The field of complex numbers is a two-dimensional vector space over A ring that is also a vector space over a field F is called an algebra over F. In the nineteenth century many attempts were made to describe the division algebras that are finite dimensional over and in particular those (like ) that are fields. Certainly the three-dimensional examples would have had applications to physics, but after looking in vain for such an algebra, the first success came in 1843 when W.R. Hamilton discovered the ring of quaternions, a four-dimensional algebra that, surprisingly, was not commutative. It was not until 1878 that G. Frobenius showed that there is no three-dimensional example, and that the only possible associative examples are , and Meanwhile, G. Grassmann described many examples that were rings but not necessarily division rings, of which the matrix algebras, constructed in 1858 by A. Cayley, were an important special case. The next event in the development of the theory came in 1907 when J.H.M. Wedderburn gave the first characterization of the simple finite dimensional algebras. Finally, in 1927, E. Artin extended Wedderburn's theorem to a result about rings by eliminating the dependence on the fact that the ring is finite dimensional as an algebra, replacing it with finiteness conditions on the set of left ideals. These seminal results mark the beginning of the theory of noncommutative ring theory, and we prove them in this chapter.

11.1 Wedderburn's Theorem

If F is a field, a ring R is called an F-algebra if it is a vector space over F and t(ab) = a(tb) for all t F and all a, b R. Hence, is a two-dimensional -algebra and the ring M_n(F) of all n × n matrices over F is an n²-dimensional F-algebra. The ring of real quaternions is a four-dimensional algebra over that has the distinction of being a division ring (every nonzero element is a unit). Wedderburn's theorem asserts that if R is a simple, finite dimensional algebra then R ≅ M_n(D) for some n ≥ 1 and some division ring D. We are going to prove an extension of this theorem that removes the restriction that R is an algebra. The first task is to find an appropriate “ finiteness condition” on R to replace the finite dimensional requirement.

Finiteness Conditions

If R is a ring, recall (Section 7.1) that a left R-module is an additive abelian group with a left R-action rx M, r R, x M, such that the axioms for a vector space are satisfied. In the nineteenth and early twentieth centuries, most of the modules studied were finite dimensional vector spaces. Emmy Noether, and later Emil Artin, realized that the right way to extend this finite dimensional condition was to use finiteness conditions on the set of submodules. Lemma 1 below identifies the most important of these.

If is a nonempty set of submodules of a module M, then is said to be maximal in if implies that K = N. Similarly N is called minimal in if implies that K = N. For example, the maximal ideals in a ring are the maximal members of is an ideal of R and A ≠ R}.

Lemma 1. If M is a module, consider the following conditions where the K_i denote submodules of M. Then (ACC) (MAX) and (DCC) (MIN).

(ACC) If K₁⊆ K₂ ⊆ then K_n = K_n+1 = for some n.

(MAX) Every nonempty set of submodules of M has a maximal member.

(DCC) If K₁⊇ K₂ ⊇ then K_n = K_n+1 = for some n.

(MIN) Every nonempty set of submodules of M has a minimal member.

Proof. If (MAX) holds then (ACC) holds where K_n is any maximal member of {K_i i ≥ 1}. Conversely, suppose that is a nonempty set of submodules of M that has no maximal member. Choose Since K₁ is not maximal in there exists such that K₁ ⊂ K₂. But K₂ is also not maximal in so we obtain K₁ ⊂ K₂ ⊂ K₃ for some This process continues indefinitely ¹²⁹ to violate (ACC). This proves that (ACC) (MAX); the proof that (DCC) (MIN) is analogous.

The notations ACC and DCC refer to the ascending and descending chain conditions, on submodules, respectively. A module M is called noetherian if it has the ACC, and M is called artinian if it has the DCC. If R is an F-algebra with unity 1 then any left module _RM becomes an F-space via the action t x = (t1)x for all t F and x M. Hence every submodule of M is a subspace, so finite dimensional modules are both noetherian and artinian.¹³⁰

Example 1. Regarded as a module over itself, is noetherian but not artinian.

Proof. is not artinian as shows. For the converse, suppose K₁⊆ K₂ ⊆ are subgroups of Then K = ∪ K_i is a subgroup and so for some by Theorem 7 §2.4. But then k K_n for some n and it follows easily that K_n = K_n+1 = = K. Hence is noetherian.

If is a prime, let . This is an additive subgroup of and one verifies that the groups are the only subgroups of X containing (Exercise 9). The factor group is called the Prüfer p-group, and this shows that the only subgroups of are

,

where for each k. Furthermore, o(x_k) = p^k and px_k+1 = x_k hold for each k. Hence is an infinite group, but every proper subgroup is finite. Clearly

Example 2. The Prüfer group is artinian but not noetherian as a -module.

The following basic properties of the ACC and the DCC will be needed.

Lemma 2. If N ⊆ M are modules then M is noetherian (artinian) if and only if the same is true of both N and M/N.

Proof. We prove the noetherian case; the other is analogous. If M has the ACC then N has the ACC because submodules of N are submodules of M. On the other hand, every submodule X of M/N has the form X = K/N for some submodule K of M containing N by Theorem 5 §8.1. Hence every ascending chain in M/N takes the form K₁/N⊆ K₂/N ⊆ where K₁⊆ K₂ ⊆ in M. It follows that M/N has the ACC.

Conversely, let K₁⊆ K₂ ⊆ be a chain from M. If both N and M/N have the ACC then the chains N∩ K₁ ⊆ N ∩ K₂ ⊆ and both terminate, so there exists n ≥ 1 such that N ∩ K_n = N ∩ K_n+1 = and (whence N + K_n = N + K_n+1 = ). Since K_i ⊆ K_i+1 for each i, we have K_i+1 ∩ (K_i + N) = K_i + (K_i+1 ∩ N) by the modular law (Theorem 2 §7.1). So if i ≥ n we have

This proves that M is noetherian.

Corollary. A sum M = M₁ + + M_n of modules is artinian (noetherian) if and only if the same is true of each M_i.

Proof. We prove only the artinian case, the other being analogous. If M is artinian, so are its submodules M_i by Lemma 2. Conversely, if n ≥ 2, assume inductively that K = M₂ + + M_n is artinian. By Corollary 2 of Theorem 1 §7.1, it follows that M/K = (M₁ + K)/K ≅ M₁/(M₁ ∩ K) is also artinian by Lemma 2, being an image of M₁. Hence, M is artinian, again by Lemma 2, as required.

