Division Algebras with Left Algebraic Commutators

Let D be a division algebra with center F and K a (not necessarily central) subfield of D. An element a ∈ D is called left algebraic (resp. right algebraic) over K, if there exists a non-zero left polynomial a0 + a1x + ⋯ + anxn (resp. right polynomial a0 + xa1 + ⋯ + xnan) over K such that a0 + a1a + ⋯ + anan = 0 (resp. a0 + aa1 + ⋯ + anan). Bell et al. proved that every division algebra whose elements are left (right) algebraic of bounded degree over a (not necessarily central) subfield must be centrally finite. In this paper we generalize this result and prove that every division algebra whose all multiplicative commutators are left (right) algebraic of bounded degree over a (not necessarily central) subfield must be centrally finite provided that the center of division algebra is infinite. Also, we show that every division algebra whose multiplicative group of commutators is left (right) algebraic of bounded degree over a (not necessarily central) subfield must be centrally finite. Among other results we present similar result regarding additive commutators under certain conditions.


Introduction
A theorem due to Jacobson [7] says that every algebraic division algebra of bounded degree over its center, is centrally finite. Akbari et al. in [10] generalized this theorem and proved that the statement holds if one assumes that D is a division algebra whose multiplicative group of commutators is algebraic of bounded degree over the center. Recently, Bell et al. in [4] using a technique based on combinatorics of words have provided a generalization of Jacobson Theorem from other aspects of view. They proved that every division algebra whose elements are algebraic of bounded degree over a (not necessarily central) subfield, must be centrally finite. More precisely, they proved that if D is a division algebra with center F such that its elements are left algebraic of bounded degree n over a subfield K, then dim F D ≤ n 2 .
For any centrally finite division algebra D over F, by Kothe's Theorem it is known that there exists a maximal subfield K of D such that the extension of fields K/F is separable [8, p. 244]. In [10], authors proved that if K/F is a finite separable extension of fields in D, then there exists an element d ∈ D , the multiplicative group of commutators of D * , such that K = F (d). Now, one may ask: whether F (d) is a maximal subfield of D for some d in additive or multiplicative groups of commutators. Chebotar et al. in [5] answered the question in affirmative in the case of centrally finite division algebras. In this note we give a new proof for these questions. Also, we concern with a generalization of a theorem due to Akbari et al. [10] and Bell et al. [4]. In fact we prove that a division algebra whose all multiplicative commutators are left (right) algebraic of bounded degree over a (not necessarily central) subfield is centrally finite, provided that has infinite center. As another result we show that if D is a division algebra with center F whose multiplicative group of commutators is left (right) algebraic of bounded degree n over a (not necessarily central) subfield, then dim F D ≤ n 2 . Among other results we prove that if D is a division algebra with infinite center F and non-central subfield K with a ∈ K \ F algebraic over F such that for every x ∈ D, ax − xa is left (right) algebraic over subfield K of bounded degree n, then D is centrally finite.

