FINITELY GENERATED SIMPLE ALGEBRAS: A QUESTION OF B. I. PLOTKIN
A. I. LICHTMAN AND D. S. PASSMAN
Abstract. In his recent series of lectures, Prof. B. I. Plotkin discussed ge- ometrical properties of the variety of associative K-algebras. In particular, he studied geometrically noetherian and logically noetherian algebras and, in this connection, he asked whether there exist uncountably many sim- ple K-algebras with a fixed finite number of generators. We answer this question in the affirmative using both crossed product constructions and HNN extensions of division rings. Specifically, we show that there exist un- countably many nonisomorphic 4-generator simple Ore domains, and also uncountably many nonisomorphic division algebras having 2 generators as a division algebra.
1. Introduction In his recent series of lectures [Pl], B. I. Plotkin discussed some geometrical properties of the variety of associative K-algebras. In particular, he considered geometrically noetherian and logically noetherian algebras and, in this connec- tion, he asked whether there exist uncountably many simple K-algebras with a fixed finite number of generators. We answer this question in the affirmative using both crossed product constructions and HNN extensions of division rings. To be precise, we show that there exist uncountably many nonisomorphic 4-generator simple Ore domains, and also uncountably many nonisomorphic division algebras having 2 generators as a division algebra. These results can be viewed as algebra analogs of the well-known theorem [N] which asserts that there exist uncount- ably many nonisomorphic 2-generator groups. Indeed, it was shown in [H] that there exist uncountably many nonisomorphic 6-generator simple groups (see [LS, Theorems IV.3.3 and IV.3.5]). Our first three results use crossed product constructions and are proved in the next two sections. As will be apparent, a key ingredient here is the fact that any field K has uncountably many nonisomorphic field extensions of transcendence degree 1.
2000 Mathematics Subject Classification. Primary 16S35, 16K40. The first author is grateful to Professor B. I. Plotkin for communicating this problem to him and for stimulating conversations. 1 2 A. I. LICHTMAN AND D. S. PASSMAN
Theorem 1.1. Suppose that K is a field having an element ζ of infinite mul- tiplicative order with its nth roots, for all n = 1, 2, 3,..., also contained in K. Then there exist uncountably many nonisomorphic K-algebras R such that i. R is a simple right and left Ore domain. ii. R is generated as a K-algebra by four elements. iii. Z(R) = K. Each such ring R is a crossed product E∗G, where E is a K-algebra integral domain with quotient field F having transcendence degree 1 over K. Furthermore, G is the wreath product G = C o C with C infinite cyclic. Since Z(R) = K, the above algebras are actually nonisomorphic as rings. As a consequence of the above construction, we also show Theorem 1.2. Let K be a field containing all nth roots of unity. Then there exist uncountably many nonisomorphic K-algebras D such that i. D is a division ring. ii. D is generated as a K-division algebra by two elements. iii. Z(D) = K(t), the rational function field over K in one variable t. Each such D is the division ring of fractions of a crossed product E∗G, where E is a K-algebra integral domain with quotient field F having transcendence degree 1 over K. Furthermore, G is the wreath product G = C o C with C infinite cyclic. In the next result, we allow the centers of the algebras to vary. Proposition 1.3. If K is an arbitrary field, then there exist uncountably many nonisomorphic K-algebras R such that i. R is a right and left Ore domain with division ring of fractions D. ii. R is generated as a K-algebra by four elements. D is generated as a K-division algebra by two elements. iii. F = Z(D) is the field of fractions of E = Z(R), and these fields F , each of transcendence degree 1 over K, are different K-algebras for the different choices of R. Each such ring R is a twisted group ring Et[G], where G = C o C with C infinite cyclic. Furthermore, if K is countable then these examples can be constructed with E = F and hence with R a simple ring. The remaining two results concern division algebras. They are proved, by means of HNN extensions, in the last section of this paper. Again, we use the fact that any field K has uncountably many nonisomorphic field extensions of transcendence degree 1. Theorem 1.4. If K is an arbitrary field, then there exist uncountably many nonisomorphic K-algebras D such that i. D is a division ring. ii. D is generated as a K-division algebra by two elements. iii. Z(D) = K(t), the rational function field over K in one variable t. FINITELY GENERATED SIMPLE ALGEBRAS 3
Each such D is the universal field of fractions of an HNN extension of the free field F
2. Crossed Product Constructions Let K be a field and let F be a countably generated field extension. Suppose F is such a countable generating set consisting of nonzero elements, so that F = K(F). We assume for convenience that either each element of F has infinite multiplicative order or that each element has finite multiplicative order, but that these orders are unbounded. In addition, let σ be a given field automorphism of F whose fixed field F σ contains K, and let E be the K-subalgebra of F generated by the set F ∪ F −1 and all its conjugates under the cyclic group hσi. The goal of this section is to construct four K-algebras based upon the above information. First, T = F t[A] is a twisted group algebra over F of the countably generated free abelian group A. Next, S = TC = F ∗G is a skew group ring over T of the infinite cyclic group C. It can also be viewed as a crossed product over F of the wreath product group G = C o C. Thirdly, R is the K-subalgebra of S given by R = E∗G ⊆ F ∗G = S. Then R is a right and left Ore domain, and it is a finitely generated K-algebra. Furthermore, if E is σ-simple, then R is a simple ring. In any case, we let D denote its division ring of fractions. Lemma 2.1. Construction of the twisted group algebra T = F t[A].
