A. I. LIKHTMAN

ON NORMAL DIVISORS OF THE MULTIPLICATIVE GROUP OF A DIVISION RING

(Presented by Academician P. S. Novikov on 18 IV 1963)

The Cartan—Brauer—Hua theorem is known (see, for example, (¹), p. 271):

The only subfields of a division ring \(\Delta\) that are invariant under all inner automorphisms are \(\Delta\) and the subfields of the center \(\Phi\) of the division ring \(\Delta\).

This theorem gives some information about normal divisors of the multiplicative group of a division ring.

In Scott’s paper (²) the following result was obtained:

The factor group of the multiplicative group of a division ring by its center contains no abelian normal divisors.

A number of recent works have been devoted to generalizations of the Cartan—Brauer—Hua and Scott theorems, in particular those of M. Sh. Khuzurbazar (⁶, ⁷).

From Jacobson’s theorem ((¹), p. 314) it follows that the multiplicative group of a noncommutative division ring cannot be periodic. It may therefore be conjectured that all periodic normal divisors of the multiplicative group of a division ring are contained in the center. For division rings locally finite over their center, this will follow from Theorem 1 of the present paper; in the general case the question remains open. In the present paper, however, it is proved that the factor group of the multiplicative group of a division ring by its center contains no locally finite normal divisors.

In what follows, \(D\) denotes an arbitrary division ring, \(Z\) its center, and \(D^*\) and \(Z^*\) the multiplicative groups of \(D\) and \(Z\), respectively.

Lemma 1. Let \(D\) be a division ring of characteristic \(p \ne 0\), and let \(F\) be a locally finite normal divisor of the group \(D^*\). Then \(F \subset Z^*\).

Proof. As noted in (³), Theorem 6, every finite subgroup of a division ring of prime characteristic is cyclic. Consequently, \(F\) is an abelian group. From Scott’s theorem (²) it follows that \(F \subset Z^*\). The lemma is proved.

Lemma 2. Let \(D\) be a division algebra over the field \(R\) of rational numbers, with dimension \((D : R)=n\). Let \(\xi_1,\ldots,\xi_l\) be roots of unity in the division ring \(D\); let \(\alpha_i\) be rational integers; and let \(\beta\) be a rational number whose denominator is relatively prime to \(n\). If the equality

\[ \sum \alpha_i \xi_i = \beta, \tag{1} \]

holds, then \(\beta\) is a rational integer.

Proof. Choose in the algebra \(D\) a basis \(e_1=1, e_2,\ldots,e_n\). To each element \(b\in D\) we put in correspondence a linear transformation \(B\) according to the following rule: \(xB=xb\) for all \(x\) in \(D\). Then we obtain an isomorphic representation of the algebra \(D\) in the matrix algebra \(R_n\). From equality (1) we obtain

\[ \sum \alpha_i B_i = \beta E_n, \tag{2} \]

where \(B_i\) is the matrix corresponding to \(\xi_i\) in our representation. Note that for each matrix \(B_i\) there exists a number \(n_i \ge 1\) such that \(B_i^{n_i}=E_n\). Therefore all characteristic roots of the matrix \(B_i\) are roots of unity. Consequently, \(\operatorname{Sp}(B_i)=q_i\), where \(q_i\) is an algebraic integer. From equality (2) it follows that

\[ \sum \alpha_i \operatorname{Sp}(B_i)=n\beta. \tag{3} \]

In equality (3), on the left there is an algebraic integer. Since a rational number is an algebraic integer if and only if it is a rational integer, \(n\beta\) is a rational integer. From the condition of the lemma it now follows that \(\beta\) is a rational integer.

Lemma 3. Let \(D\) be a division algebra of dimension \(n\) over the field \(R\) of rational numbers; let \(F\) be a periodic normal subgroup of the group \(D^*\); let \(p\) be a prime number not dividing \(n\); and let \(C_p\) be the ring of \(p\)-integral rational numbers. Consider the subring \(A_p\) of the division ring \(D\) consisting of elements \(x=\sum \alpha_i g_i\), where \(\alpha_i \in C_p\), \(g_i \in F\). Then \(A_p\) does not contain the element \(p^{-1}\).