Preliminary
Let K be a subdivision algebra of D with center F. An element a ∈ D is called left algebraic (resp. right algebraic) over K, if there exists a non-zero left polynomial a 0 +a 1 x+· · ·+a n x n (resp. right polynomial a 0 + xa 1 + · · · + x n a n ) over K such that a 0 + a 1 a + · · · + a n a n = 0 (resp. a 0 + aa 1 + · · · + a n a n ). In other words, a is left (resp. right) algebraic over K if the set {1, a, a 2 , . . .} is a dependent set of left (resp. right) vector space D over K. In this case, assume that n is the smallest degree of left polynomials f (x) (resp. right polynomial) such that f (a) = 0. We denote by deg K (a) = n (resp. K deg(a) = n) the left degree (resp. right degree) of a over K. If K = F is the center, then the concept of left and right algebraic coincides. It is easy to see that if a is left algebraic over K, then the left degree of a over K is the smallest integer n such that {1, a, . . . , a n } is left dependent over K. If a ∈ D, then F (a) denotes the subfield of D generated by F and {a}. If A is a subset of D, then K(A) denotes the subdivision algebra generated by K ∪ A. Also, for a subset A of D, we use K[A] to denote the subalgebra generated by K ∪ A. When A = {a}, then we use K[a] to denote subalgebra generated by K ∪ {a}. If A is a subset of D we use A * to denote A \ {0}. A subfield K of D is called a maximal subfield if K is its own centralizer in D * . We denote by dim F D, Z(D) and char(D) the dimension of D over F, center of D and characteristic of D, respectively. If dim F D = n 2 , then n is called the degree of division algebra D. By M n (K) and GL n (K) we mean all square matrices and all invertible matrices of order n with entries from K, respectively.
Before stating our results we need to recall some theorems. The first one is a generalization of celebrated theorem due to Brauer, Cartan and Hua.
Theorem 1 [6] Let D be a division algebra with center F and H a proper subdivision algebra of D such that H is invariant under a subnormal non-central subgroup G in D * , that is gHg −1 ⊆ H for every g ∈ G * . Then H ⊆ F. Theorem 2 [8, p. 242] If D is a division algebra of degree n over a field F and K is a subfield of D containing F, then dim F K ≤ n. The quality holds if and only if K is a maximal subfield of D. Conversely, if K is a maximal subfield of a division algebra D such that dim F K = n < ∞, then D is of degree n over F.
Let X be a set of non-commutating indeterminates and F be an arbitrary field. The Laurent polynomial ring F [X, X −1 ] is the ring of all finite sums of form . . x m t n t , where α n ∈ F, x n i ∈ X and m i ∈ {1, −1}. Every element of F [X, X −1 ] is called a Laurent polynomial. A Laurent polynomial f ∈ F [X, X −1 ] is called a Laurent polynomial identity of a ring R if it vanishes on all permissible substitutions from R. In this case, we say that R satisfies the Laurent polynomial identity f = 0. We consider the following example which is important in this paper. Given a positive integer n, and n + 1 non-commutating indeterminates y 0 , y 1 , . . . , y n , define g n (y 0 , y 1 , y 2 , . . . , y n ) = δ∈S n+1 where S n+1 is the symmetric group over {0, 1, . . . , n} and sign(δ) is the sign of permutation δ. This is a Laurent polynomial over any field and it is defined in [3] to connect an algebraic element of degree n and a polynomial of n + 1 indeterminates.

Lemma 3 Let F be a field and A be a central simple algebra over F . For any element
a ∈ A, the following conditions are equivalent: 1. The element a is algebraic over F of degree less than n. 2. g n (a, r 1 , . . . , r n ) = 0 for any r 1 , . . . , r n ∈ A.
Proof This is a corollary of [3, Corollary 2.3.8].
In particular, a centrally finite division algebra of degree m satisfies the identity g m since whose elements are algebraic over the center of degree m. In other words, g m is a Laurent polynomial identity of any centrally finite division algebra of degree m.
Let I(A) denote the set of all Laurent polynomial identities of the centrally finite division algebra A. The following theorem gives us a relation between the set of all Laurent polynomial identities of central simple algebras.

Theorem 4 Let F be an infinite field and
A be a central simple algebra of degree n over F . Assume that L is an extension field of F . Then Proof This result is just a corollary of [2,Theorem 11].
In what follows we show that every centrally finite division algebra has a maximal subfield that is obtained by adjunction of a multiplicative or additive commutator to the center of division algebra. First we need to prove the following lemma.

Lemma 5
Let K be a field and n ≥ 2 be an integer.
Proof (i) It is well-known that every matrix of determinant one is a single multiplicative commutator [12]. Now, consider n × n-matrix T = (t ij ), where Then T has determinant one and it is algebraic of degree n. This completes the proof of part (i).
(ii) It is well-known that every matrix with trace zero is a single additive commutator [1].
Then T has trace zero and it is algebraic of degree n. This completes the proof.
Proof If F is finite, then D is also finite, so there is nothing to prove. Suppose that F is infinite and D is of degree n over F . By Theorem 2, it suffices to show that there exist Applying Lemma 3 we find that g (rsr −1 s −1 , r 1 , . . . , r ) = 0, for any r 1 , . . . , r ∈ D and r, s ∈ D * . In other words, g (xyx −1 y −1 , y 1 , . . . , y ) is a Laurent polynomial identity of D. Hence, by Theorem 4 it is also a Laurent polynomial identity of M n (F ). This yields that g (XY X −1 Y −1 , X 1 , . . . , X ) = 0, for all matrices X, Y ∈ GL n (F ) and X i ∈ M n (F ). By the first part of the previous lemma one can find matrices A and B in GL n (F ) such that Now, maximality of yields that n ≤ . This completes the proof.
Theorem 7 Let D be a centrally finite division algebra with center F. There exist x, y ∈ D such that F (xy − yx) is a maximal subfield of D.
Proof If F is finite then D is also finite, so that there is nothing to prove. Suppose that F is infinite and D is of degree n. By Theorem 2, it suffices to show that there exist