Let M₁, . . ., M_n be modules. The external direct sum M = M₁ ⊕ ⊕ M_n is artinian (noetherian) if and only if the same is true of each M_i by the Corollary because M is a sum of submodules isomorphic to the M_i.

Because the -action on an abelian group is naturally written on the left, we discussed only left R-modules in Chapter 7. However, an additive abelian group M is called a right R-module (written M_R) if R acts on the right: That is, for any r R and x M, an element xr M is defined such that

(x + y)r = xr + yr,  x(r + s) = xr + xs,  (xr)s = x(rs),  and x1 = x

hold for all x, y M and all r, s R.¹³¹ With this, the definition of submodules and homomorphisms of right modules are analogous to those for left modules. Moreover, the analogues of theorems about left modules in Section 7.1 go through verbatim for right modules. Note that the distinction does not matter for a commutative ring R because a left R -module M becomes a right module if we define the action by x · r = rx for all r R and x M.

A ring R is called left artinian (left noetherian) if _RR is artinian (noetherian), with similar definitions on the right. The submodules of _RR (R_R) are the left (right) ideals.

Example 3. In each case consider the subring R of

(1) is left noetherian but not right noetherian.

(2) is left artinian but not right artinian.

Solution.

(1) If then A is an ideal of R and is noetherian as a ring by the Corollary to Lemma 2 (verify). Hence, R/A is noetherian as a left R-module (the left R-submodules of R/A are just the left ideals of the ring R/A). Since _RA is also noetherian (verify), it follows by Lemma 2 that R is left noetherian. But the sequence of right ideals shows that R is not right noetherian. This proves (1).

(2) Observe that is not finite dimensional as a -space (otherwise would be countable contradicting Cantor's theorem from set theory). Hence, there exist -subspaces X₁⊃ X₂ ⊃ in But then are right ideals of R, proving that R is not right artinian. The proof that R is left artinian is analogous to the argument in (1).

We note in passing that, despite Example 3, every left artinian ring is automatically left noetherian; this result is called the Hopkins–Levitzky theorem, proved independently in 1939 by Charles Hopkins and Jacob Levitzki.

A left R-module _RM is said to be simple if it is nonzero and satisfies the following equivalent conditions:

Thus, the simple -modules are the cyclic groups where is a prime. Every simple module _RM is principal, that is, M = Rx for some x M. If R = D is a division ring the converse holds: If _DM = Dx is simple, then M ≅ _DD via dx ↔ d. In particular, the simple modules over a field are the one-dimensional vector spaces.

As for groups, a series M = M₀ ⊃ M₁ ⊃ ⊃ M_n = 0 of submodules of a module M is called a composition series of length n for M if all the factors M_i/M_i+1 are simple modules (see Section 9.1).

Theorem 1. Let M ≠ 0 be a module.

Proof.

(1) If M = M₀ ⊃ M₁ ⊃ ⊃ M_n = 0 is a composition series then M_n−1 and M_n−2/M_n−1 are both simple, so M_n−2 is noetherian and artinian by Lemma 2. Then the same is true of M_n−3 because M_n−3/M_n−2 is simple. Continuing we see that every M_k (including M₀ = M) is noetherian and artinian.

Conversely, let M be noetherian and artinian. Since M is artinian, let K₁ ⊆ M be a simple submodule. If K₁ ≠ M, choose K₂ minimal in the set {K K ⊃ K₁}. Then K₂ ⊃ K₁ ⊃ 0 and K₂/K₁ is simple. If K₂ ≠ M, let K₃ ⊃ K₂ ⊃ K₁ ⊃ 0 where K₃/K₂ simple. Since M is noetherian, this process cannot continue indefinitely, so some K_n = M and we have created a composition series.

(2) The analogue of the proof of the Jordan-Hölder theorem for arbitrary groups (Theorem 1 §9.1) goes through.

The length n of any composition series for M is called the composition length of M and denoted length (M). Note that Lemma 2 and Theorem 1 combine to show that, if K ⊆ M are modules, then M has a composition series if and only if both K and M/K have composition series. Moreover, in this case the proof of Theorem 2 §9.1 goes through to show that

length (M) = length (K) + length (M/K), (*)

and that the composition factors of M are exactly those of K and M/K.

The finitely generated vector spaces over a field are called finite dimensional, and an Theorem 6 §6.1 shows that they all have a finite basis. The same is true for any division ring, but we give a different proof using the Jordan–Hölder theorem.

Corollary. Let D be a division ring and let _DM be a module. Then

Proof. (1) Since M is finitely generated, let n ≥ 1 be minimal such that M has a set {x₁, , x_n} such that M = Σ_iDx_i. If Σ_id_ix_i = 0 and d_k ≠ 0 for some k, then (since it follows that M = Σ_i≠kDx_i contradicting the minimality of n. So {x₁, . . ., x_n} is independent and hence is a basis of M. This proves (1).

(2) If {x₁, . . ., x_n} is a basis, then M = Dx₁ ⊕ ⊕ Dx_n where Dx_i is simple, and we obtain a composition series

M ⊃ Dx₂ ⊕ ⊕ Dx_n ⊃ ⊃ Dx_n−1 ⊕ Dx_n ⊃ Dx_n ⊃ 0

for M. Hence n is the composition length of M, proving (2).

(3) If K⊆ _DM is a submodule, then K has a composition series by Lemma 2 and Theorem 1, and the composition length is at most n by (*).

If D is a division ring, the number of elements in any finite basis of a module _DM is called the dimension of M and denoted dim M.¹³² With this, most of the theorems about finite dimensional vector spaces (Section 6.1) go through for finitely generated modules over D.

Endomorphism Rings

If R is a ring and M and N are two R-modules, recall that a map α: M → N is called R-linear (or an R-morphism) if α(x + y) = α(x) + α(y) and α(rx) = rα(x) for all x, y M and all r R. Many results about rings (in particular, Wedderburn's theorem) arise from representing them as rings of R-linear maps.

If M and N are two modules write

hom (M, N) = {α α: M → N, α is R-linear}.

If α, β hom (M, N), define α + β and −α by

(α + β)(x) = α(x) + β(x)  and (− α)(x) = − α(x), for all x M.

These are R-linear and make hom (M, N) into an abelian group. Furthermore, composition of maps distributes over this addition:

  γ(α + β) = γα + γβ whenever M → ^α,β N → ^γ K are R-linear,

  (α + β)δ = αδ + βδ whenever are R-linear.

All these routine verifications are left to the reader.