Proof. Suppose F = {g0, g1, g2,...} and define F1 = {f1, f2, f3,...} ⊆ F so that f1 = g0, f2 = g1, f4 = g2 and in general f2n = gn. Furthermore, set fi = 1 if i is not a power of 2. Then this new sequence contains all the given generators of F along with arbitrarily long subsequences of 1’s. Now let H be the free class 2 nilpotent group on countably many generators {ai | i ∈ Z}, where Z is the set of all integers. Then the commutator subgroup 0 H of H is central and generated by the commutators ui,j = [ai, aj] for all i, j ∈ Z 0 0 0 with i > j. Indeed, H/H is free abelian with generators {aiH | i ∈ Z} and H is free abelian with generators {ui,j | i > j}. Form the group ring F [H] and let P 0 j be the kernel of the homomorphism ϕ: F [H ] → F given by ϕ(ui,j) = σ (fi−j). Since H0 is central, it follows that Q = P ·F [H] is an ideal of F [H] and then T = F [H]/Q is clearly a twisted group algebra of A = H/H0 over F . Indeed, if 4 A. I. LICHTMAN AND D. S. PASSMAN ai denotes the image of ai in F [H]/Q, then T has as an F -basis all expressions Q ki a = i ai , with the product in the natural order. Furthermore, [ai, aj] = −1 −1 j t (ai) (aj) aiaj = σ (fi−j) for all i > j. Hence we can write T = F [A], and we let A = {a | a ∈ A} denote the corresponding basis. Lemma 2.2. T is a simple right and left Ore domain with center F .
• Proof. If a ∈ A, then a determines a map λa : A → F , the multiplicative group −1 −1 −1 of F , given by x ax = λa(x)a for all x ∈ A or equivalently λa(x) = a x ax = [a, x]. It is easy to see that each such λa is a group character, namely a group • homomorphism to F . Furthermore, λab = λaλb for all a, b ∈ A. We claim now Qn ki that if a 6= 1, then λa 6= 1. To this end, let a = m ai with kn 6= 0. Then the nature of the F1-sequence guarantees that we can find a sufficiently negative subscript j < m so that fn−j has multiplicative order larger than |kn|, while fi−j = 1 for the remaining subscripts i satisfying m ≤ i < n. Recall that the generators in F are assumed to either all have infinite multiplicative order or they all have finite multiplicative order, but that these orders are unbounded. j In particular, since λai (aj) = σ (fi−j) for i > j, we conclude that λa(aj) = Qn ki kn n kn λa (aj) = λa (aj) = σ (fn−j) 6= 1. Thus λa 6= 1, as required. i=m i n P Now suppose α = a αaa is central in T , with each αa ∈ F . Then for P −1 P all x ∈ A, we have a αaa = α = x αx = a αaλa(x)a. In particular, if αa 6= 0, then λa = 1 and a = 1. In other words, α = α1 ∈ F , and hence Z(T ) = F . Next, suppose I is a nonzero ideal of T and choose 0 6= β ∈ I having minimal support size, that is with the minimal number of A terms occurring. By multiplying β by some element of A if necessary, we can assume that 1 ∈ supp β. P Then β = a βaa with βa ∈ F and β1 6= 0. Note that, for each x ∈ A, we have P −1 P −1 a βaλa(x)a = x βx ∈ I and hence γ(x) = a βa(λx(x)−1)a = x βx−β ∈ I. But 1 is no longer in the support of γ(x), so γ(x) has smaller support than β and hence γ(x) = 0 for all x. In particular, β is central in T , so 0 6= β ∈ F , I = T , and T is simple. Finally, since T = F t[A] with A a torsion free abelian group, it is clear that T is locally noetherian and hence an Ore domain. Lemma 2.3. Construction of the skew group ring S = TC = F ∗G. Furthermore, if u and 1 + u are units of S, then u ∈ F . Proof. Let C = hci be an infinite cyclic group and define the map τ : F t[A] → t F [A] to extend σ on F and to satisfy τ(¯ai) =a ¯i+1. Since (i + 1) − (j + 1) = j i − j, it follows that τ preserves the relations [ai, aj] = σ (fi−j). Thus τ is an automorphism of T = F t[A], and we can form the skew group ring S = TC = F t[A]C with c acting like τ. Since AC is an F -basis for S, it is easy to see that S = F ∗G is a crossed product over F of the group G = C o C. Indeed, G has a base group B =∼ A which is the direct product of countably many copies of C, and G = B o hzi with z acting as the shift operator. We write G = AC. Now it is easy to see that if G1 and G2 are ordered groups, then so is G1 o G2. Thus G = C o C is ordered, and it follows that every unit u of F ∗G is trivial, FINITELY GENERATED SIMPLE ALGEBRAS 5 that is of the form u = fg with f ∈ F • and g ∈ G. In particular, if 1 + u is also a unit, then g = 1 and u ∈ F . Lemma 2.4. S is a simple right and left Ore domain with center F σ, the fixed field of F under the action of σ. Proof. Since T is a right and left Ore domain and S = TC with C an infinite cyclic group, it is clear that S is also a right and left Ore domain. Now let α P i −i i be central in S and write α = i c αi with αi ∈ T . Since c a0c = ai, we P i P i have i c αia0 = αa0 = a0α = i c aiαi