Combinatorics of Words
The theory of identities gives us a technique to study algebraic elements of an algebra R of bounded degree over a base field F . However, the base field in results are assumed to be contained in the center of R. The idea of [4] provides another technique to work on algebraic elements of bounded degree over any field. It is the theory of combinatorics on words. Before stating our next result we need to remind the following theorems. Theorem 9 [4] Let D be a division algebra with center F whose elements are left algebraic of bounded degree n over (not necessarily central) subfield K. Then dim F D ≤ n 2 .

Theorem 11
Let D be a division algebra with infinite center F and let K be a subfield (not necessarily central) of D. Assume that there exists non-central element a ∈ D such that for every x ∈ D * , xax −1 a −1 is left algebraic over K of bounded degree n. Then D must be centrally finite.
Proof Consider c / ∈ C D (a). Hence for every t ∈ F we have Put u = b −1 a and observe that u(c+t) is left algebraic over K of bounded degree n. Hence, previous lemma implies that dim K K[c] ≤ 3 2n−1 + 1 = n . In short, we could show that all elements of D which are outside of C D (a) are left algebraic over K of bounded degree n . Let x ∈ C D (a) and z / ∈ C D (a). Then for any r ∈ F we have z(x + r) / ∈ C D (a), and by foregoing argument we have dim K K[z(x + r)] ≤ n . Again, using previous lemma we obtain that dim K K[x] ≤ 3 2n −1 + 1 = n . Thus every element of D is left algebraic over K of bounded degree n . Now, using Theorem 9 we are done.
Remark 1 Let D be division algebra of degree n over its center F. Also, assume that there exists element a ∈ D such that F (a) is a maximal subfield in D. Then for any maximal Proof Without loss of generality assume that K is a maximal subfield of D. If N ⊆ K, then K is N -invariant and by Theorem 1 either K = F or K = D. In both cases, D is commutative, a contradiction. Hence, N K. Let a ∈ N \ K. Since N is normal in D * for any x ∈ D * we have axa −1 x −1 ∈ N, which implies that axa −1 x −1 is algebraic over K of bounded degree n. So, using previous theorem D is centrally finite. Now, assume that char(D) = 0 and let dim F D = l 2 . By [5,Theorem 7], there exists x ∈ D * such that axa −1 x −1 is algebraic over F of degree l. Therefore, F (axa −1 x −1 ) is a maximal subfield of D. In the view of Remark 1, there exists b ∈ D * such that baxa −1 x −1 b −1 is algebraic over K of degree l. But baxa −1 x −1 b −1 ∈ N and by the assumption is algebraic over K of bounded degree n. This implies that l ≤ n. Thus, dim F D ≤ n 2 .
In the sequel we deal with a result in which the normal subgroup in the above theorem is the special one, the multiplicative group of commutators D . In this case we could obtain the result without additional restrictions. First we need to recall some definitions.
Let X be a set of non-commuting indeterminates. For convenience, we assume that X contains m elements x 1 , . . . , x m . Let M be the free monoid generated by X (it is obvious that M is a subset of the Laurent polynomial ring F [X, X −1 ]). Then, M is a totally ordered monoid with the lexicographic order x 1 > · · · > x m . Every element w = x i 1 . . . x i t of M is called a word on X and t is called the length of w. Denote by (w) = t the length of w. Let u, v ∈ M. The element u is called a subword of v if v = v 1 uv 2 for some v 1 , v 2 ∈ M. Let q be a natural number. We say that u is q-decomposable if there exist u 1 , . . . , u q ∈ M such that u = u 1 . . . u q and for any non-trivial permutations f of {1, . . . , q}, one has u 1 . . . u q > u f (1) . . . u f (q) .
In addition, if (q − 1) (u i ) < (u), then we say that u is strongly q-decomposable.