Our interest here is in a special case: If M is any module, an R-linear map α: M → M is called an endomorphism of M, and we write

end M = hom (M, M).

The additive abelian group end M becomes a ring, called the endomorphism ring of M, if we define addition as above and use composition of maps as the multiplication. Again we leave to the reader the routine verifications of the ring axioms, and that end M ≅ end N whenever M ≅ N. Note that the unity of end N is the identity map 1_M.

If S is a ring, denote the ring of all n × n matrices over S by M_n(S). Wedderburn's theorem asserts that certain rings are isomorphic to M_n(D) where D is a division ring. The next result shows how matrix rings arise as endomorphism rings. If M is a module, recall that Mⁿ denotes the external direct sum of n copies of M.

Lemma 3. Let _RM be a module and write S = end M for the endomorphism ring. Then end(Mⁿ) ≅ M_n(S) as rings for each integer n ≥ 1.

Proof. Identify Mⁿ with the set of n-tuples from M, and let be the standard maps: σ_i(x) = (0, . . ., x, . . ., 0), where x is in position i, and π_i(x₁, x₂, . . ., x_n) = x_i. Then one verifies that

  and  π_jσ_i = δ_ij1_M, for all j and i,

where δ_ij = 0 or 1 according as i ≠ j or i = j. ¹³³ Given α end (Mⁿ), we have π_iασ_j S for all i and j, so we define

θ: end (Mⁿ) → M_n(S) by θ(α) = [π_iασ_j].

It routine to see θ(α + β) = θ(α) + θ(β) for all α and β, and that the identity matrix. The (i, j) entry of the matrix θ(α)θ(β) is

Σ_k(π_iασ_k)(π_kβσ_j) = π_iα(Σ_kσ_kπ_k)βσ_j = π_iαβσ_j,

so θ(α)θ(β) = θ(αβ). Thus, θ is a ring homomorphism; we claim it is an isomorphism.

If θ(α) = 0 then π_iασ_j = 0 for all i and j, so

This shows that θ is one-to-one, and it remains to show that θ is onto. To this end, let [γ_ij] M_n(S), and define α: Mⁿ → Mⁿ by

.

Then, for every x M we have

π_iασ_j(x) = π_i(α(0, . . ., x, . . ., 0)) = π_i(γ_1j(x), γ_2j(x), . . ., γ_nj(x)) = γ_ij(x).

It follows that π_iασ_j = γ_ij for all i and j, that is θ(α) = [γ_ij]. This shows that θ is onto, and so proves the lemma.

If V is an n-dimensional vector space over a field F, then V ≅ Fⁿ so Lemma 3 shows that end (V) ≅ M_n(F), a familiar fact from linear algebra.

We say that a mapping acts as a left operator x β(x) because we write β on the left of its argument. If we have the composite map αβ: M → K where αβ(x) = α[β(x)] for all x M. Hence, because α and β are left operators, the notation αβ means “ first β then α”. This is somewhat unfortunate because the order gets reversed, but the use of left operators is very common. On the other hand, we could write a map on the right of its argument, so that x xβ. In this case the composite is given by x(βα) = (xβ)α, so we write it as βα: M → K. In particular, βα and means “ first β then α ” in the same order as the arrows. Hence composition means a different thing depending on whether the maps act on the left or the right, and we must always be clear which we are using. Lemma 4 below gives a good illustration of the use of right operators.

An element e R is called an idempotent if e² = e. Note that 0 and 1 are idempotents in any ring, and they are the only ones in a division ring (in fact in a domain). If e R is an idempotent, we define eRe ⊆ R as follows:

eRe = {ere r R} = {s R es = s = se} = eR ∩ Re.

Then eRe is a ring with unity e, called the corner ring corresponding to e. The name comes from the following example: If where F is a field, and if then e² = e and

These corner rings arise as endomorphism rings in a natural way.

Lemma 4. If R is a ring and e² = e R, then eRe ≅ end(Re) where endomorphisms of Re act as right operators.

Proof. Given a eRe, define a right operator

ρ_a: Re → Re by xρ_a = xa, for all x Re.

Note that xa Re because a eRe ⊆ Re. Then ρ_a preserves addition and, in fact ρ_a end(Re) because (rx)ρ_a = (rx)a = r(xa) = r(xρ_a) for all r R and x Re.¹³⁴ Hence, we may define

θ: eRe → end(Re) by θ(a) = ρ_a for each a eRe.

We show that θ is a ring isomorphism. We have θ(e) = ρ_e = 1_Re, so θ preserves the unity. If a, b eRe, we have ρ_a+b = ρ_a + ρ_b (verify) so θ preserves addition. Also ρ_ab = ρ_aρ_b because xρ_ab = x(ab) = (xa)b = (xρ_a)ρ_b = x(ρ_aρ_b) for all x Re, so θ preserves multiplication. Hence θ is a ring homomorphism. Moreover, θ is one-to-one because θ(a) = 0 implies that ρ_a = 0, that is, xa = 0 for all x Re. Taking x = e gives that ea = 0, so a = 0 because a eRe.

To show that θ is onto, let λ: Re → Re be R -linear, and define a = eλ. Then ae = a because a Re, and ea = e(eλ) = (e²)λ = eλ = a. Hence a eRe. Finally, if x Re then x = xe, so xλ = (xe)λ = x(eλ) = xa = xρ_a. Since this holds for all x Re, we have λ = ρ_a = θ(a), so θ is onto. This completes the proof.

Taking e = 1 in Lemma 4 gives a very important special case.

Corollary. If R is a ring then R ≅ end (_RR) as rings where the endomorphisms of _RR are right multiplications by elements of R.

Wedderburn's Theorem

Wedderburn's theorem asserts that every simple, left artinian ring is isomorphic to a matrix ring over a division ring. The division ring comes from the next result, due in 1905 to Issai Schur.

Lemma 5. Schur's Lemma. Let M and N be simple modules.

Proof. (1) Observe that ker α and im α = α(M) are submodules of M and N, respectively. If α ≠ 0 then ker α ≠ M and α(M) ≠ 0, so ker α = 0 and α(M) = N by simplicity. Hence α is an isomorphism, proving (1). Now (2) follows if N = M.

In 1907, J.H.M. Wedderburn proved the following important theorem for finite dimensional algebras. A ring R is called simple if the only ideals are 0 and R.

Theorem 3. Wedderburn's Theorem. The following conditions are equivalent for a ring R:

Furthermore, the integer n is uniquely determined by R, as is the division ring D up to isomorphism.

Proof. (1)⇒(2) is clear, and (4) follows by the left–right symmetry in (3).