and hence aiαi = αia0. In particular, if Wi = supp αi, then aiWi = Wia0 and hence the finite set Wi ⊆ A is closed −1 under multiplication by aia0 . Now, if αi 6= 0, then Wi 6= ∅ and therefore −1 −1 aia0 Wi = Wi implies that Wi is a union of cosets of the cyclic group haia0 i. −1 Hence aia0 must have finite order. But this can occur only when i = 0, so α = α0 ∈ T and, since α is now central in T , it follows from Lemma 2.2 that α ∈ F . Furthermore, c acts on F like σ does, so α = c−1αc = ασ and α ∈ F σ. Since the reverse inclusion is obvious, we conclude that Z(S) = F σ. Now let I be a nonzero ideal of S = TC. Since I is closed under multiplication −1 Pn i by c and c , we can clearly choose 0 6= β = i=0 βic ∈ I, with βi ∈ T , β0 6= 0, Pn i and with n minimal. Set J = {γ0 ∈ T | γ = i=0 γic ∈ I with γi ∈ T }. Then J is easily seen to be an ideal of T , and J 6= 0 since β0 ∈ J. Thus, Lemma 2.2 implies that J = T , and we can now assume that β0 = 1. With this, it is easy to see, for any element δ = fg ∈ S with f ∈ F and g ∈ G, that (δβ −βδ)c−1 ∈ I is a polynomial in c of degree at most n − 1. The minimality of n now yields δβ = βδ and, since this holds for all such elements δ = fg, we have β ∈ Z(S) ⊆ F . Thus β = 1, I = S, and S is a simple ring. Lemma 2.5. R = E∗G ⊆ S is an Ore domain with center Eσ. It is generated −1 −1 as a K-algebra by a0, a0 , c and c . Furthermore, if E is σ-simple, that is if E has no proper σ-stable ideal, then R is a simple ring.
j Proof. Note that E contains all elements of the form σ (fi−j) = [ai, aj], for i > j, and their inverses. Thus Et[A] is a subring of F t[A]. Since E is σ-stable, Et[A] is τ-stable, and hence R = Et[A]C is a subring of S = F t[A]C. Indeed, since R = EAC = EG, we see that R = E∗G ⊆ F ∗G = S. Of course, E is a right and left Ore domain, so the same is true of Et[A] and of R = Et[A]C in turn. 0 −1 −1 Let R be the K-subalgebra of R generated by a0, a0 , c and c . Since −1 −1 0 c aic = ai+1 and caic = ai−1, we see that R contains all ai, and similarly −1 it contains all ai . Next, for i > 0, we see that [ai, a0] = fi and [a0, ai] = −1 −1 0 [ai, a0] = fi are contained in R . Thus, since c acts on E like σ does, we 0 k k −1 0 conclude that R contains all σ (fi) and σ (fi) . Hence R ⊇ E,A and C, so R0 = R, as required. Finally, suppose E is σ-simple and let I be a nonzero ideal of R. Since S = FR = RF , it follows that FIF is a nonzero ideal of S. Hence, by Lemma 2.4, we have FIF = S, and in particular 1 ∈ FIF . But F is the field of fractions of 6 A. I. LICHTMAN AND D. S. PASSMAN
−1 −1 E ⊆ R, so by taking common denominators, we see that e1 αe2 = 1 for some • • α ∈ I and e1, e2 ∈ E . In other words, α = e1e2 ∈ E , so we conclude that I ∩ E is a nonzero ideal of E. Now I ∩ E is closed under conjugation by c, and c acts like σ on E, so I ∩ E is actually a σ-stable ideal of E. Thus, when E is assumed to be σ-simple, we have I ∩ E = E, so 1 ∈ I ∩ E ⊆ I, I = R, and R is a simple ring. Lemma 2.6. If D is the division ring of fractions of S, then D is the division ring of fractions of R, it is generated as a K-division algebra by a0 and c, and Z(D) = F σ. Furthermore, if L is a field extension of F σ of finite degree, then L ⊗F σ D has zero divisors if and only if L ⊗F σ F has zero divisors. Proof. Let D be the division ring of fractions of S. Then D ⊇ R, so every nonzero element of R is invertible in D. Furthermore, since S = FR = RF and since F is the field of fractions of E ⊆ R, we see that every element of D is a quotient of two elements of S. Hence D is also the division ring of fractions of R. In particular, Lemma 2.5 implies that D is generated as a K-division algebra by a0 and c. Now S = F ∗G with G = C o C, and 1 ∈ G is the unique element of this group having only finitely many G-conjugates. Thus it follows, as in [Pa, The- orem 4.4.5], that the center of D is the quotient field of the center of S. In particular, Lemma 2.4 implies that Z(D) = F σ. σ Finally, let L be a field extension of F of finite degree. If L ⊗F σ F has zero divisors, then so does L ⊗F σ D since F ⊆ D. Conversely, assume that 0 0 L = L ⊗F σ F has no zero divisors. Then L is a commutative domain, finite dimensional over the field F , and hence L0 is a field. Furthermore, since S = F ∗G, 0 0 0 we see that S = L ⊗F σ S = (L ⊗F σ F )∗G = L ∗G, and hence S is also an Ore Domain with division ring of fractions D0. Now D0 ⊇ S, so D0 ⊇ D, and of course D0 ⊇ L. Furthermore, L is central in D0 and any linear relation between L and D gives rise, by taking common denominators, to a linear relation between 0 L and S. With this, we see that D ⊇ L ⊗F σ D and consequently L ⊗F σ D has no zero divisors.
