V. P. Elizarov

RELATIONS BETWEEN GENERALIZED RINGS OF QUOTIENTS

(Presented by Academician A. I. Mal'cev on 6 VI 1960)

Let \(R\) be an associative ring, and let \(S_1\) and \(S_2\) be multiplicatively closed systems of its nonzero elements, with respect to which there exist left generalized rings of quotients \(R_{(S_1)}\) and \(R_{(S_2)}\) \((^2)\). It is natural to raise the question of how these rings are related to one another. If \(S_1\) and \(S_2\) are not connected by any relations, then it is hard to expect that any relations will arise between \(R_{(S_1)}\) and \(R_{(S_2)}\). Therefore we shall consider the case when \(S_1 \supset S_2\).

Denote by \(I_1\) and \(I_2\) the kernels of the homomorphic mappings of the ring \(R\) into \(R_{(S_1)}\) and \(R_{(S_2)}\). The ring \(R_{(S_i)}\), \(i=1,2\), consists of equivalence classes of pairs of the form \((r,s)\), where \(r \in R\), \(s \in S_i\), and two pairs \((r',s')\) and \((r'',s'')\) are equivalent if the relations \(\bar{s}r' - \bar{r}s' \in I_i\) and \(\bar{s}s' - \bar{r}s'' \in I_i\) hold, where \(\bar{r}\in R\), \(\bar{s}\in S_i\). An element of the ring \(R_{(S_i)}\) represented by the pair \((r,s)\) will be denoted by \(\left[\frac{r}{s}\right]_i\). In the ring \(R_{(S_2)}\) consider the set of elements represented by pairs of the form \((s',s'')\), where \(s'\in S_1\), \(s''\in S_2\), and denote it by \(S_{1(S_2)}\). \(S_{1(S_2)}\) is a multiplicatively closed system of nonzero elements of the ring \(R_{(S_2)}\).

Theorem 1. If \(I_{(S_2)}\) is an arbitrary \(S_{1(S_2)}\)-prime ideal of the ring \(R_{(S_2)}\), then the set \(I\) of numerators of its elements forms an \(S_1\)-prime ideal of the ring \(R\).

Proof. Let \(\left[\frac{i}{s}\right]_2 \in I_{(S_2)}\), i.e. \(i\in I\). For an arbitrary element \(s_2\in S_2\), by the definition of multiplication in the ring \(R_{(S_2)}\), the relation

\[ \left[\frac{s}{s_2}\right]_2\cdot \left[\frac{i}{s}\right]_2 = \left[\frac{si}{ss_2}\right]_2 = \left[\frac{i}{s_2}\right]_2 \in I_{(S_2)}. \]

Consequently, an arbitrary element of \(S_2\) can serve as the denominator of an element from \(I_{(S_2)}\), in the sense that changing the denominator does not affect the membership of the element in the ideal \(I_{(S_2)}\). If \(i_1,i_2\in I\), then, by the definition of addition in the ring \(R_{(S_2)}\), for any element \(s_2\in S_2\) the relation

\[ \left[\frac{i_1}{s_2}\right]_2 - \left[\frac{i_2}{s_2}\right]_2 = \left[\frac{s_2 i_1 - s_2 i_2}{s_2s_2}\right]_2 = \left[\frac{i_1-i_2}{s_2}\right] \in I_{(S_2)}, \]

holds, i.e. \(i_1-i_2\in I\). By the definition of multiplication in the ring \(R_{(S_2)}\), for any elements \(r\in R\), \(i\in I\), and \(s\in S_2\) the relation

\[ \left[\frac{i}{s}\right]_2\cdot \left[\frac{sr}{s}\right]_2 = \left[\frac{\bar{r}sr}{\bar{s}s}\right]_2 \in I_{(S_2)}, \]

holds, where \(\bar{s}\in S_2\), \(\bar{r}\in R\), and \(\bar{s}i-\bar{r}s\in I_2\). But \(I_2\subset I\), since elements with numerators from \(I_2\) are zero elements in the ring \(R_{(S_2)}\). Then \(\bar{s}ir-\bar{r}sr\in I\), \(\bar{r}sr\in I\), and \(\bar{s}ir\in I\). But