(2)⇒(3). Let L be a simple left ideal of R, and recall that LR is the set of all finite sums of products ba where b L and a R. Then LR is an ideal of R containing L, so LR = R because R is a simple ring. In particular, let 1 = b₁a₁ + b₂a₂ + + b_na_n where n ≥ 1, and where b_i L and a_i R for each i. Assume that n is the smallest positive integer with this property. The reader can verify that

R = La₁ + La₂ + + La_n. (*)

Note that La_i ≠ 0 for each i because b_ia_i ≠ 0 by the minimality of n. Hence L ≅ La_i because the map x xa_i is a nonzero, onto, R-morphism L → La_i, and L is simple. In particular each La_i is simple.

Now we claim that the sum in (*) is direct. By Theorem 3 §7.1, we must show that La_k ∩ (Σ_i≠kLa_i) = 0 for each k. Suppose not. Then La_k ∩ (Σ_i≠kLa_i) is a nonzero left deal contained in La_k. Hence, La_k ∩ (Σ_i≠kLa_i) = La_k because La_k is simple. But then La_k ⊆ Σ_i≠kLa_i, and it follows that R = Σ_i≠kLa_i, contrary to the minimality of n. Hence (*) is a direct sum.

Since La_i ≅ L for each i, (*) gives But then Lemma 3 and the Corollary to Lemma 4 and give

Since end L is a division ring by Schur's lemma, (3) follows with D = end L.

(3)⇒(1). The ring M_n(D) is simple by the Corollary to Theorem 7 §3.3 so, given (3), it remains to show that M_n(D) is left artinian. But M_n(D) is a finite dimensional vector space over D (in fact the dimension is n²), and so is artinian as a D-space. Hence, the ring M_n(D) is left artinian because every left ideal is a D -subspace. This proves (1).

Uniqueness. The fact that _RR ≅ Lⁿ shows that n is the composition length of _RR and so is uniquely determined by R. To show that D is uniquely determined, we prove that D ≅ end K for any simple left ideal K of R. But the proof of (2)⇒(3) shows that R ≅ Kⁿ, and hence that Lⁿ ≅ Kⁿ. We have maps , where τ is an isomorphism, σ₁ is the inclusion, and the π_i are the projections. Then π_iτσ₁ ≠ 0 for some i because τσ₁(L) ≠ 0. Hence L ≅ K by Schur's lemma, and so end K ≅ end L = D, as required.

Remark. The “left artinian” condition in (1) of Theorem 3 cannot be replaced with “ left noetherian”. In fact, there exists a simple, left and right noetherian domain that contains no simple left ideal. It is called the first Weyl algebra, after Hermann Weyl, and can be described roughly as the ring of polynomials over in noncommuting indeterminates x and y which satisfy the condition that xy − yx = 1. This ring first arose in quantum mechanics as an algebra generated by position and momentum operators.

Exercises 11.1

11.2 The Wedderburn–Artin Theorem

The proof of Wedderburn's theorem (Theorem 3 §11.1) shows that if R is a simple ring containing a simple left ideal L, then R = L₁ ⊕ L₂ ⊕ ⊕ L_n where the L_i are isomorphic, simple left ideals (in fact L_i ≅ L for each i). Wedderburn actually treated the case where the L_i are not necessarily all isomorphic, albeit in the special case when R is a finite dimensional algebra. We deal with this general case in this section, in the context of rings.

The work necessitates a look at modules that are the sum of a (possibly infinite) family of simple submodules. This investigation provides an extension of the well-known theory of vector spaces over any division ring. However, not surprisingly, transfinite methods are required.

The technique we need is called Zorn's lemma. Let be a nonempty family of subsets of some set. A chain from is a (possibly infinite) set where either X_i ⊆ X_j or X_j ⊆ X_i for any i, j I. We say that is inductive if the union of any chain from is again in Zorn's lemma asserts that every inductive family has a maximal member, that is a set such that implies Y = Z. A more general version of Zorn's lemma is discussed in Appendix C.

Semisimple Modules

Let R be a ring, and let {M_i i I} be a (possibly infinite) family of submodules of a module M. The sum Σ_iIM_i of these submodules is defined to be the set of all finite sums of elements of the M_i; more formally

Σ_iIM_i = {x₁ + + x_m m ≥ 1, x_j M_j for some j I}.

This is a submodule of M that contains every M_i. The following lemma is the extension of Theorem 3 §7.1 to the case of infinite sets of submodules.

Lemma 1. The following are equivalent for submodules {M_i i I} of a module:

Proof. (1) ⇒ (2). If x₁ + + x_m = 0 = 0 + + 0 then each x_i = 0 by (1).

(2) ⇒ (3). If where the i_j are distinct, then N is direct by (2) and Theorem 3 §7.1.

(3) ⇒ (1). If x M let x = x₁ + + x_m = y₁ + + y_k where and Inserting zeros where necessary, we may assume that x_i and y_i are in for each i, and that these are distinct. Hence, x₁ + + x_m = y₁ + + y_k in for distinct i_jby (3). Thus, x_i = y_i for each i by Theorem 3 §7.1, proving (1).

In this case we say that Σ_iIM_i is a direct sum, and we write it as ⊕_iIM_i. One reason for introducing these infinite direct sums here is that they provides the language needed to generalize some of the results about vector spaces.

If D is a division ring and _DM is an module, let B = {b_i i I} be a (possibly infinite) set of nonzero elements of M. Then B is said to span M if M = Σ_iIDb_i. The set B is called independent if Σr_ib_i = 0, r_i D, implies that r_i = 0 for each i, equivalently if M = ⊕ _iIDb_i (since D is a division ring). A set B that is both independent and spans M is called a basis of M. Furthermore, each principal module Db_i is simple (again because D is a division ring). In this form, these facts can be stated much more generally.

It is shown in Appendix C that every module M over a division ring D has a basis. By the above discussion this means that M is a direct sum of simple submodules. In general, a module _RM over a ring R is called semisimple if it is the direct sum of a (possibly infinite) family of simple submodules. The following example shows that _RR is semisimple if R is the 2 × 2 matrix ring over a division ring.

Example 1. Let D be a division ring, and let If and , show that L₁ and L₂ are each simple left ideals of R, that R = L₁ ⊕ L₂, and that L₁ ≅ L₂ as R-modules.