3. Fields of Transcendence Degree 1 We will, of course, use the constructions of the previous section to prove our first two main results, namely Theorems 1.1 and 1.2. Since K is given, we can only vary F and the field automorphism σ in this process. As will be apparent here, we use the same family of fields F for both of these results, but we use two essentially different types of automorphisms. In the case of Proposition 1.3, we use the same field F , but uncountably many distinct field automorphisms. To start with, let K(t) be the rational function field over K in the one variable t, and let P be the set of all prime numbers different from the characteristic of K. Then, working in a fixed algebraic closure of K(t), we let tp be a fixed pth root of t for each prime p ∈ P. Of course, K(tp) is also a rational function field in the FINITELY GENERATED SIMPLE ALGEBRAS 7 single variable tp, and |K(tp): K(t)| = p. It is easy to see that if p1, p2, . . . , pk are finitely many distinct primes with p1p2 ··· pk = n, then K(tp1 , tp2 , . . . , tpk ) = n K(tn) for some element tn with (tn) = t. Consequently,
|K(tp1 , tp2 , . . . , tpk ): K(t)| = |K(tn): K| = n = p1p2 ··· pk.
Now, if I is a nonempty subset of P, then we let FI be the field extension of K(t) generated by all tp with p ∈ I. It is easy to see that FI determines I provided we know how K(t) is embedded in the field. Specifically, we have Lemma 3.1. Let I be a nonempty subset of P.
i. I is precisely the set of all primes p ∈ P such that FI ⊇ L ⊇ K(t) for some field L with |L : K(t)| = p. ii. If p ∈ P \ I and if L is a field extension of K(t) with |L : K(t)| = p, then L ⊗K(t) FI has no zero divisors. iii. If p ∈ I, then there exists a field extension L of K(t) with |L : K(t)| = p and such that L ⊗K(t) FI has zero divisors.
Proof. (i) This is clear since any such field L must be contained in some K(tn) with n = p1p2 ··· pk and p1, p2, . . . , pk ∈ I. (ii) It suffices to observe that, for each n as above, L ⊗K(t) K(tn) is a field. Equivalently, we need to show that the polynomial Xn − t ∈ L[X] is irreducible. To this end, let s be a root of the polynomial and consider the field L[s]. Then L[s] ⊇ K(s) ⊇ K(t) and |K(s): K(t)| = n. Thus, the relatively prime integers n and p both divide |L[s]: K(t)|. It therefore follows that |L(s): K(t)| = np, and hence |L[s]: L| = n, as required. (iii) If p ∈ I, take L = K(tp) so that |L : K(t)| = p. Then L ⊗K(t) K(tp) is not a domain and hence neither is L ⊗K(t) FI . Since P is an infinite set, there are uncountably many nonempty subsets I ⊆ P and hence there are potentially uncountably many fields FI , viewed as K- algebras. We show below that this is indeed the case. For convenience, if I and J are subsets of P, then we write I ∼ J if and only if I and J differ by just finitely many elements. In other words, I ∼ J if and only if there exist finite subsets I0 and J0 of P with I ∪ I0 = J ∪ J0. It is clear that ∼ is an equivalence relation and that each equivalence class is countable, since P has only countably many finite subsets. Of course, the collection of all finite subsets of P forms one equivalence class, and obviously, there are uncountably many other such classes. We have
Lemma 3.2. Let I and J be nonempty subsets of P. Then the fields FI and FJ are K-isomorphic if and only if I ∼ J . Proof. Let I be given and suppose q ∈ P\I. Then, for each prime p ∈ I, it is easy 0 0 p 0 to see that K(tq, tp) = K(tq, tp), where (tp) = tq. In particular, if I = I ∪ {q}, 0 ∼ then FI0 = K(tq, tp | p ∈ I) = K(tq, tp | p ∈ I). But K(t) = K(tq) via the K-isomorphism given by t 7→ tq, and then this map extends to a K-isomorphism 8 A. I. LICHTMAN AND D. S. PASSMAN
∼ 0 FI = FI0 given by tp 7→ tp for all p ∈ I. It now follows by induction on the size ∼ of the change that I ∼ J implies FI = FJ . Conversely, suppose FI is K-isomorphic to FJ . Then we can write FI = FJ provided we rename the generators of FJ as s and sq for all q ∈ J . Now there exist primes p1, p2, . . . , pk ∈ I, with n = p1p2 ··· pk, such that
