∈ 𝔟, ab ∈ sA ⟹ (sA : a) ∩ 𝔟 ≠ 0. If a𝔟 = 0, then 𝔟 ⊂ (sA : a). Now (sA : a) is essential ⟹ ∃ regular t ∈ (sA : a). Thus at = sb for some b ∈ A.

(2) Show that Q = A[S_{reg}^{−1}] is semi-simple. By lemma 3.1, it suffices to prove that Q has no proper essential right ideals. For any

right ideal 𝔞 of Q, we have (𝔞 ∩ A)Q ⊂ 𝔞. On the other hand, ∀q ∈ 𝔞, q = as^{−1}, so a = qs ∈ 𝔞 ∩ A ⟹ 𝔞 ⊂ (𝔞 ∩ A)Q.

So we have (𝔞 ∩ A)Q = 𝔞. If 𝔞 is an essential right ideal in Q, then 𝔞 ∩ A is essential in A, so ∃ regular t ∈ 𝔞 ∩ A,

therefore 𝔞 = (𝔞 ∩ A)Q = Q.

Theorem 3.6 The following properties of A are equivalent:

(a) A is a right order in a simple ring.

(b) A has finite right rank, satisfies ACC on right annihilators and is a prime ring. Remark If G is a compact p-adic analytic group, then Λ(G) is a semi-prime ring.

References

[CFKSV] Coates, J., Fukaya, T., Kato, K., Sujatha, R., and Venjakob, O., The GL_{2} main conjecture for elliptic curves without complex multiplication, preprint (2004).

[CSS] Coates, J., Schneider, P., and Sujatha, R., Modules over Iwasawa algebras, J. Inst. Math. Jussieu 2 (2003) no. 1, 73-108.

[MR] McConnell, J. C., and Robson, J. C., Noncommutative Noetherian Rings, Graduate Studies in Math. 30 AMS (1987).

[S] Stenström B, Rings of Quotients, Springer-Verlag, New York-Berlin (1975).