Solution. L₁ and L₂ are left ideals by the definition of matrix multiplication, and it is clear that R = L₁ + L₂ and L₁ ∩ L₂ = 0. Hence R = L₁ ⊕ L₂. We show that L₁ is simple; the proof for L₂ is similar. So let say a ≠ 0. Given we have so L₁ = Rx. A similar argument shows that L₁ = Rx when b ≠ 0.

Finally, if and then it is a routine matter to show that the maps L₁ → L₂ given by x xa and L₂ → L₁ given by x xb are mutually inverse R-isomorphisms. Hence L₁ ≅ L₂.

We are going to give several characterizations of semisimple modules, and the following notion is essential. A module M is said to be complemented if every submodule K is a direct summand, that is if M = K ⊕ N for some submodule N. Hence every simple module is complemented, as is every finite dimensional vector space (Theorem 8(3) §6.1). A submodule N of a module M is called proper if N ≠ M, and a maximal proper submodule is called simply a maximal submodule.

Lemma 2. Let M be a complemented module. Then

Proof. (1) If K ⊆ N let K ⊕ K₁ = M. Then N = N ∩ (K ⊕ K₁) = K ⊕ (N ∩ K₁) by the modular law (Theorem 2 §7.1).

(2) If x ∉ N, write is a submodule, N ⊆ P and x ∉ P}. Then is nonempty and inductive (verify) so, by Zorn's lemma, choose a submodule P maximal in Since M is complemented, let M = P⊕ Q; we show that P is maximal by showing that Q is simple. If not, let A ⊆ Q where A ≠ 0, Q. By (1) write Q = A ⊕ B. Then P ⊕ A and P ⊕ B both strictly contain P, so neither is in Hence x lies in both by the maximality of P, say x = p₁ + a = p₂ + b where p₁, p₂ P, a A and b B. Since a Q and b Q, and since P ⊕ Q is direct, it follows that a = b. But then a = b A ∩ B = 0, so x = p₁ P, a contradiction.

Lemma 3. Let M = Σ_iIM_i where each M_i is simple. If N is any submodule of M there exists J ⊆ I such that M = N ⊕ (⊕ _jJM_j). In particular, M is complemented.

Proof. If N = M take J = ∅. If N ≠ M then some M_j N so N ∩ M_j = 0 (because M_j is simple). Hence, is nonempty where

is a direct sum}.

Let {J_i} be a chain from and write J = ∪ J_i. To see that J is in let

,

where x N and each Since {J_i} is a chain, there exists some J_k containing all these i_t so (since it follows that for each t. Hence N + Σ_iJM_i is direct, which shows that

Hence, by Zorn's lemma, has a maximal member J₀. If it remains to show that L = M. Since M = ΣM_i, we show that M_i ⊆ L for each i I. This is clear for all i J₀. If i₀ I − J₀, then so is not direct, that is, Hence, as is simple, so as required.

With this we can characterize the semisimple modules.

Theorem 1. The following conditions are equivalent for a module M ≠ 0:

Proof. (1)⇒(2) and (2)⇒(3). These are by Lemma 3, the first with N = 0.

(3)⇒(4). This is clear by Lemma 2.

(4)⇒(1). Let S denote the sum of all simple submodules of M (take S = 0 if there are none). Suppose S ≠ M. By (4) let S ⊆ N where N is maximal in M, and write M = N ⊕ K for some submodule K, again by (4). Then K is simple (because K ≅ M/N) so K ⊆ S ⊆ N, contrary to K ∩ N = 0. So S = M and (1) follows.

A module M ≠ 0 is called semisimple if it satisfies the conditions in Theorem 1, We also regard 0 as a semisimple module. Hence every module over a division ring is semisimple, and the ring in Example 1 is semisimple as a left module over itself. Lemma 3 gives

Corollary 1. If M is semisimple so also is every submodule and image of M. More precisely, if N ⊆ M = ⊕ _iIM_i where each M_i is simple, then

N ≅ ⊕ _iJM_i  and  M/N ≅ ⊕ _iI−JM_i for some J ⊆ I.

Note that we need not have N = ⊕ _iJM_i in Corollary 1. As an example, let where F is a field. Then M = M₁ ⊕ M₂ where and But if N = {(a, a) a F} then M_i6 ⊆ N for i = 1, 2.

Corollary 2. If M is finitely generated then M is semisimple if and only if every maximal submodule is a direct summand.

Proof. By Theorem 1, it remains to show that every proper submodule K ⊂ M is contained in a maximal submodule of M. To this end, let we show that contains maximal members (they are then maximal in M). So let {X_i k I} be a chain from and write X = ∪ _iIX_i. By Zorn's lemma it suffices to show that X ≠ M. But if X = M, let M = Rm₁ + + Rm_k (as M is finitely generated). Then each for some i_t I. But the X_i are a chain so there exists k I such that for each i_t. Hence each m_t X_k, so M ⊆ X_k, a contradiction. Hence X ≠ M after all.

Lemma 4. The following are equivalent for a semisimple module M:

Finally the decomposition in (2) is unique: If M = N₁ ⊕ N₂ ⊕ ⊕ N_m, where each N_i is simple then m = n and (after relabeling) N_i ≅ M_i for each i.

Proof. (1)⇒(2). If M = ⊕ _iM_i where the M_i are simple, the generators of M all lie in a finite sum by (1).

(2)⇒(3) and (2)⇒(4). By (2), M ⊃ M₂ ⊕ ⊕ M_n ⊃ ⊃ M_n ⊃ 0 is a composition series for M so (3) and (4) follow from Theorem 1 §11.1.

(3)⇒(1) and (4)⇒(1). If M = ⊕ _iIM_i where I is infinite, we may assume that {1, 2, 3, } ⊆ I. Then (M₁⊕ M₂ ⊕ M₃ ⊕ ) ⊃ (M₂ ⊕ M₃ ⊕ ) ⊃ contradicts (3) and M₁⊂ M₁ ⊕ M₂ ⊂ contradicts (4).

Finally, as we saw above, (2) gives rise to a composition series M with factors M₁, M₂, . . ., M_n. Hence, the last statement in the corollary follows from the Jordan–Hölder theorem (Theorem 1 §11.1).

Note that Lemma 4 proves the uniqueness of the number of elements in a finite basis of a module over any division ring (so we can speak of dimension). In fact, a version of the uniqueness actually goes through for arbitrary infinite direct sum decompositions of a semisimple module, a result beyond the scope of this book.

Homogeneous Components

Before proceeding we need a technical lemma.

Lemma 5. Let M = ⊕ _iIM_i with each M_i simple and let K ⊆ M be a simple submodule, Then there exist i₁, i₂, . . ., i_m in I such that and for each t = 1, 2, . . ., m.