s ∈ K(tp1 , tp2 , . . . , tpk ) = K(tn), and thus |K(tn): K(s)| = m < ∞. Now if q ∈ J , then FI = FJ ⊇ K(sq) ⊇ K(s) with |K(sq): K(s)| = q. Thus either q|m or K(tn) has an extension of degree q in FI and hence q ∈ I. In other words, J ⊆ I ∪ I0 for some finite subset I0 of P. Similarly, I ⊆ J ∪ J0, and then clearly I ∼ J . Now suppose that K contains an element ζ 6= 0 of infinite multiplicative order such that all mth roots of ζ are also in K. For example, K could be the algebraic closure of a nonabsolute field. For each p ∈ P, let ζp ∈ K be a fixed pth root of ζ. If I is any nonempty subset of P, we can then define a field automorphism 0 0 0 σ of FI over K by σ : t 7→ ζt and σ : tp 7→ ζptp for all p ∈ I. As usual, if p1, p2, . . . , pk are finitely many distinct primes in I with p1p2 ··· pk = n, then 0 K(tp1 , tp2 , . . . , tpk ) = K(tn) and σ : tn 7→ ζntn, where ζn ∈ K is some nth root of ζ. With this, it is clear that σ0 is indeed a field automorphism.
σ0 Lemma 3.3. For each nonempty subset I ⊆ P, we have (FI ) = K. Further- 0 −1 more, if E is a σ -stable K-subalgebra of FI containing all tp and tp with p ∈ I, then E is a σ0-simple ring.
σ0 0 Proof. If s ∈ (FI ) , then s ∈ K(tn) for some n as above. But σ is an au- tomorphism of infinite order on K(tn), so |K(tn): K(s)| = ∞ and s ∈ K, as required. Now let I be a nonzero σ0-stable ideal of E. Then there exists an integer n = −1 p1p2 ··· pk with I ∩ K(tn) 6= 0. Furthermore, since tn, tn ∈ E and since K(tn) is −1 the field of fractions of the K-subalgebra N = K[tn, tn ], we see that J = I ∩ N 0 is a nonzero σ -stable ideal of N. Note that N = K[htni] is the group algebra of 0 the infinite cyclic group htni and that σ : tn 7→ ζntn. Now, using group algebra notation, choose 0 6= α ∈ J to be an element of minimal support size. Multiplying Pm i by some element of htni if necessary, we can assume that α = i=0 ki(tn) with 0 0 Pm i i k0 6= 0. Then σ (α) − α ∈ J and σ (α) − α = i=1 ki(ζn − 1)(tn) has smaller 0 support. Thus σ (α) − α = 0 and, since ζn ∈ K has infinite multiplicative order, • we conclude that ki = 0 for all i 6= 0. Hence α = k0 ∈ K , so 1 ∈ J ⊆ I, I = E, 0 and E is a σ -simple ring. We can now verify our first main result. Proof of Theorem 1.1. Since K is assumed to have a nonzero element ζ of infinite multiplicative order with all its mth roots also contained in K, we can use the above notation. For each infinite subset I of P, we construct a K-algebra RI 0 using the results of the preceding section. Specifically, we take F = FI , σ = σ FINITELY GENERATED SIMPLE ALGEBRAS 9 and F = {tp, 1 + tp | p ∈ I}, with the primes listed in their natural order, and then we let RI be the K-algebra given by Lemma 2.5. In particular, we know that RI is an Ore domain, generated as a K-algebra by four elements, and that σ0 Z(RI ) = E . Furthermore, by Lemma 3.3 and the fact that E is generated by 0 −1 σ0 all hσ i-conjugates of F ∪ F , we see that Z(RI ) = E = K and that E is 0 σ -simple. Thus, by Lemma 2.5 again, we conclude that RI is a simple ring. Finally, we claim that RI determines FI . To this end, let LI be the K- subalgebra of RI generated by all units u of RI with u + 1 also a unit. By Lemma 2.3, we know that LI ⊆ FI . Furthermore, since F contains both tp and 1 + tp, we see that each tp, with p ∈ I, is contained in LI . Thus FI is the field of fractions of LI , and hence this field is indeed determined by RI . In particular, if RI and RJ are K-isomorphic, then FI and FJ are K-isomorphic, so I ∼ J by Lemma 3.2. But there are uncountably many equivalence classes under ∼ consisting of infinite subsets of P, so in this way we obtain uncountably many nonisomorphic K-algebras RI , as required. Now suppose, instead, that K contains all mth roots of unity and, for each prime p ∈ P, let εp ∈ K be a primitive pth root of 1. Recall that, by definition, P avoids the characteristic of K. If I is a nonempty subset of P, we define the field 00 00 automorphism σ of FI over K(t) by σ : tp 7→ εptp for all p ∈ I. As above, it is 00 clear that σ is indeed a K(t)-field automorphism. Furthermore, if p1, p2, . . . , pk are distinct primes in I with p1p2 ··· pk = n, then K(tp1 , tp2 , . . . , tpk ) = K(tn) 00 and σ : tn 7→ εntn, where εn is a suitable primitive nth root of unity in K. In 00 particular, σ has order n in its action on K(tn) and, since |K(tn): K(t)| = n, σ00 it follows immediately that (FI ) = K(t). With this, we have Proof of