\[ \left[\frac{ir}{s_2}\right]_2 = \left[\frac{\bar{s}ir}{ss_2}\right]_2 \]

and \(ir\in I\). From the fact that \(\left[\frac{i}{s}\right]_2 \in I_{(S_2)}\), it follows that \(\left[\frac{si}{ss}\right]_2\in I_{(S_2)}\) and \(\left[\frac{si}{s}\right]_2\in I_{(S_2)}\). Then from the relation

\[ \left[\frac{sr}{s}\right]_2\cdot \left[\frac{si}{s}\right]_2 = \left[\frac{\bar{r}si}{ss}\right]_2, \]

where \(\bar{s}\in S_2\), \(\bar{r}\in R\), and \(\bar{s}sr-\bar{r}s\in I_2\), it follows that \(\bar{r}si\in I\) and \(\bar{s}sri-\bar{r}si\in I_2\subset I\). Hence \(\bar{s}sri\in I\) and \(ri\in I\). Consequently, \(I\) is a two-sided ideal of the ring \(R\).

From the definition of an \(S_{1(S_2)}\)-prime ideal it follows that \(I_{(S_2)}\cap S_{1(S_2)}\) is empty, and hence \(I\cap S_1\) is empty. Let \(r\in R\), \(s_1\in S_1\), and \(rs_1\in I\). For an arbitrary element \(s_2\in S_2\) we have the equality

\[ \left[\frac{s_2r}{s_2}\right]_2\cdot \left[\frac{s_2s_1}{s_2}\right]_2 = \left[\frac{\bar r s_2s_1}{\bar s s_2}\right]_2, \]

where \(\bar r\in R\), \(\bar s\in S_2\), and \(\bar s s_2r-\bar r s_2\in I_2\subset I\). Then \(\bar s s_2rs_1-\bar r s_2s_1\in I\). But \(rs_1\in I\), i.e. \(\bar s s_2rs_1\in I\) and \(\bar r s_2s_1\in I\). Therefore,

\[ \left[\frac{rs_1s_2}{\bar s s_2}\right]_2\in I_{(S_2)} \quad\text{and}\quad \left[\frac{s_2r}{s_2}\right]_2\in I_{(S_2)}, \]

since \(I_{(S_2)}\) is an \(S_{1(S_2)}\)-prime ideal. Hence we obtain that \(r\in I\). In a similar way, from the equality

\[ \left[\frac{s_2s_1}{s_2}\right]_2\cdot \left[\frac{s_2r}{s_2}\right]_2 = \left[\frac{\bar r s_2r}{s_2\bar s}\right]_2, \]

where \(\bar s\in S_2\), \(\bar r\in R\), and \(\bar s s_2s_1-\bar r s_2\in I_2\subset I\), and from the assumption that \(s_1r\in I\), it follows that \(r\in I\). To complete the proof it remains to show that, for any elements \(r\in R\) and \(s_1\in S_1\), there exist elements \(r'\in R\) and \(s'\in S_1\) such that \(s'r-r's_1\in I\). For an arbitrary element \(s_2\in S_2\), consider the elements

\[ \left[\frac{s_2r}{s_2}\right]_2\in R_{(S_2)} \quad\text{and}\quad \left[\frac{s_2s_1}{s_2}\right]_2\in S_{1(S_2)}. \]