Proof. Choose 0 ≠ y K, and write , where for each t. Then so, because K is a simple module, Because this is a direct sum, we can define as follows: If x K take , where and for each j. Then α_t is R-linear for each t, and α_t ≠ 0 because So α_t is an isomorphism by Schur's lemma.

If K ⊆ M are modules with K simple, define the homogeneous component H(K) of M generated by K as follows:

H(K) = Σ{X ⊆ M Xis a submodule and X ≅ K}.

We say that H(K) = 0 if M contains no copy of K. Hence, if K ⊆ M is simple then H(K) is a submodule of M containing every submodule isomorphic to K (and hence K itself). In fact every simple submodule of H(K) is isomorphic to K.

Lemma 6. Let K ⊆ M be modules with K simple. The following are equivalent for a simple submodule L ⊆ M:

Proof. (1)⇒(2). This is because X ≅ L if and only X ≅ K by (1).

(2)⇒(3). Here L ⊆ H(L) = H(K) by (2).

(3)⇒(1). If L ⊆ H(K) = Σ{X ⊆ M X ≅ K}, then L ≅ X for some X ≅ K by Lemma 5. This proves (1).

A submodule N of M is said to be fully invariant in M if α(N) ⊆ N for every endomorphism α: M → M. Clearly 0 and M are fully invariant for every module M. Recall that the endomorphisms of _RR are just the right multiplications by elements of R (Corollary to Lemma 4 §11.1). It follows that a left ideal L of R is fully invariant in _RR if and only if L is an ideal of R.

We can now prove the main structure theorem for semisimple modules. For convenience, we say that submodules K and N meet if K ∩ N ≠ 0.

Theorem 2. Let M be a semisimple module. Then the following hold:

Proof. Let {H(K_i) i I} be the distinct homogeneous components of M.

(1) M = Σ_iIH(K_i) because M is semisimple; to see that this sum is direct we show that H(K_t) ∩ [Σ_i≠tH(K_i)] = 0 for each t I, and invoke Lemma 1. But if H(K_t) ∩ [Σ_i≠tH(K_i)] ≠ 0 then it contains a simple submodule L (being semisimple). Then L ≅ K_t by Lemma 5 because L⊆ H(K_t); and L≅ K_i for some i ≠ t by Lemma 5 because L ⊆ Σ_i≠tH(K_i). Hence, K_t ≅ L ≅ K_i and so H(K_t) = H(K_i) by Lemma 6, contrary to the choice of the H(K_i).

(2) Consider H(K), where K ⊆ M is simple. If α end M we must show that α[H(K)] ⊆ H(K), that is α(L) ⊆ H(K) whenever L ⊆ M and L ≅ K. But since L is simple, either α(L) = 0 or α(L) ≅ L ≅ K by Schur's lemma. Either way, α(L) ⊆ H(K).

(3) Let the K_i be as above, and let N ⊆ M be fully invariant. We show that N = Σ{H(K_i) H(K_i) ∩ N ≠ 0}. Every simple submodule of M is in some H(K_i) by Lemma 5, so N ⊆ Σ{H(K_i) H(K_i) ∩ N ≠ 0} because N is semisimple. Conversely, if H(K_i) ∩ N ≠ 0 then there exists U ⊆ H(K_i) ∩ N such that U ≅ K_i. If L ≅ K_i is arbitrary then L ≅ U by Lemma 5, so let σ: U → L be an isomorphism. Since M is complemented (Theorem 1), write M = U ⊕ U₁, and define α: M → M by α(u + u₁) = σ(u). Then L = σ(U) = α(U) ⊆ α(N) ⊆ N, where α(N) ⊆ N because N is fully invariant. Since L ≅ K_i was arbitrary, it follows that H(K_i) ⊆ N.

A semisimple module M is called homogeneous if it has only one homogeneous component, that is (by Lemma 6) if all simple submodules of M are isomorphic. In particular if D is a division ring and R = M₂(D), then Example 1 shows that _RR is a homogeneous, semisimple module. In fact much more is true as we shall see below.

Free and Projective Modules

The Wedderburn–Artin theorem shows that rings R such that _RR is semisimple as a left R-module are isomorphic to finite direct products of matrix rings over division rings. Other equivalent conditions on R are also considered.

Finitely generated free and projective modules were discussed in Section 7.1. However, the Wedderburn–Artin theorem involves arbitrary projective modules, so we pause to review these notions. Let _RW be a module, and let B be a set of nonzero elements of W. We make three definitions:

B is said to generate W if  

B is called independent  if   implies each r_i = 0.

B called a basis of W if it is independent and generates W.

A module _RW that has a basis is called a free module. Note that the second condition above implies that Rw_i ≅ _RR for each i B (see the corollary to Theorem 1 §7.1). Hence, every free module is isomorphic to a direct sum of copies of R. The finitely generated free modules in Section 7.1 are examples, but free modules with bases of arbitrary size can easily be constructed.

If I is any nonempty set and R is any ring, there exists a free R -module with a basis indexed by I. If i r_i is any function I → R, write it as Hence if and only if r_i = s_i for each i I, and we call r_i the ith component of We call an I-sequence from R. Define

for all but finitely many i I }.

If I has n elements it is clear that R^(I) = Rⁿ. In general, R^(I) becomes a left R-module with componentwise operations:

If e_i denotes the I-sequence with ith component 1 and all other components 0, then it is a routine matter to check that {e_i i I} is a basis of R^(I), called the standard basis. Hence, R^(I) = ⊕ _iIRe_i, where Re_i≅ _RR for each i.

Now let {x_i i I} be a generating set for a module _RM, that is, M = Σ_iIRx_i. Let W be a free module with a basis indexed by I (for example W = R^(I)). Given r₁, r₂, . . ., r_n in R, define

β: W → M  by 

Because B is a basis, this map is well defined, and it is evidently onto. Since every module M has a generating set (the set of all nonzero elements, for example), this proves the first part of

Lemma 7. Let M denote any left R -module.

Proof. We proved (1) above. As to (2), let be a basis of W. As α is onto, let , x_i M for each i. By the above discussion, there exists β: W → M such that for each i. Then for each i, so αβ = 1_W because the generate W. But then M = ker α ⊕ β(W) by Lemma 8 below, proving (2).

Lemma 8. If α: M → P is onto, the following conditions are equivalent:

In this case the map α is said to split.