Theorem 1.2. We use the above notation. For each infinite subset I of P, we construct a K-division algebra DI using the results of the preceding 00 section. Specifically, we take F = FI , σ = σ and F = {tp | p ∈ I}, with the primes in their natural order, and we let DI be the division ring given by Lemma 2.6. Then we know that DI is generated as a K-division algebra by two σ00 elements and furthermore that Z(DI ) = F = K(t). It remains to show that DI determines I. To this end, let L be a field contain- ing Z(DI ) = K(t) with |L : K(t)| = p ∈ P. Then, by Lemma 2.6, L ⊗K(t) DI contains a zero divisor if and only if L ⊗K(t) FI has a zero divisor. Furthermore, by Lemma 3.1(ii)(iii), the latter cannot occur if p∈ / I and it can and does oc- cur if p ∈ I. Thus I is indeed determined by DI , and consequently we obtain uncountably many nonisomorphic division algebras in this way. We close this section with the
Proof of Proposition 1.3. Let K be an arbitrary field and again use the above notation. For each infinite subset I of P, let RI be the K-algebra given by Lemma 2.5 with F = FI , σ = 1, and F = {tp | p ∈ I}, where the primes t are listed in their natural order. Since σ = 1, we know that RI = (EI ) [G] 10 A. I. LICHTMAN AND D. S. PASSMAN
−1 is a twisted group ring with Z(RI ) = EI = K[tp, tp | p ∈ I], and with four generators as a K-algebra. Furthermore, RI is a right and left Ore domain with field of fractions DI given by Lemma 2.6. Thus DI is generated as a K-division algebra by two elements, and Z(DI ) = FI , the field of fractions of EI . Finally, ∼ ∼ if RI = RJ is a K-isomorphism, then FI = Z(DI ) = Z(DJ ) = FJ is also a K-isomorphism. In particular, Lemma 3.2 implies that I ∼ J , and hence we obtain uncountably many different K-algebras RI and K-division algebras DI . If K is a countable field, then so is each FI , and we change the above con- struction by taking F to be the set of all nonzero elements of FI having infinite −1 multiplicative order. Then EI = K[F ∪ F ] = FI , since if 0 6= f ∈ FI has finite order, then f ∈ K[tf, t−1] and both tf and t−1 have infinite order. In this case, EI is σ-simple, since it is a field, and hence RI is a simple ring by Lemma 2.5. 4. HNN Constructions We start with a brief summary of several results and concepts from the theory of skew fields which will be used in the proofs of Theorem 1.4 and Proposition 1.5. We refer the reader to [C1, C2] for a more detailed exposition. Let R be a arbitrary ring and recall that an n × n R-matrix α is said to be full if every factorization α = βγ, with β an n × m matrix and with γ of size m × n, implies that m ≥ n. It is not true, in general, that every invertible matrix is full, but we do have Lemma 4.1. Let R be a ring. i. If R is locally noetherian, then every invertible R-matrix is full. ii. If D is a K-division algebra and if L is a commutative K-algebra, then R = L ⊗K D is locally noetherian. Proof. (i) Let α be an invertible n × n matrix which can be factored as α = βγ with β of size n×m and γ of size m×n. Then 1 = β(γα−1) yields a factorization of the identity matrix with the matrix factors having the same parameters n and m. In other words, we can assume that α is the identity matrix. Furthermore, since R is locally noetherian and since α, β, and γ have only finitely many entries, we can now also assume that R is noetherian. Next, let R be a prime homomorphic image of R. Then the map R → R yields a factorization α = βγ of R-matrices with the same sizes. Thus, we can now assume that R is a prime noetherian ring. But then, by Goldie’s theorem, R embeds in Mt(D), the t × t matrix ring over a division ring D, and we can view α, β, and γ as matrices over this larger ring. Indeed, since an n × m matrix over ∗ ∗ ∗ Mt(D) can be viewed as an nt × mt matrix over D, we obtain α = β γ , where α∗ is the nt × nt identity matrix, β∗ is a D-matrix of size nt × mt, and γ∗ has size mt × nt. The latter matrices translate to D-linear transformations Dnt → Dmt → Dnt with composite map the identity. In particular, the map Dmt → Dnt is onto, so dimension considerations yield mt ≥ nt and hence m ≥ n, as required. FINITELY GENERATED SIMPLE ALGEBRAS 11
