UDC 591.44

MATHEMATICS

Z. I. BOREVICH, Corresponding Member of the Academy of Sciences of the USSR D. K. FADDEEV

A NOTE ON ORDERS WITH CYCLIC INDEX

1°. Let \(\mathfrak{o}\) be a Dedekind ring, \(k\) its field of fractions, \(K/k\) a finite separable extension, and \(\mathfrak{D}\) the integral closure of the ring \(\mathfrak{o}\) in the field \(K\). By a full \(\mathfrak{o}\)-module in \(K\) one means a finitely generated \(\mathfrak{o}\)-submodule of the field \(K\) containing \((K:k)\) elements linearly independent over \(k\). A full \(\mathfrak{o}\)-module in \(K\) containing the identity element of the field and being a ring is called an \(\mathfrak{o}\)-order of the field \(K\). Every \(\mathfrak{o}\)-order \(\Lambda\) of the field \(K\) is contained, as is known, in the maximal \(\mathfrak{o}\)-order \(\mathfrak{D}\).

In the paper \((^1)\) it is shown that if the \(\mathfrak{o}\)-order \(\Lambda\) has cyclic index in \(\mathfrak{D}\) (see the definition below), then every finitely generated torsion-free \(\Lambda\)-module decomposes into a direct sum of \(\Lambda\)-modules that are \(\Lambda\)-isomorphic to ideals of the ring \(\Lambda\). On the other hand, in Bass’s paper \((^2)\) it is established that if \(R\) is a Noetherian integral domain whose integral closure (in its field of fractions) is a finitely generated \(R\)-module, then every finitely generated torsion-free \(R\)-module decomposes into a direct sum of \(R\)-ideals if and only if every ideal of the ring \(R\) is generated by no more than two elements. (The integral closure of such a ring \(R\) is a Dedekind ring.)

It turns out that the following theorem is true (in the notation and assumptions given above).

Theorem. In order that every finitely generated torsion-free \(\Lambda\)-module decompose into a direct sum of \(\Lambda\)-ideals, it is necessary and sufficient that the \(\mathfrak{o}\)-order \(\Lambda\) have cyclic index in the maximal \(\mathfrak{o}\)-order \(\mathfrak{D}\) of the field \(K\).

The sufficiency of the condition, as already noted, was proved in \((^1)\). The necessity follows from the result of Bass \((^2)\) cited above and Lemma 1 below.

2°. Definition. Let \(\mathfrak{D}\) be a commutative ring with identity and \(\Lambda\) its subring whose identity coincides with the identity of the ring. We say that \(\Lambda\) has cyclic index in \(\mathfrak{D}\) if the factor-module \(\mathfrak{D}/\Lambda\), as a \(\Lambda\)-module, is cyclic (i.e., generated by one element).

The cyclicity of the index of \(\Lambda\) in \(\mathfrak{D}\) is evidently equivalent to the existence of an element \(\omega \in \mathfrak{D}\) such that \(\mathfrak{D} = \Lambda + \Lambda \omega\).

Lemma 1. If the ring \(\mathfrak{D}\), as a \(\Lambda\)-module, admits a system of \(\Lambda\)-generators consisting of two elements, then \(\Lambda\) has cyclic index in \(\mathfrak{D}\).

The proof of Lemma 1 is based on the following assertion.

Lemma 2. Suppose that the ring \(\mathfrak{D}\) is a finitely generated \(\Lambda\)-module. If elements \(\lambda_1, \ldots, \lambda_m\) of \(\Lambda\) are relatively prime in \(\mathfrak{D}\) (i.e., generate the unit ideal in \(\mathfrak{D}\)), then they are relatively prime also in the ring \(\Lambda\).

Proof. Let \(\mathfrak{D} = \Lambda \omega_1 + \cdots + \Lambda \omega_n\) and let \(\mathfrak{a} = \Lambda \lambda_1 + \cdots + \Lambda \lambda_m\). Since, by hypothesis, \(\mathfrak{a}\mathfrak{D} = \mathfrak{D}\), we have

\[ \omega_i = \sum_{j=1}^{n} \alpha_{ij}\omega_j \quad (1 \leq i \leq n), \tag{1} \]

where the coefficients \(\alpha_{ij}\) belong to the ideal \(\mathfrak{a}\). Denote by \(\Delta\) the deter-

divisor of the matrix \((a_{ij}) - E\), where \(E\) is the identity matrix of order \(n\). From (1) it follows easily that \(\Delta \omega_i = 0\) for all \(i = 1,\ldots,n\), and hence \(\Delta = 0\). But, since \(a_{ij} \in \mathfrak a\), we have \(\Delta \equiv 1 \pmod{\mathfrak a}\). Thus \(1 \in \mathfrak a\), and Lemma 2 is proved.

Proof of Lemma 1. Let \(\mathfrak D = \Lambda \omega_1 + \Lambda \omega_2\). Then \(1 = \lambda_1 \omega_1 + \lambda_2 \omega_2\), where \(\lambda_1, \lambda_2 \in \Lambda\). By Lemma 2, in the ring \(\Lambda\) there exist elements \(\lambda\) and \(\mu\) such that \(1 = \lambda_1 \mu - \lambda_2 \lambda\). Put \(\omega = \lambda \omega_1 + \mu \omega_2\). Since the systems \(\omega_1, \omega_2\) and \(1, \omega\) are related by a unimodular transformation, it follows that \(\mathfrak D = \Lambda + \Lambda \omega\), as was required to prove.

References

¹ Z. I. Borevich, D. K. Faddeev, Tr. Mat. Inst. im. V. A. Steklova AN SSSR, 80, 51 (1965). ² H. Bass, Trans. Am. Math. Soc., 102, No. 2, 319 (1962).