Proof. (1) ⇒ (2). If αβ = 1_P as in (1), let m M. As α(m) P, we have α(m) = 1_P[α(M)] = αβα(m). Thus m − βα(m) ker α, so M = ker α + β(P). But if m ker α ∩ β(P), let m = β(p), p P. Then 0 = α(m) = αβ(p) = p, so m = β(p) = β(0) = 0. This proves that ker α ∩ β(P) = 0 and so proves (2).

(2) ⇒ (1). Given (2), let M = ker α ⊕ Q. Observe that P = α(M) = α(Q), so we define β: P → M as follows: if p P and p = α(q), q Q, define β(p) = q. This is well defined because if p = α(q₁), q₁ Q, then q − q₁ Q ∩ ker α = 0. Now given p = α(q) in P then αβ(p) = α(q) = p, proving (1).

A module _RP is called projective if it satisfies the condition:

is R -linear and onto then ker α is a direct summand of M.

Hence all free modules are projective by Lemma 7 (but see Example 3 below). Also, P is projective if and only if every onto R-morphism M → P splits. We need

Lemma 9. If P is projective and Q ≅ P, then Q is projective.

Proof. Let α: M → Q be onto. If σ: Q → P is an isomorphism then σα: M → P is onto so ker α = ker σα is a direct summand of M. Hence, Q is projective.

Theorem 3. The following conditions on a module _RP are equivalent:

Proof. (1)⇒(2). If is onto, then W = ker α ⊕ Q for some Q by (1). Hence, P = α(P) ≅ W/ker α ≅ Q, proving (2).

(2)⇒(3). Since being projective is preserved under isomorphism, we may assume that W = P ⊕ Q is free. Let π: W → P be the projection defined by π(p + q) = p for p P and q Q, and let be a basis of W. Given α as in the diagram, we must construct γ: P → M such that αγ = β.

Note that for each i I. Since α: M → N is onto, there exists m_i M such that But the are a basis of W, so there exists λ: W → M such that for each i. Hence,

,  for each i.

It follows that αλ = βπ because the generate W. Finally, let γ: P → M be the restriction of λ to P, that is, γ(p) = λ(p) for all p P. Compute

αγ(p) = αλ(p) = βπ(p) = β(p), for all p P,

because π(p) = p. Hence, αγ = β, as required.

(3)⇒(4). Take N = P and β = 1_P in the diagram.

(4)⇒(1). This is Lemma 8.

Example 2. A module _RP is a principal projective module if and only if for some e² = e R.

Solution. If P = Rx is projective, define α: R → Rx by α(r) = rx for all r R. Then α is onto so _RR = ker α ⊕ L for some left ideal L. Hence P ≅ R/ker α ≅ L. But direct summands of _RR have the form Re for some idempotent e by Example 6 §7.1, so we have P ≅ Re for some e² = e R.

Conversely, if e² = e then Re ⊕ R(1 − e) = R is free, so Re is projective by Theorem 3.

Example 3. There exist projective modules that are not free.

Solution. Let F be a field, and let R = F × F. Then e = (1, 0) is an idempotent in R, so Re = {(a, 0) a R} is projective. But dim _F(Re) = 1 so Re is not free since any free R -module has F-dimension at least 2 (because dim _FR = 2).

The Wedderburn–Artin Theorem

Let A be a left ideal of a ring R. If X is a nonempty subset of a module _RM, define AX to be the set of all finite sums of elements ax where a A and x X. This is a submodule of M, and it is easy to verify that

(A + B)X = AX + BX and A(BX) = (AB)X

hold for all left ideals A and B. Note that RX = X if X is a submodule of M. Taking M = R in the above discussion shows that multiplication of left ideals is associative: A(BC) = (AB)C for any left ideals A, B, and C.

An ideal A is called nilpotent if Aⁿ = 0 for some n≥ 1; equivalently if any product of n elements of A is zero. The next result, proved in 1942 by Richard Brauer, characterizes the non-nilpotent, simple left ideals.

Lemma 9. Brauer's Lemma. Let K be a simple left ideal of a ring R. Then either K² = 0 or K = Re for some nonzero idempotent e² = e K.

Proof. Assume that K² ≠ 0, so that Ka ≠ 0 for some 0 ≠ a K. Since Ka ⊆ K is a left ideal, it follows that Ka = K because _RK is simple. In particular ea = a for some 0 ≠ e K. Hence e²a = ea, so e² − e B = {b K ba = 0}. Moreover, B is a left ideal and B ≠ K because Ka ≠ 0, so B = 0, again by simplicity. This means that e² = e, and so Re ⊆ K because e K. But e ≠ 0 so Re = K by a third appeal to the simplicity of _RK. This is what we wanted.

We now turn to another property of rings that plays a prominent role in the Wedderburn-Artin theorem. A ring R is called semiprime if it satisfies the following equivalent conditions (the routine verifications are left to the reader):

Hence, every ring with no nonzero nilpotent elements is semiprime, and the converse is true for commutative rings. A product R = R₁ × R₂ × × R_n of rings is semiprime if and only if each R_i is semiprime (using condition (3)). A matrix ring M_n(R) is semiprime if and only if R is semiprime because the ideals of M_n(R) all have the form M_n(A) for some ideal A of R by Lemma 3 §3.3.

The next theorem gives several useful characterizations of when a ring R is semisimple as a left module over itself.

Theorem 4. If R is a ring the following conditions are equivalent:

Proof. (1) ⇒ (2). Each module _RM is an image of a free module, and free modules are semisimple by (1). So (2) follows from Corollary 1 of Theorem 1.

(2) ⇒ (3). Given a module M, let α: N → M be an onto R-linear map. Then ker α is a summand of N because N is semisimple by (2). This proves (3).

(3) ⇒ (1). Let L be any left ideal of R, and consider the coset map ϕ: R → R/L. Then L = ker ϕ so L is a direct summand of _RR because R/L is projective by (3). Hence _RR is complemented, and so is semisimple by Theorem 1.

(1) ⇒ (4). First, R is left artinian by Lemma 4 because _RR is finitely generated. Suppose A is an ideal of R with A² = 0. By Theorem 1 R = A ⊕ L, L a left ideal, so AL ⊆ A ∩ L = 0. Hence, A = AR = A² + AL = 0 + 0 = 0, so R is semiprime.

(4) ⇒ (1). If _RR is not semisimple, let L be minimal among nonzero left ideals of R that are not semisimple (by (4)). Since L ≠ 0, let K ⊆ L be a simple left ideal, again by (4). Then K² ≠ 0 because R is semiprime, so K = Re for some 0 ≠ e² = e by Brauer's lemma. Since R = K ⊕ R(1 − e) and K ⊆ L, we obtain

L = K ⊕ (L ∩ R(1 − e)) by the modular law (Theorem 2 §7.1).

