Advanced Algebra II

Advanced Algebra II

structure of rings

Definition 0.1. For M ∈ _RM, we said that M is simple if RM 6= 0 and M has no proper submodule.

A ring is R said to be simple if R² 6= 0 and R has no proper ideal.

Example 0.2. A simple ring R maybe not a simple left _RM.

Let R := Mat₂(k), the 2 × 2 matrices over a field k. One can check that ideal of R is of the form Mat₂(J) for some J C k. Hence R has no proper ideal, hence a simple ring. However, R has two left ideals Ii := {(ajk)|ajk = 0∀k 6= i}, for i = 1, 2. Therefore, as an left RM, R has proper submodule Ii.

Example 0.3. A division ring is a simple ring.

Proposition 0.4. A simple _RM is cyclic, i.e. M = Ra for some a ∈ M.

Proof. For each a ∈ M, we consider Ra < M . Since M is simple, Ra = 0 or Ra = M. Since RM 6= 0, it follows that Ra 6= 0 hence Ra = M for some a ∈ M.

We can actually prove more. Let B = {a ∈ M|Ra = 0}. It’s easy to check that B < M . As we have seen that B 6= M. Therefore, B = 0.

That is, M = Ra for all a 6= 0 ∈ M. ¤

Proposition 0.5. Let M be a simple _RM, then M ∼= R/I for some maximal left ideal I. (Note that R/I is not necessarily a ring, we just consider it as a left _RM.)

Proof. Since M is cyclic, one has that the natural map ϕ : R → M by ϕ(r) = ra for a fixed a 6= 0 ∈ M is surjective. Consider all these as left _RM, then by the first isomorphism theorem, M ∼= R/I for some submodule I := Kerϕ = A(a) ⊂ R. It’s clear that I is a left ideal. M

is simple implies that I is maximal. ¤

Note that M = Ra, then a = ea for some e ∈ R. e play the role as an ”idempotent”. It follows that ra = rea, i.e r − re ∈ A(a) = I for all r ∈ R. A left ideal I with the property that there is e ∈ R with r − re ∈ I for all r ∈ R is called regular.

We can actually prove the following:

Theorem 0.6. A left R-module M is simple if and only if M ∼= R/I for some regular maximal left ideal I.

Proof. We have seen one direction (only if part). Suppose now that M ∼= R/I for some regular maximal left ideal I. Since I is maximal, there is no proper submodule in M. We still need to check that RM ∼= R(R/I) 6= 0. Since there is e ∈ R with r − re ∈ I for all r ∈ R. We

1

2

have r(e + I) = r + I 6= 0 if r 6∈ I. Hence R(R/I) 6= 0. Thus R/I is

simple. ¤

A similar and important notion is primitive. We need some more notions before we get into primitive.

Proposition 0.7. Let A(M) := {r ∈ R|rM = 0} be the annihilator of M. Then A(M) C R is an ideal. Moreover, M is a faithful R/A(M) module, i.e. A_R/A(M) = 0 as R/A(M) module.

Proof. For r ∈ R, s ∈ A(M), we have (rs)M = r(sM) = 0, hence rs ∈ A(M). Also sr(M) = s(rM) < sM = 0, hence sr ∈ A(M).

The map R × M → M induces a natural map R/A(M) × M → M.

If for some r ∈ R such that ¯r ∈ A_R/A(M) ⊂ R/A(M). ¯rM = rM = 0,

r ∈ A(M) and thus ¯r = 0. ¤

For M ∈ RM, we can consider End(M) the endomorphims of the abelian group M. It’s clear that End(M ) is a ring. There is a natural map Φ : R → End(M) define by Φ(r) = T_r such that T_r(x) := rx for x ∈ M. Then KerΦ = A(M). By the isomorphism theorem, we have R/A(M) is isomorphic to a subring of End(M).

Definition 0.8. A ring R is primitive if there exists a simple faithful left R-module.

If R is primitive, let M be a simple faithful R-module, then R is isomorphic to a subring of End(M).

Before we proceed to the study of the structure of primitive rings, we take a break by looking at some examples.

Proposition 0.9. A simple ring with identity is primitive.

Proof. Since R has identity, R has maximal left ideal, say I. Again, R with identity implies that every left ideal of R is regular. Thus R/I is a simple R-module.

It remains to show that A(R/I) = 0. This is the case since 1 6∈

A(R/I) and A(R/I) C R which is simple. ¤

Proposition 0.10. A commutative ring R is primitive if and only if R is a field.

Therefore, if R is a commutative ring with identity, then being sim- ple, primitive and being a field are equivalent. However, if R is com- mutative without identity, then it’s possible to be simple but not a field.

Proof. A field is simple with identity, hence is primitive.

Conversely, let M be a faithful simple R-module. Then M ∼= R/I for some regular maximal left ideal I. R is commutative, so I is in fact an ideal. M is faithful, thus A(M) = A(R/I) = 0. It’s clear that I ⊂ A(R/I) = 0. Moreover, I is regular, that is, there exist e ∈ R such

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that r − re = 0. It follows that e is the identity. We have that R is a commutative ring with identity such that 0 is a maximal ideal. Thus

R is a field. ¤

The following example also exhibit the difference between simple and primitive.

Example 0.11. Let V be a infinite dimensional vector space over a di- vision ring D. Let R := End_D(V ) be the ring of linear transformations.

Id clear that V is a left R-module.

Claim 1. V has no proper submodule. And thus V is a simple R-module.

Claim 2. A(V ) = 0, thus R is primitive.

Claim 3. R is not simple. This is because that we have an ideal I = {ρ ∈ R|dim(Imρ) < ∞}.

Remark 0.12. If V is finite dimensional, then R is simple and prim- itive.

One notice hat End(M) plays an important role here. Inside End(M), there is yet another interesting object

Definition 0.13. The commuting ring of R on M is C(M) = {ψ ∈ End(M)|T_rψ = ψT_r, ∀r ∈ R}.

One notice that if ψ ∈ C(M), then

ψ(rx) = ψT_r(x) = T_rψ(x) = rψ(x),

for all x ∈ M. In other words, ψ is an _RM-homomorphism. So C(M) is actually End_R(M).

Theorem 0.14 (Schur’s Lemma). If M is a simple module over R, then EndR(M) is a division ring.

Proof. Suppose that ψ 6= 0 ∈ End_R(M). Let W := ψM. One sees that W 6= 0 < M. By the simplicity of M, one has W = M. Thus ψ is surjective.

Similarly, one checks that Kerψ  M . Thus Kerψ = 0, ψ is injective.

Hence ψ⁻¹ ∈ End(M). It’s direct to check that ψ⁻¹ ∈ End_R(M). ¤ Example 0.15. Let k be a field and V be a n-dimensional vector space over k. Let R = Mat_n(k) = End_k(V ). If S ⊂ R, let ¯S be the subalgebra generated by S in R. Then we can view V as a faithful ¯S-module.

We say that S is irreducible if V is an simple ¯S-module. Then by Schur’s Lemma, C( ¯S) is a division ring.

1. Let n = 2, and S =

µ 0 −1 1 0

¶

. One can check that S is irreducible and C( ¯S) ∼= C.

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2. Let n = 4, and S =















0 −1 0 0

1 0 0 0

0 0 0 −1

0 0 1 0





 ,







0 0 −1 0

0 0 0 1

1 0 0 0

0 −1 0 0













 .

One can check that S is irreducible and C( ¯S) ∼= H the real quaternions.