By the definition of an \(S_{1(S_2)}\)-prime ideal, there exist elements

\[ \left[\frac{r'_1}{s'_2}\right]_2\in R_{(S_2)} \quad\text{and}\quad \left[\frac{s'_1}{s''_2}\right]_2\in S_{1(S_2)} \]

such that the relation holds

\[ \left[\frac{s'_1}{s''_2}\right]_2\cdot \left[\frac{s_2r}{s_2}\right]_2 - \left[\frac{r'_1}{s'_2}\right]_2\cdot \left[\frac{s_2s_1}{s_2}\right]_2 = \left[\frac{\bar r s_2r}{\bar s s'_2}\right]_2 - \left[\frac{\bar{\bar r}s_2s_1}{\bar{\bar s}s'_2}\right]_2 = \left[ \frac{\beta'\bar r s_2r-\alpha'\bar{\bar r}s_2s_1}{\beta'\bar s s''_2} \right]_2, \]

where \(\dot s,\bar s,\beta'\in S_2\), \(\bar r,\bar{\bar r},\alpha'\in R\),

\[ \bar s s'_1-\bar r s_2\in I_2,\qquad \dot s r'_1-\bar{\bar r}s_2\in I_2,\qquad \beta'\dot s s''_2-\alpha'\bar s s_2\in I_2. \]

By the condition,

\[ \beta'rs_2r-\alpha'\bar{\bar r}s_2s_1\in I. \]

But \(\bar r s_2=\bar s s'_1-i_2\), where \(i_2\in I_2\). Then

\[ \beta'\bar s s'_1r-\beta'i_2r-\alpha'\bar{\bar r}s_2s_1\in I, \]

or

\[ \beta'\bar s s'_1r-\alpha'\bar{\bar r}s_2s_1\in I. \]

But \(\beta'\bar s s'_1\in S_1\), and \(\alpha'\bar{\bar r}s_2\in R\), and these elements may be taken as \(s'\) and \(r'\). The theorem is proved.

Assume now that, along with the condition \(S_1\supset S_2\), the relation \(I_1\supset I_2\) also holds.

Lemma 1. The correspondence \(\alpha\), established by the formula

\[ \alpha\left(\left[\frac{r}{s}\right]_2\right)=\left[\frac{r}{s}\right]_1, \]

is an \(S_{1(S_2)}\)-reducing mapping of the ring \(R_{(S_2)}\) into the ring \(R_{(S_1)}\). The kernel of this mapping is \(I_{1(S_2)}\).

Proof. Let the pairs \((r,s)\) and \((r',s')\), where \(r,r'\in R\) and \(s,s'\in S_2\), be representatives of one and the same element in the ring \(R_{(S_2)}\). Since \(S_1\supset S_2\) and \(I_1\supset I_2\), these pairs are representatives of one and the same element in the ring \(R_{(S_1)}\). Consequently, the mapping \(\alpha\) is single-valued and does not depend on the choice of the representative of the element

\[ \left[\frac{r}{s}\right]_2. \]

By direct verification it is established that the mapping \(\alpha\) is a homomorphism of the ring \(R_{(S_2)}\) into the ring \(R_{(S_1)}\) with kernel \(I_{1(S_2)}\). Let

\[ \left[\frac{s_1}{s_2}\right]_2 \]

be an arbitrary element of \(S_{1(S_2)}\). By the definition of the ring \(R_{(S_1)}\), in it there exists an element

\[ \left[\frac{s_2}{s_1}\right]_1, \]

which is a two-sided inverse for

\[ \left[\frac{s_1}{s_2}\right]_1. \]

But

\[ \alpha\left(\left[\frac{s_1}{s_2}\right]_2\right)=\left[\frac{s_1}{s_2}\right]_1, \]

i.e.

\[ \left(\alpha\left(\left[\frac{s_1}{s_2}\right]_2\right)\right)^{-1} = \left[\frac{s_2}{s_1}\right]_1 \in R_{(S_1)}. \]

Let

\[ \left[\frac{r}{s}\right]_1 \]

be an arbitrary element of \(R_{(S_1)}\). For any \(s_2\in S_2\) the equality holds

\[ \left[\frac{s_2}{s}\right]_1\cdot \left[\frac{r}{s_2}\right]_1 = \left[\frac{r}{s}\right]_1. \]

Then

\[ \left[\frac{r}{s}\right]_1 = \left(\alpha\left(\left[\frac{s}{s_2}\right]_2\right)\right)^{-1}\cdot \alpha\left(\left[\frac{r}{s_2}\right]_2\right). \]

The lemma is proved.