Proof. Suppose that \(\sum \alpha_i g_i=p^{-1}\). Then, reducing the \(\alpha_i\) to a common denominator \(s\), we obtain \(\sum \beta_i g_i=sp^{-1}\), where \(sp^{-1}\) is an irreducible fraction, and \(\beta_i\) and \(s\) are rational integers. From Lemma 2 it now follows that \(sp^{-1}\) is a rational integer, which is impossible. The contradiction obtained proves the lemma.

Lemma 4. Suppose that the conditions of Lemma 3 are satisfied. Then in the ring \(A_p\) all elements \(y=\sum \gamma_i g_i\), where \(\gamma_i \equiv 0 \pmod p\) in the ring \(C_p\), form a proper ideal \(T\). In the factor ring by this ideal all nilpotent elements form a nilpotent ideal.

Proof. If the ideal \(T\) coincided with the ring \(A_p\), then the equality \(1=\sum \gamma_i g_i\) would hold, where \(\gamma_i \equiv 0 \pmod p\) in the ring \(C_p\). It would follow that the element \(p\) is invertible in the ring \(A_p\), which is impossible in view of Lemma 3. Consider now the quotient ring of the ring \(A_p\) by the ideal \(T\). It is clear that the ring \(\mathcal L \cong A_p/T\) is a module over the ring \(C_p\), and the ideal \(p\) of the ring \(C_p\) annihilates the ring \(\mathcal L\). Therefore \(\mathcal L\) is a module over the quotient ring \(C_p/(p)\), which we shall denote by \(\Omega\). Since in the ring \(A_p\) any \(n+1\) elements are linearly dependent over \(C_p\), and \(\Omega\) is contained in the center of \(\mathcal L\), it follows that \(\mathcal L\) is a finite-dimensional algebra over the field \(\Omega\) of \(p\) elements.

We now prove that for any elements \(v,u\) of the ring \(A_p\) the equalities
\[ (uv)^m=u_1v^m,\qquad (vu)^m=v^m u_2, \tag{4} \]
hold, where \(m\) is an arbitrary positive integer, \(u_1,u_2\in A_p\). Indeed, the ring \(A_p\) is invariant with respect to all inner automorphisms of the division ring \(D\). Suppose that for all \(k<m\) (4) has already been proved. We may assume that \(v\ne0\). Then
\[ (uv)^m=(uv)(uv)^{m-1}=(uv)u'v^{m-1}=u(vu'v^{-1})v^m= \]
\[ =uu''v^m=u_1v^m,\qquad \text{where } u_1=uu'',\ u_1\in A_p. \]
The second equality is proved in exactly the same way.

It follows from (4) that every nilpotent element in the ring \(\mathcal L\) generates a nilpotent ideal. Consequently, the Jacobson radical of the ring \(\mathcal L\) coincides with the set of all nilpotent elements.

Lemma 5. Let \(D\) be a division algebra, finite-dimensional over the field of rational numbers, and let \(F\) be a periodic normal subgroup of the group \(D^*\). Then \(F\subset Z^*\).

Proof. Consider the preimage of the radical of the ring \(\mathcal L\) (see the proof of Lemma 4) under the natural homomorphism \(A_p\to\mathcal L\). Denote this ideal by \(T^*\). The ideal \(T^*\supseteq T\), and the quotient ring \(Q\cong A_p/T^*\) contains no nilpotent elements. Moreover, the algebra \(Q\) is finite-dimensional over the field \(\Omega\). From (1), p. 315, it follows that the algebra \(Q\) is commutative. Therefore the commutator ideal \(K\) of the algebra \(A_p\) is contained in \(T^*\). There exists a positive integer \(r\) such that \((T^*)^r\subset T\). In particular, \((g_i g_j-g_j g_i)^r\subset T\), \(g_i,g_j\in F\). Hence we obtain that the element \(g_i g_j-g_j g_i\) is not invertible in the ring \(A_p\). Now denote the element \(g_j^{-1}g_i^{-1}g_jg_i\) by \(q\). The element \(1-q\) is not invertible in the ring \(A_p\). We shall show that it follows from this that \(q=1\).

Indeed, otherwise there exists a polynomial irreducible over \(R\),
\(\varphi(x)=x^r+\lambda_1x^{r-1}+\cdots+\lambda_r\) with integral coefficients, such that the element \(1-q\) is its root. Recall that the element \(q\) is a root of unity, and therefore \(1-q\) is an integral algebraic number. Substituting the element \(1-q\) into the polynomial \(\varphi(x)\), we obtain

\[ (1-q)\psi(1-q)=-\lambda_r, \tag{5} \]

where \(\psi(x)\) is a polynomial with integral rational coefficients.