Hence, there are two cases: either L ∩ R(1 − e) = 0 (in which case L = K is simple) or L ∩ R(1 − e) is semisimple (by the minimality of L). Either way, L is semisimple, a contradiction. So _RR is semisimple.

Note that we cannot replace “left artinian” by “left noetherian” in (4) of Theorem 4. In fact the ring of integers is noetherian and semiprime, but it is not semisimple.

The first, and possibly the most important application of the theory of semisimple modules is to prove the following fundamental theorem: If R is a ring then _RR is semisimple if and only of R is a finite direct product of matrix rings over division rings. This result was first proved in 1908 by Wedderburn for finite dimensional algebras. Then in 1927, Artin replaced the finite dimensional hypothesis by the descending chain condition on left ideals.

Theorem 5. Wedderburn–Artin Theorem. The following conditions are equivalent for a ring R:

Moreover, the integers k, n₁, . . ., n_k in (3) are uniquely determined by R as are the division rings D_i up to isomorphism.

Proof. We need prove only (1) (3) by the right–left symmetry of (3).

(1)⇒(3). Let H₁, H₂, . . ., H_m be the homogeneous components of _RR. Hence

by Theorem 2. Moreover, each H_i is an ideal of R (being fully invariant), so R ≅ H₁ × H₂ × × H_m as rings by Theorem 7 §3.4 (and induction). So, by Wedderburn's theorem, it remains to show that H_i is left artinian and simple for each i. Write H = H_i for convenience. The ring R is left artinian (by Lemma 4 since it is semisimple), so the same is true of _RH. But the left ideals of the ring H are exactly the left ideals of R that happen to be contained in H (verify). Hence _HH is artinian. Finally, if A ≠ 0 is an ideal of the ring H, then A is an ideal of R because H = Re where e² = e is central in R (verify). Hence, A is fully invariant in _RR. But then Theorem 2 shows that A = Σ{H_i A ∩ H_i ≠ 0}, and it follows that A = H. Thus, the ring H is left artinian and simple, and (3) follows by Wedderburn's theorem.

(3)⇒(1). If D is a division ring, the ring M_n(D) = K₁ ⊕ K₂ ⊕ ⊕ K_n where K_j is the left ideal of all matrices with only column j nonzero. It is a routine verification (see Example 1) that each K_j is a simple left ideal of M_n(D), so M_n(D) is semisimple. Hence, (1) follows from (3).

Uniqueness. Suppose is another such decomposition where each B_i is a division ring. Lemma 10 below shows that k = l and, after relabeling, that for each i = 1, 2, , k. We can now apply the uniqueness in Wedderburn's theorem (Theorem 3 §11.1).

The rings in Theorem 5 are called semisimple rings. The next result shows that this property is inherited by several related rings—the proofs are Exercises 10, 11, and 12.

Corollary. If R is a semisimple ring, so also is every matrix ring M_n(R), every factor ring R/A, and every corner ring eRe where e² = e.

Note that subrings of semisimple rings need not be semisimple (consider

Lemma 10. Let R₁ × R₂ × × R_k ≅ S₁ × S₂ × × S_l as rings, where each R_i and each S_j is a simple ring with a left composition series. Then k = l and, after relabeling, R_i ≅ S_i for each i.

Proof. Write and We may assume that k ≤ l. Let σ: R → S be a ring isomorphism. The ideals of S are all of the form ΠA_j, where A_j is an ideal of S_j for each j (Exercise 4). Hence, since σ(R₁) is a simple ideal of ΠS_j, it must be one of the S_j; by relabeling assume that σ(R₁) = S₁. Similarly σ(R₂) = S_j for some j, and j ≠ 1 because σ(R₁) ≠ σ(R₂), σ is one-to-one, and R₁ ∩ R₂ = 0. So, after relabeling, let σ(R₂) = S₂. Continue to obtain S = σ(R₁) × × σ(R_k) × S_k+1 × × S_l. Hence, S ≅ R × (S_k+1 × × S_l) so, since S and R have the same left composition length, the Jordan–Hölder theorem (Theorem 1 §11.1) shows that S_k+1 × × S_l has length zero. Hence, k = l and S_i = σ(R_i) ≅ R_i for each i. This completes the proof.

Wedderburn's 1908 version of Theorem 5 was a breakthrough. ¹³⁵ To quote Emil Artin, “ This extraordinary result has excited the fantasy of every algebraist and still does so to this day. Very great efforts have been directed toward a deeper understanding of its meaning.” ¹³⁶ When the Wedderburn-Artin theorem appeared in 1927 it was a landmark in algebra. It influenced a generation of ring theorists and has inspired many generalizations.

Wedderburn's theorem asserts that a left artinian simple ring is a matrix ring over a division ring. In 1945, Nathan Jacobson, and independently Claude Chevalley, extended Wedderburn's theorem by dropping the artinian hypothesis and showing that a simple ring with a simple left ideal must be isomorphic to a “ dense” subring of the ring of endomorphisms of a vector space over a division ring. This result is called the density theorem.

Also in 1945, Jacobson showed that the intersection J of all the maximal left ideals of R always equals the intersection of the maximal right ideals of R. He then proved that J is the largest ideal with the property that 1 + a is a unit for all a J, extending work of Sam Perlis done in 1942 for finite dimensional algebras over a field. The ideal J is called the Jacobson radical of the ring R. It is known that, if R is left (or right) artinian, the factor ring R/J is semisimple, and idempotents can be lifted modulo J in the sense that if a² − a J then there exists e² = e such that a − e J. In 1960 this led Hyman Bass to carry the theory further. He called a ring R semiperfect if R/J is semisimple and idempotents can be lifted modulo J, and he showed that many properties of left artinian rings carry over to these semiperfect rings.¹³⁷

Finally, the Wedderburn–Artin theorem shows that a left artinian semiprime ring is semisimple, and another natural question is what happens if we replace the left artinian condition by the requirement that the ring be left noetherian. In 1960, Alfred Goldie proved a fundamental structure theorem for the semiprime left noetherian rings. There is a way of embedding certain rings into a ring of left quotients, a noncommutative version of the construction of the field of quotients of an integral domain. Goldie showed that every left noetherian, semiprime ring has a semisimple ring of left quotients.

Exercises 11.2