Theorem 17 Let D be a division algebra with center F and K be a subfield of D. Assume that every element of the multiplicative group of commutators D of D * is left algebraic over K of bounded degree n, then dim F D ≤ n 2 .
Proof Without loss of generality, we may assume that K is a maximal subfield of D. There is nothing to do if K = F , so assume that F = K. Choose a 1 , . . . , a m ∈ D arbitrarily and let D m = F (a 1 , . . . , a m ) denote the subdivision algebra of D generated by a 1 , . . . , a m over F . Let A m = F [a 1 , . . . , a m ] be the subalgebra of D generated over F and K m be the vector space generated by A m over K. Also, let M be the free monoid generated by m indeterminate X = {x 1 , . . . , x m }, and N = N (m, n, n) be as in Lemma 14. By Lemma 14, Kw(a 1 , . . . , a m ).
Observe that the cardinality of {w(x 1 , . . . , x m ) ∈ M | (w) = i} is m i . Thus, K m is a vector space of dimension d(m) ≤ 1 + m + · · · + m N over K. This implies that K * D is algebraic over K of bounded degree, say t, because K * D is a subset of K m and K m generated by A m . Now, consider the factor group D * K * D which is an abelian group. By the Brauer-Cartan-Hua Theorem we know that D = F (D ). If d 1 and d 2 are two elements of D such that d 1 + d 2 = 0, then by Lemma 16 and abelian property of D * K * D , (d 1 d 2 )K * D has order at most n. Now, implies that d 1 + d 2 is left algebraic over K of degree at most n. By induction, for elements d 1 , . . . , d s ∈ D with d 1 + · · · + d s = 0, we find that (d 1 + · · · + d s )K * D has order at most n in factor group. This implies that every element of D * is algebraic over K of degree at most tn. So by Theorem 9, D is centrally finite. Using Theorem 6, we find that dim F D ≤ n 2 . This completes the proof.
Our next result provides analogous theorem with respect to additive commutators of a division algebra.
Theorem 18 Let D be a division algebra with infinite center F. Let K be a non-central subfield of D that contains an element a ∈ K \ F algebraic over F. Assume that for every x ∈ D, ax − xa is left algebraic over subfield K of bounded degree n. Then D is centrally finite.
Proof Since a is non-central, there exists b ∈ D such that u = ab − ba = 0. Thus for every c ∈ C D (a), a(bc) − (bc)a = uc is left algebraic over K. In particular for every r ∈ F, a(b(c + r)) − (b(c + r))a = u(c + r) is left algebraic over K. Now, using Lemma 10 we find that dim K K[c] ≤ k + 1 = n . In short, we proved that all elements of C D (a) are algebraic over K ⊆ C D (a) of bounded degree. Hence using Theorem 9 we obtain that C D (a) is centrally finite. Notice that a is algebraic over F, so we have dim F F (a) < ∞. Hence, using Double Centralizer Theorem [8,Theorem 15.4,p. 240] we find that Z(C D (a)) = F (a). Therefore, dim F C D (a) = dim F (a) C D (a).dim F F (a) < ∞.
Again applying Double Centralizer Theorem yields dim F D < ∞. This completes the proof.
At the end of this note we pose following conjectures.

Conjecture 19
Let D be a division algebra with center F and let K be a subfield (not necessarily central) of D. Then D is left algebraic over K if and only if D is left algebraic over K.
Conjecture 20 Let D be a division algebra with center F and let K be a subfield (not necessarily central) of D. Assume that for every x, y ∈ D * , xyx −1 y −1 is left algebraic over K of bounded degree n. Then dim F D ≤ n 2 .

Conjecture 21
Let D be a division algebra with center F and let K be a subfield (not necessarily central) of D. Then D is left algebraic over K if and only if [D, D] is left algebraic over K.

Conjecture 22
Let D be a division algebra with center F and let K be a subfield (not necessarily central) of D. Assume that for every x, y ∈ D * , xy − yx is left algebraic over K of bounded degree n. Then dim F D ≤ n 2 .