Recall now that the number \(p\) was subject to the sole condition \((p,n)=1\), where \(n\) is the degree of \(D\) over \(R\). We now require also that \(p\) satisfy the condition \((p,\lambda_r)=1\). Then it follows from (5) that the element \(1-q\) is invertible in the ring \(A_p\). This contradiction shows that \((1-q)=0\). Since the elements \(g_i,g_j\) were arbitrary, the group \(F\) is abelian. It now follows from Scott’s theorem that \(F\subset Z^*\).

Lemma 6. Let \(D\) be an arbitrary division ring, and let \(F\) be a locally finite normal divisor of the group \(D^*\). Then \(F\subset Z^*\).

Proof. In view of Lemma 1, we may assume that \(D\) is a division ring of characteristic zero. Consider a finite subgroup \(F_1\) of the group \(F\), generated by the elements \(g_1=1,g_2,\ldots,g_s\). The ring \(P\), consisting of elements
\(\tau=\sum \alpha_i f_i\), where \(\alpha_i\in R,\ f_i\in F\), is a division ring, since \(P\) is a finite-dimensional algebra over \(R\) without zero divisors. The group \(P^*\) contains the normal divisor \(P^*\cap F\). In view of Lemma 5, \(P^*\cap F\) is an abelian group. Hence \(F_1\subset P^*\cap F\) is also an abelian group. Since the elements \(g_1=1,g_2,\ldots,g_s\) are arbitrary, the group \(F\) is abelian. This proves the lemma.

Theorem 1. Let \(D\) be a division ring locally finite over its center \(Z\). Then the factor group \(D^*/Z^*\) contains no nontrivial periodic normal divisors.

Proof. Suppose first that the division ring \(D\) is finite-dimensional over its center. Let \(G\) be a normal divisor in \(D^*\), each element \(g\) of which satisfies the condition \(g^k=z_g\), where \(k\) is some positive number depending on \(g\), and \(z_g\in Z^*\).

Consider the representation of the group \(G\), as was done in the proof of the lemma. Recall that to an element \(g\) there corresponds the linear transformation \(B_g\) according to the following law: \(xg=xB_g\), where \(x\) is an arbitrary element of \(D\). In this case an element \(z\) of \(G\cap Z^*\) passes into the matrix \(zE_n\), where \(n\) is the dimension of \(D\) over \(Z\). From the equality \(g^k=z_g\) it follows that

\[ (B_g)^k=z_gE_n. \tag{6} \]

From (6) we conclude that \(B_g^{kk_1}=E_n\) if and only if \(z_g^{k_1}=1\), where \(k_1\) is any positive number. But \((\det B_g)^k=z_g^n\). Therefore \(\det B_g\) is a root of unity if and only if \(z_g\) is a root of unity. Hence \((B_g)^t=E_n\) for some positive \(t\) if and only if \(\det B_g\) is a root of unity. It follows from this, by virtue of the faithfulness of the representation, that an element \(g\) has finite order if and only if \(\det B_g\) is a root of unity. Consequently, all elements of finite order in the group \(G\) form a subgroup \(F\), which is a normal divisor in \(D^*\). But the group \(F\) is isomorphic to a matrix group. From (4) we obtain that \(F\) is locally finite. Lemma 6 now permits us to conclude that \(F\subset Z^*\). Note that if an element of the group \(G\) is a commutator of elements \(f_1\) and \(f_2\), i.e. \(f=f_1^{-1}f_2^{-1}f_1f_2\), then \(B_f\) is a matrix with determinant equal to 1. From the preceding considerations it is clear that the commutant of the group \(G\) is contained in \(F\), and hence in \(Z^*\). Thus the group \(G\) is metabelian. From (2) it follows that \(G\subset Z^*\).

If the division ring \(D\) is locally finite over its center, then consider the subgroup \(G_1\),

generated by the elements \(g_1=1, g_2,\ldots,g_s\) and the subfield \(D_1=Z(g_2,\ldots,g_s)\). The division ring \(D_1\) is finite-dimensional over its center. The group \(D_1^*\) contains the normal divisor \(G_2=D_1^*\cap G \supset G_1\). From the preceding arguments it follows that \(G_2\) is contained in the center of \(D_1^*\). Consequently, \(G_1\) is an abelian group. In view of the arbitrariness of \(G_1\), we obtain that \(G\) is an abelian normal divisor in \(D^*\), which proves the theorem.