(ii) If R = L ⊗K D, then any finite subset of R is contained in a subring of the form K[z1, z2, . . . , zk] ⊗K D with z1, z2, . . . , zk in the commutative algebra L. Since this subring is a homomorphic image of the ordinary polynomial ring D[x1, x2, . . . , xk], and since D is a division ring, the result follows. A ring R is said to be a fir (semifir) if every left ideal (finitely generated left ideal) is free of unique rank. It is known that every semifir has a universal field of fractions U(R) which is a skew field containing R and generated by it. Furthermore, every isomorphism between two semifirs extends uniquely to an isomorphism of their universal fields of fractions. However, an embedding of semifirs need not give rise to an embedding of their universal fields of fractions. We will need the following observations. Part (i) is standard and part (ii) comes from the proof of [C2, Proposition 6.4.4]. Lemma 4.2. Let R ⊆ R0 be semifirs, let Σ be the set of full matrices of R, and let Σ0 be the set of full matrices of R0. i. Suppose S ⊇ U(R) is a ring with the property that all invertible S- matrices are full. If the embedding R → U(R) extends to a homomor- phism R0 → S, then Σ ⊆ Σ0 and U(R) embeds naturally in U(R0). 0 ii. If R is a K-algebra and R = L ⊗K R, where L is a field extension of K, then Σ ⊆ Σ0 and U(R) ⊆ U(R0). Furthermore, if Z = Z(U(R)) and if 0 L ⊗K Z is a field, then L ⊗K U(R) ⊆ U(R ) and hence L ⊗K U(R) is a domain. Proof. (i) If α ∈ Σ, then α becomes invertible over the ring U(R) and hence over the larger ring S. In particular, α is full when viewed as a matrix over S. Thus, since R0 → S extends the embedding R → U(R), it follows that α, when viewed as a matrix over R0, must also be full. In other words, Σ ⊆ Σ0 and hence all matrices in Σ are invertible in U(R0). This implies that there exists a homomorphism U(R) → U(R0) and, since U(R) is a division ring, we obtain the required embedding. 0 (ii) The embedding R → U(R) clearly extends to a map R = L ⊗K R → L ⊗K U(R), and L ⊗K U(R) is locally noetherian by Lemma 4.1(ii). Thus, by Lemma 4.1(i), all invertible matrices of L ⊗K U(R) are full, and consequently part (i) above implies that Σ ⊆ Σ0 and U(R) ⊆ U(R0). Furthermore, since U(R) 0 is the localization RΣ, it is clear that L ⊗K U(R) is obtained from R = L ⊗K R 0 by inverting the matrices in Σ, or equivalently L ⊗K U(R) = RΣ. Thus, since 0 0 0 0 Σ ⊆ Σ , we have a homomorphism RΣ → RΣ0 = U(R ). Now observe that U(R) is a division algebra, so every nonzero ideal of L ⊗K U(R) must meet L ⊗K Z nontrivially. In particular, if L ⊗K Z is a field, then L ⊗K U(R) must be a 0 simple ring. In this case, the homomorphism L ⊗K U(R) → U(R ) is necessarily one-to-one.
Now let X = {xi | i ∈ I} be a set, and let K
L
B = {[x, y]0, [x, y]1, [x, y]2,...}. Then R0 = K
Proof. By symmetry, we can assume that D 6= A, so that {d1, d2,...}, an A- basis for DA, has size at least 2. Since {d1 ⊗ 1, d2 ⊗ 1,...} is a D-basis for MD, it is easy to see that the nonzero right ideals (d1 ⊗ 1)T (M) and (d2 ⊗ 1)T (M) are disjoint. In particular, T (M) is not an Ore domain and hence it is not a principal ideal domain. [C1, Theorem 7.8.4] now implies that the center Z of U(S) = U(T (M)) coincides with the center of T (M). Furthermore, Z is certainly a field, so Z = Z(T (M)) consists of invertible elements in T (M). But T (M) is a graded ring, so all invertible elements must be contained in its 0-component, namely D. Indeed, since T (M) is generated by D and the element 1 ⊗ 1 ∈ M, it follows that Z is the set of elements of Z(D) that commute with 1 ⊗ 1. Now if r ∈ Z, then r ⊗ 1 = r(1 ⊗ 1) = (1 ⊗ 1)r ∈ (r ⊗ 1)T (M) ∩ (1 ⊗ 1)T (M), so the above basis information implies that r must be contained in A and conse- quently, since a(1 ⊗ 1) = (1 ⊗ 1)ϕ(a) for all a ∈ A, the result follows. Next, we consider the behavior of HNN extensions under a field extension. Lemma 4.5. Let K be a field and let L be a field extension of K of finite degree. Suppose D is a division algebra, let A and B be proper K-subdivision algebras, 0 0 and let ϕ: A → B be a K-isomorphism. Write A = L ⊗K A, B = L ⊗K B and let ϕ0 = 1 ⊗ ϕ: A0 → B0 be the natural extension of ϕ to an L-isomorphism. If D0 is a domain, then D0, A0 and B0 are all division algebras. Furthermore, if S is the HNN extension of D determined by the map ϕ, and if S0 is the HNN 0 0 0 extension of D determined by the map ϕ , then L ⊗K U(S) ⊆ U(S ), and hence L ⊗K U(S) is a division algebra.. Proof. Since D0 = L ⊗ D is a domain and |L : K| < ∞, it follows that D0 is a division algebra. Similarly, A0,B0 ⊆ D0 are also division algebras. Now let 0 0 0 M = D⊗AD be the (D,D)-bimodule constructed from ϕ, and let M = D ⊗A0 D be the (D0,D0)-bimodule constructed from ϕ0. Since A0 ⊇ L and ϕ0 is the 14 A. I. LICHTMAN AND D. S. PASSMAN