Lemma 2. If the ring \(R_{(S_2)}\) is mapped into some ring \(R'\) by means of some \(S_{1(S_2)}\)-reducing mapping \(\beta\), then the mapping \(\psi\), defined by the formula

\[ \psi\left(\left[\frac{r}{s}\right]_1\right) = \left(\beta\left(\left[\frac{s}{s_2}\right]_2\right)\right)^{-1}\cdot \beta\left(\left[\frac{r}{s_2}\right]_2\right), \]

where \(r\in R\), \(s\in S_1\), \(s_2\in S_2\), is such a homomorphic mapping of the ring \(R_{(S_1)}\) into the ring \(R'\) that

\[ \beta\bigl(R_{(S_2)}\bigr)=\psi\bigl(\alpha(R_{(S_2)})\bigr). \]

Proof. Let \(s_2' \in S_2\). By the definition of multiplication in the ring \(R_{(S_2)}\), we have
\(\beta\left(\left[\dfrac{s_2}{s_2'}\right]_2\cdot\left[\dfrac{r}{s_2}\right]_2\right) =\beta\left(\left[\dfrac{s_2}{s_2'}\right]_2\right)\cdot \beta\left(\left[\dfrac{r}{s_2}\right]_2\right) =\beta\left(\left[\dfrac{s_2r}{s_2^{0}s_2'}\right]_2\right) =\beta\left(\left[\dfrac{r}{s_2'}\right]_2\right)\), where \(s_2^{0}\in S_2\), \(\overline{s}_2\in R\), and \(s_2^{0}s_2-\overline{s}_2s_2\in I_2\). Since \(\beta\) is an \(S_{1(S_2)}\)-reducing homomorphism, it follows that
\(\beta\left(\left[\dfrac{r}{s_2}\right]_2\right) =\left(\beta\left(\left[\dfrac{s_2}{s_2'}\right]_2\right)\right)^{-1}\cdot \beta\left(\left[\dfrac{r}{s_2'}\right]_2\right)\), and
\(\beta\left(\left[\dfrac{s}{s_2}\right]_2\right) =\left(\beta\left(\left[\dfrac{s_2}{s_2'}\right]_2\right)\right)^{-1}\cdot \beta\left(\left[\dfrac{s}{s_2'}\right]_2\right)\).
Then
\(\psi\left(\left[\dfrac{r}{s}\right]_1\right) =\left(\beta\left(\left[\dfrac{s}{s_2'}\right]_2\right)\right)^{-1}\cdot \beta\left(\left[\dfrac{s_2}{s_2'}\right]_2\right)\cdot \left(\beta\left(\left[\dfrac{s_2}{s_2'}\right]_2\right)\right)^{-1}\cdot \beta\left(\left[\dfrac{r}{s_2'}\right]_2\right) =\left(\beta\left(\left[\dfrac{s}{s_2'}\right]_2\right)\right)^{-1}\cdot \beta\left(\left[\dfrac{r}{s_2'}\right]_2\right)\).
Consequently, the formula does not depend on the choice of the element \(s_2\in S_2\).