Theorem 2. Let \(D\) be an arbitrary division ring, and let \(Z\) be its center. The factor group \(D^*/Z^*\) contains no nontrivial locally finite normal divisors.

Let \(\bar G\) be a nontrivial locally finite normal divisor in the group \(D^*/Z^*\): \(Z^*, g_2Z^*,\ldots,g_sZ^*\) is some collection of its elements. Then they generate a finite subgroup \(G_i\), consisting of the cosets \(Z^*, g_2Z^*,\ldots,g_sZ^*\), where \(g_1=1, g_2,\ldots,g_s,\ldots,g_{s_1}\) is a system of representatives of the cosets. Consider the subset of the division ring \(D\) consisting of elements

\[ h=\sum_{i=1}^{s_1} z_i g_i, \]

where \(z_i\in Z,\ i=1,\ldots,s_1\). These elements form a subfield which is finite-dimensional over \(Z\). Therefore, if \(G\) is the full inverse image of the group \(\bar G\) in \(D^*\) under the natural homomorphism \(D^*\to D^*/Z^*\), then the set of elements \(a=\sum z_j g_j\), where \(z_j\in Z,\ g_j\in G\), forms a subfield, locally finite-dimensional over its center. In view of the Cartan–Brauer–Hua theorem, this subfield coincides with \(D\). Hence \(D\) is locally finite-dimensional over \(Z\). From Theorem 1 it now follows that \(G\subset Z^*\). This contradiction proves the theorem.

Remark. Theorem 2 can be obtained from Lemma 6 by applying a group-theoretic theorem proved in \((^5)\), p. 78.

Theorem 3. Let \(D\) be a division ring of characteristic \(p\ne 0\), and let \(G\) be a periodic normal divisor in the group \(D^*\). Then \(G\subset Z^*\).

Proof. Denote by \(\Omega\) the prime subfield of the division ring \(D\). Consider the subring \(Y\) of the division ring \(D\) consisting of elements

\[ y=\sum_{i=1}^{n}\alpha_i g_i, \tag{7} \]

where \(\alpha_i\in\Omega,\ g_i\in G\). The minimal one among all possible \(n\) for which (7) holds will be called the length of the element \(y\), if \(y\ne 0\). We shall show that for every \(y\ne 0\) there is a number \(t=t(y)\), with \(t>0\), such that \(y^{t(y)}=1\).

Suppose it has been proved for all \(y\ne 0\) whose lengths are less than \(n\) that \(y^{t(y)}=1\), where \(t(y)>0\). Assume that the element \(y\) has length \(n>1\). Then

\[ y=\sum \alpha_i g_i=\alpha_1g_1\left(1+\sum \alpha'_i g'_i\right)=\alpha_1g_1(1+y_1), \]

where \(y_1\), by assumption, satisfies the condition \(y_1^{t_1(y_1)}=1\). Since the field \(\Omega\) is finite, there exists a number \(t_2>0\) such that \((1+y_1)^{t_2}=1\). Then, in the factor group \(D^*/G\), the coset \(yG\) has finite order. It follows that the element \(y\) has finite order. Thus, all elements \(y\) of the ring \(Y\) satisfy the condition \(y^{t(y)+1}=y\). From Jacobson’s theorem (\((^1)\), p. 314) it follows that \(Y\) is commutative. Since \(Y\supset G\), we have \(G\subset Z^*\). The theorem is proved.

The author expresses deep gratitude to A. G. Kurosh for supervising the work.

Moscow State University
named after M. V. Lomonosov

Received
18 IV 1963

References

\(^1\) I. Jacobson, Structure of Rings, Moscow, 1961.
\(^2\) W. R. Scott, Proc. Am. Math. Soc., 8, 303 (1957).
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\(^4\) M. I. Kargapolov, Siberian Math. J., 3, 834 (1962).
\(^5\) I. Kaplansky, Introduction to Differential Algebra, Moscow, 1959.
\(^6\) M. Sh. Khuzurbazar, DAN, 131, 1268 (1960).
\(^7\) M. Sh. Khuzurbazar, DAN, 137, No. 1, 42 (1961).