Advanced Algebra II

Advanced Algebra II

structure of rings

Let M be a simpleRM, by Schur’s Lemma, EndR(M) = HomR(M, M ) is a division ring, call it D. Then M is naturally a D-module as we have

D = End_R(M) × M → M.

With this, we are able to analyze the structure of primitive rings. Recall that R is primitive if there exist a faithful simple module M. Since M is faithful, one has an embedding Φ : R ,→ End(M) by Φ(r) = Tr, where Tr(x) := rx. For ψ ∈ D = C(M) = EndR(M), we have

ψT_r(x) = T_rψ(x)∀x ∈ M.

Therefore, T_r is D-linear, i.e. Φ : R ,→ End_D(M). If M is a finite dimensional vector space over D of dimension n, then End_D(M) is nothing but Mat_n(D).

A natural question is that whether a primitive ring R = End_D(M)?

The answer is YES under some finiteness condition. This is the content of Wedderburn-Artin Theorem. However, this is not the case in general.

Take the example as in Example 0.11, I is a ring and V is faithful simple I-module. Thus I is primitive but not equal EndD(M). Therefore, we need the refined notion of density to characterize primitive ring.

Definition 0.1. Let V be a vector space over a division ring D. A subring R ⊂ End_D(V ) is called a dense subring if for every positive integer n, every linearly independent subset {u1, ..., un} of V and every arbitrary subset {v1, ..., vn} of V , there is θ ∈ R such that θ(ui) = vi

for all i = 1, ..., n.

Theorem 0.2 (Jacobson Density Theorem). Let R be a primitive ring with a faithful simple module M. And let D be the division ring End_R(M). Then R is isomorphic to a dense subring of End_D(M).

Conversely, let M be a vector space over a division ring D, then a dense subring of EndD(M) is primitive.

Proof. We have seen that Φ : R ,→ End_D(M). It suffices to show that Φ(R) is dense.

Claim. Let V be a finite dimensional subspace of M, then for x ∈ MV , there exist r ∈ R such that rx 6= 0 and rV = 0.

Grant this for the time being. For a linearly independent subset {u1, ..., un}. Let V be the vector space spanned by it, and Vi be the space spanned by {u₁, ., u_i−1, u_i+1, .., u_n}. By the claim, there is r_i such that r_iu_i 6= 0, r_iu_j = 0 for all j 6= i. Consider now Rr_iu_i < M . Apply the claim to V = 0, then Rr_iu_i 6= 0. By the simplicity, Rr_iu_i = M. In particular, for v_i ∈ M, there is s_i ∈ R such that s_ir_iu_i = v_i. We now have θ :=P

s_ir_i satisfying the required property.

1

2

The converse statement is easy.

We now prove the claim by induction on dimDV . First of all, it’s clearly true if dimV = 0. Suppose that V = V⁰+ Dw for some w 6∈ V⁰ and the claim is true for V⁰. Let A(V⁰) be the annihilator of V⁰ in R, which is a left ideal of R. The induction hypothesis asserts that there is r ∈ A(V⁰) and rw 6= 0. In particular, A(V⁰)w < M is a non-zero submodule. By the simplicity of M, A(V⁰)w = M.

Let S := {m ∈ M, m 6∈ V |A(V )m = 0}. We need to show that S = ∅. Say if m ∈ S, we can define τ : M → M by: if x ∈ M hence x = aw for some a ∈ A(V⁰). We define τ (x) := am. Note that if x = a₁w = a₂w, then a₁ − a₂ ∈ A(w) ∩ A(V⁰) = A(V ). One has a₁m − a₂m ∈ A(V )m = 0. Thus τ is well-defined.

If x = aw with a ∈ A(V⁰), then rx = raw. So τ (rx) = ram = r(am) = rτ (x). So τ ∈ D. Therefore, for all a ∈ A(V⁰), a(m − τ (w)) = 0. So m − τ (w) ∈ V⁰, otherwise, by induction hypothesis, there exist b ∈ A(V⁰) which does not annihilates m − τ (w). Thus, m ∈ τ (w) + V⁰ ⊂ Dw + V⁰ = V . This is the required contradiction. ¤ Definition 0.3. A (left) module M ∈_RM is said to be Artinian (resp.

Noetherian) if it satisfies descending chain condition (resp. ascending) on submodules.

A ring R is said to be left Artinian (resp. Noetherian) if R is Ar- tinian (resp. Noetherian) as left R-module.

R is said to be Artinian (resp. Noetherian) if R is both left and right Artinian (resp. Noetherian).

Theorem 0.4. The following condition on left Artinian rings are equiv- alent:

(1) R is simple.

(2) R is primitive.

(3) R ∼= EndD(V ) for a non-zero finite dimensional vector space over D.

(4) R ∼= Matn(D) for some positive integer n and division ring D.

Proof. (1) ⇒ (2) Let I := {r ∈ R|Rr = 0} the right annihilator of R. It’s clear that I C R. R is simple, thus I = 0. R is left Artinian, thus there is a minimal left non-zero ideal, say J. Now A(J) C R. If A(J) = R, then J ⊂ I = 0 which is absurd. Thus A(J) = 0. So J is a faithful simple _RM, and hence R is primitive.

(2) ⇒ (3) By density theorem, R is isomorphic to a dense subring of T ⊂ EndD(V ). We claim that dimD(V ) is finite. Suppose not, then there is an infinite sequence of independent subset {u1, u2, ...}. Let In

be the left annihilator of {u₁, ..., u_n}. One sees that I₁ ) I₂ ) ... give a proper descending chain, which is the required contradiction. One next show that a dense subring of End_D(V ) is the whole ring if dim_D(V ) is finite.

(3) ⇔ (4) trivial. And we have seen (4) ⇒ (1) before.

3

¤