Suppose that the pair \((r_1,s_1)\) is equivalent to the pair \((r,s)\), i.e., there exist elements \(r'\in R\) and \(s'\in S_1\) such that \(s's-r's_1\in I_1\) and \(s'r-r'r_1\in I_1\). Then for any element \(s_2\in S_2\) we have
\(\beta\left(\left[\dfrac{s_2s's}{s_2s_2s_2}\right]_2\right) =\beta\left(\left[\dfrac{s_2r's_1}{s_2s_2s_2}\right]_2\right)\) and
\(\beta\left(\left[\dfrac{s's_2}{s_2s_2}\right]_2\right)\cdot \beta\left(\left[\dfrac{s}{s_2}\right]_2\right) =\beta\left(\left[\dfrac{r's_2}{s_2s_2}\right]_2\right)\cdot \beta\left(\left[\dfrac{s_1}{s_2}\right]_2\right)\).
Analogously,
\(\beta\left(\left[\dfrac{s's_2}{s_2s_2}\right]_2\right)\cdot \beta\left(\left[\dfrac{r}{s_2}\right]_2\right) =\beta\left(\left[\dfrac{r's_2}{s_2s_2}\right]_2\right)\cdot \beta\left(\left[\dfrac{r_1}{s_2}\right]_2\right)\).
Hence it follows that
\(\beta\left(\left[\dfrac{r's_2}{s_2s_2}\right]_2\right) =\beta\left(\left[\dfrac{s's_2}{s_2s_2}\right]_2\right)\cdot \beta\left(\left[\dfrac{s}{s_2}\right]_2\right)\cdot \left(\beta\left(\left[\dfrac{s_1}{s_2}\right]_2\right)\right)^{-1}\), i.e.
\(\beta\left(\left[\dfrac{r}{s_2}\right]_2\right) =\beta\left(\left[\dfrac{s}{s_2}\right]_2\right)\cdot \left(\beta\left(\left[\dfrac{s_1}{s_2}\right]_2\right)\right)^{-1}\cdot \beta\left(\left[\dfrac{r_1}{s_2}\right]_2\right)\).
Finally,
\(\left(\beta\left(\left[\dfrac{s}{s_2}\right]_2\right)\right)^{-1}\cdot \beta\left(\left[\dfrac{r}{s_2}\right]_2\right) =\left(\beta\left(\left[\dfrac{s_1}{s_2}\right]_2\right)\right)^{-1}\cdot \beta\left(\left[\dfrac{r_1}{s_2}\right]_2\right)\).
Consequently, the formula does not depend on the choice of representative of the element \(\left[\dfrac{r}{s}\right]_1\). The remainder of the proof of the lemma is carried out by direct verification, since from Lemma 1 it follows that \(I_{1(S_2)}\) is an \(S_{1(S_2)}\)-prime ideal of the ring \(R_{(S_2)}\), and from Theorem 1 it follows that \(I_{1(S_2)}\) belongs to every \(S_{1(S_2)}\)-prime ideal that serves as the kernel of an \(S_{1(S_2)}\)-reducing homomorphism \((^{2})\). The lemma is proved.

As an immediate consequence of the above, we obtain

Theorem 2. If \(S_1, S_2\) are multiplicatively closed systems of elements of an associative ring \(R\), with respect to which the left generalized rings of quotients \(R_{(S_1)}\), \(R_{(S_2)}\) exist, and \(S_1\supset S_2\), \(I_1\supset I_2\), then the ring \(R_{(S_1)}\) is the left generalized ring of quotients of the ring \(R_{(S_2)}\) with respect to the system \(S_{1(S_2)}\).

An analogous theorem also holds for right generalized rings of quotients.

If \(R\) is a commutative ring, then the condition \(I_1\supset I_2\) already follows from the condition \(S_1\supset S_2\), and the rings \(R_{(S_1)}\) and \(R_{(S_2)}\) always exist. Consequently, \((^{1})\) holds:

Theorem 3. If \(R\) is a commutative ring; \(R_{(S_1)}\), \(R_{(S_2)}\) are its generalized rings of quotients with respect to multiplicatively closed systems \(S_1, S_2\) of its elements, and \(S_1\supset S_2\), then the ring \(R_{(S_1)}\) is the generalized ring of quotients of the ring \(R_{(S_2)}\) with respect to the system \(S_{1(S_2)}\).

We note that if \(I_1=I_2\), then the ring \(R_{(S_1)}\) contains a subring isomorphic to the ring \(R_{(S_2)}\).

Received
26 V 1960

CITED LITERATURE

\(^{1}\) A. I. Uzkov, Matem. sborn., 22 (64), no. 3, 439 (1948).
\(^{2}\) V. P. Elizarov, Izv. Akad. Nauk SSSR, ser. matem., 24, no. 2, 153 (1960).