UDC 519.48

MATHEMATICS

V. P. ELIZAROV

ON THE MULTIPLICATIVE CLOSEDNESS OF A SYSTEM OF ELEMENTS OF A RING OF QUOTIENTS

(Presented by Academician V. M. Glushkov, 24 IV 1968)

Professor H. J. Weinert drew the author’s attention to the fact that the system of elements (S_{1(S_2)}) of the ring (R_{(S_2)}) need not be multiplicatively closed (see ((^1))). Thus, the results of ((^1)) are valid only under the condition that the system (S_{1(S_2)}) is multiplicatively closed. Below we shall give a condition that is sufficient for this.*

A multiplicatively closed system (S) of elements of a ring (R), not containing zero (an m.c. system), is called a maximal m.c. system if it is not contained in any other m.c. system. For an m.c. system (S), by ({S,r}), where (r \in R \setminus S), we shall denote the least m.c. system containing (S) and (r), if such a system exists. The elements of the system ({S,r}) have the form (s_0 r s_1 r \ldots r s_n), where (s_i \in S), and some of the (s_i) may be absent.

Let (S) be an m.c. system. A two-sided ideal (I) of the ring (R) is called almost (S)-prime if (I \cap S = \varnothing) and from (rs \in I) or (sr \in I), where (r \in R) and (s \in S), it follows that (r \in I) ((^2)). We shall call an m.c. system (S) maximal with respect to an almost (S)-prime ideal (I) if the ideal (I) is not almost ({S,r})-prime for any (r \in R \setminus S) ((^3)).

A maximal m.c. system (S) is maximal with respect to any almost (S)-prime ideal. The converse is not true. In the ring (Z/(6)), take two m.c. systems (S={\overline{1},\overline{5}}) and (S_1={\overline{1},\overline{2},\overline{4},\overline{5}}). The ideal 0 is almost (S)-prime, since there are no zero divisors in (S). Any m.c. system ({S,r}), where (r \in R \setminus S), already contains zero divisors, i.e. 0 is not an almost ({S,r})-prime ideal, and the m.c. system (S) is maximal with respect to the ideal 0. At the same time (S \subset S_1), i.e. the m.c. system (S) is not maximal.

Lemma. An m.c. system (S) of a ring (R) is maximal with respect to an almost (S)-prime ideal (I) if and only if the set of all non-zero-divisors in (\varphi(R) \cong R/I) coincides with (\varphi(S)) (see ((^3))).

Proof. Let the system (S) be maximal with respect to the ideal (I), and let (\varphi(r)) be a non-zero-divisor in (\varphi(R)). Then from (r_1 r \in I) or (r r_1 \in I), where (r_1 \in R), it follows that (r_1 \in I). Suppose (r \notin S) and there exists an m.c. system ({S,r}). Since the ideal (I) is no longer almost ({S,r})-prime, either ({S,r}\cap I \ne \varnothing), or from the relation (s' r \in I) (or (r s' \in I)), where (s' \in {S,r}), it does not follow that (r \in I).

Let ({S,r}\cap I \ne \varnothing), i.e. (s_0 r s_1 r \ldots r s_n \in {S,r}\cap I). Since (I) is an almost (S)-prime ideal, (r s_1 r \ldots s_{n-1} r \in I) or (r r' \in I), where (r' = s_1 r \ldots r s_{n-1}). By the preceding, (r' \in I), and again (r s_2 \ldots s_{n-1} r \in I). Proceeding similarly, we obtain (r \in I), which contradicts the condition.

In exactly the same way, from (s' r \in I) or (r s' \in I) it follows that (r \in I). Thus (I) is an almost ({S,r})-prime ideal, which is impossible if (r \in R \setminus S). Consequently, (\varphi(r) \in \varphi(S)).

If, however, the m.c. system ({S,r}) does not exist, then (s_0 r s_1 r \ldots s_{n-1} r s_n = 0), and we can argue as above.

* Taking this occasion, the author expresses his gratitude to Professor H. J. Weinert.

Let now (\varphi(S)) be the set of all non-zero-divisors in (\varphi(R)). If (r \in R \setminus \bar S), then there exists an element (r_1 \in R \setminus I) such that (rr_1 \in I) or (r_1r \in I). If (r \notin I), then the second condition in the definition of an almost ({S,r})-prime ideal is violated. If, however, (r \in I), then ({S,r}\cap I=\varnothing). Thus (S) is maximal relative to the almost (S)-prime ideal (I). The lemma is proved (see (3)).

Assertion. If (I_1 \supset I_2), (S_1 \supset S_2), and the system (S_1) is maximal relative to the ideal (I_1), then the system (S_{1(S_2)}) is multiplicatively closed (see (1)).

Proof. Let (\dfrac{s_1}{s_2}, \dfrac{s'1}{s'_2}\in S). Then
[
\frac{s_1}{s_2}\frac{s'1}{s'_2}=\frac{rs'_1}{s'_2\alpha},
]
where (s_1\alpha-s'_2r\in I_2), (r\in R), and (\alpha\in S_2). Since (I_2\subset I_1) and (S_2\subset S_1), we have (s_1\alpha\in S_1), (s'_2\in S_1), and (s-\bar s r\in I), where (s=s_1\alpha), (\bar s=s'_2). In the ring (\varphi_1(R)\cong R/I_1) we have the equality (\varphi_1(s)=\varphi(\bar s)\varphi_1(r)), and in the ring (R\varphi_1(s)) is invertible, i.e. (\varphi_1(r)) is a non-zero-divisor in (\varphi_1(R)). By the lemma, (\varphi_1(r)\in \varphi_1(S_1)), and (r\in S_1). Thus (rs'}) the element (\varphi_1(r)=\varphi_1(\bar s)^{-11\in S_1) and
[
\frac{s_1}{s_2}\frac{s'_1}{s'_2}\in S.
]
The assertion is proved.

REFERENCES

({}^{1}) V. P. Elizarov, DAN, 135, No. 2, 252 (1960).
({}^{2}) V. P. Elizarov, A. I. Pilatovskaya, Siberian Mathematical Journal, 5, No. 5, 1191 (1964).
({}^{3}) V. P. Elizarov, A. I. Pilatovskaya, Fifth All-Union Colloquium on General Algebra (summary of communications and reports), Novosibirsk, 1963, p. 20.