0 0 identity on L, it follows that all L-factors in D ⊗A0 D can be moved to the 0 0 front. With this, it is clear that M = L ⊗K M. Furthermore, since D contains 0 0 0 L, it follows that all L-factors of M ⊗D0 M ⊗D0 · · · ⊗D0 M can also be moved to 0 ⊗n ⊗n the front. Again this says that (M ) = L ⊗K (M) for all n ≥ 0 and hence 0 T (M ) = L ⊗K T (M). 0 0 We wish to apply Lemma 4.2(ii) with R = T (M) and R = T (M ) = L ⊗K R, and we already know that both T (M) and T (M 0) are firs. Furthermore, if Z is the center of U(R) then, by Lemma 4.4 and the fact that D properly contains A, we see that Z is a field contained in D. In particular, since L ⊗K D has no zero divisors, L ⊗K Z has no zero divisors and hence it is a field. Lemma 4.2(ii) now 0 implies that L ⊗K U(R) ⊆ U(R ), as required. We can now offer our basic HNN construction. The inclusion of the interme- diate field E allows us to control the center of R.
Lemma 4.6. Let F ⊇ E ⊇ K be fields and let F = {0 = f0, f1, f2,...} be a countable sequence of elements of F . Then there exists a K-division algebra R = U(S), where S is an HNN extension of the free field F
0 0 F = L ⊗E F is a domain. Then F is a field, and consequently we know that 0 0 F
F = {0 = f0, f1, f2,...} be given by fi = tpi , where pi is the ith prime contained in I. Then we can let DI be the E-division algebra given by Lemma 4.6. It follows that Z(DI ) = E = K(t) and that DI has two generators as a K-division algebra, since K(F) = FI . Since there are uncountably many such subsets I, it remains to show that DI determines I. To this end, let L be a field containing Z(DI ) = K(t) of degree |L : K(t)| = p ∈ P. Then, by Lemma 4.6(iii), L ⊗K(t) DI contains a zero divisor if and only if L ⊗K(t) FI has a zero divisor. Furthermore, by Lemma 3.1(ii)(iii), the latter cannot occur if p∈ / I and it can and does occur if p ∈ I. Thus I is indeed determined by DI , and consequently we obtain uncountably many nonisomorphic division algebras in this way.
Proof of Proposition 1.5. Here, for each such I ⊆ P, we take E = F = FI , and we let F = {0 = f0, f1, f2,...} be given by fi = tpi , where pi is the ith prime contained in I. Again, let DI be the E-division algebra given by Lemma 4.6. It follows that Z(DI ) = E = FI and that DI has two generators as a K-division ∼ algebra, since K(F) = FI . Finally, if DI = DJ is a K-isomorphism, then ∼ FI = Z(DI ) = Z(DJ ) = FJ is also a K-isomorphism. In particular, Lemma 3.2 implies that I ∼ J , and hence we conclude that there are uncountably many different K-division algebras DI .
References [C1] P. M. Cohn, Free Rings and Their Relations, Academic Press, London, 1985. [C2] P. M. Cohn, Skew Fields, (Theory of General Division Rings), Cambridge University Press, Cambridge, 1995. [D] W. Dicks, The HNN construction for rings, J. Algebra 81 (1983), 434–487. [H] P. Hall On the embedding of a group in a join of given groups, J. Austral. Math. Soc. 17 (1974), 434–495. [L] A. I. Lichtman, On universal fields of fractions for free algebras, J. Algebra 231 (2000), 652–676. [LS] R. C. Lyndon and P. E. Schupp, Combinatorial Group Theory, Springer-Verlag, Berlin- New York, 1977. [N] B. H. Neumann, Some remarks on infinite groups, J. London Math. Soc. 12 (1937), 120– 127. [Pa] D. S. Passman, The Algebraic Structure of Group Rings, Wiley-Interscience, New York, 1977. 16 A. I. LICHTMAN AND D. S. PASSMAN
[Pl] B. I. Plotkin, Seven Lectures on Universal Algebraic Geometry, The Hebrew University, Jerusalem, preprint, 2002.
Department of Mathematics, University of Wisconsin-Parkside, Kenosha, Wiscon- sin 53141 E-mail address: [email protected]
Department of Mathematics, University of Wisconsin-Madison, Madison, Wisconsin 53706 E-mail address: [email protected]