UDC 519.48

MATHEMATICS

Academician of the Academy of Sciences of the MSSR V. A. ANDRUNAKIEVICH, Yu. M. RYABUKHIN

RINGS WITHOUT NILPOTENT ELEMENTS AND COMPLETELY PRIME IDEALS

In the present note it is proved that an associative ring is a ring without nonzero nilpotent elements if and only if it is isomorphic to a subdirect sum of rings without zero divisors. This assertion generalizes the analogous well-known result of Krull for commutative rings (see (¹); (²), Theorem 24).

Let \(K\) be an arbitrary associative ring. Recall that an ideal \(P \ne K\) is called prime if from \(AB \subseteq P\), where \(A\) and \(B\) are ideals in \(K\), it follows that \(A \subseteq P\) or \(B \subseteq P\). It is known (³, ⁴) that \(P\) is a prime ideal if and only if from \(aKb \subseteq P\), where \(a\) and \(b\) are elements of the ring \(K\), it follows that \(a \in P\) or \(b \in P\). A set \(M\) of elements of the ring \(K\) is called an \(m\)-system if for any \(a, b \in M\) there exists an \(x \in K\) such that \(axb \in M\). It is clear that if \(P\) is a prime ideal, then the complement \(C(P)=K\setminus P\) is an \(m\)-system. In some cases the converse assertion is also true. Namely, with the aid of the Kuratowski–Zorn lemma it is easy to show that any \(m\)-system \(M\) not intersecting an ideal \(A \ne K\) is contained in some maximal \(m\)-system \(M^*\) not intersecting \(A\). The complement \(C(M^*)\) will be a prime ideal—the minimal prime ideal containing the ideal \(A\) (see (³, ⁴)). Indeed, by the Kuratowski–Zorn lemma, among the ideals of the ring \(K\) containing \(A\) and not intersecting \(M^*\), there exists a maximal ideal \(P\). Let us show that \(P\) is a prime ideal. In fact, if \(D_1\) and \(D_2\) are such ideals in \(K\) that \(D_1 \supset P\) and \(D_2 \supset P\), then, by the maximality of \(P\), in \(K\) there exist elements \(d_1\) and \(d_2\) such that \(d_1 \in D_1 \cap M^*\), \(d_2 \in D_2 \cap M^*\). Since \(M^*\) is an \(m\)-system, there exists an \(x \in K\) such that \(d_1xd_2 \in M^*\). But \(d_1xd_2 \in D_1D_2\). By the primeness of the ideal \(P\), the complement \(C(P)\) is an \(m\)-system. From the relations \(P \supset A\), \(P \cap M^*=\varnothing\) we obtain \(C(P)\cap A=\varnothing\), \(M^* \subseteq C(P)\). By the maximality of \(M^*\), \(M^*C(P)=M^*\), and therefore \(C(M^*)=P\). It is clear that \(P\) is a minimal prime ideal containing \(A\). Indeed, if there existed a prime ideal \(P'\) with \(P \supset P' \supset A\), then, on the one hand, \(C(P')\cap A=\varnothing\), and, on the other hand, \(C(P')\cap A\ne\varnothing\), since \(C(P') \supset C(P)=M^*\). Recall now that an ideal \(P \ne K\) is called completely prime if from \(ab \in P\) it follows that \(a \in P\) or \(b \in P\), i.e., if the factor ring \(K/P\) contains no zero divisors. An ideal \(Q \ne K\) will be called completely semiprime if from \(a^n \in Q\) it follows that \(a \in Q\), i.e., if the factor ring \(K/Q\) is a ring without nonzero nilpotent elements. Obviously, a ring without nonzero nilpotent elements is a ring in which zero is a completely semiprime ideal. It is clear that a completely prime ideal is also a completely semiprime ideal. Moreover, the intersection of any set of completely semiprime ideals is a completely semiprime ideal.

Lemma on the existence of completely prime ideals.
If \(Q\) is a completely semiprime ideal and \(B\) is such an ideal that \(B \supset Q\), then there exists a completely prime ideal \(P\) such that \(P \supset Q\), but \(P \nsupseteq B\).

Proof. Let \(b\) be such an element of the ring \(K\) that \(b \in B\), \(b \notin Q\). Then, by the condition of the lemma, the \(m\)-system \(M=\{b^i\},\ i=1,2,\ldots,\) does not intersect the ideal \(Q\). Denote by \(M^*\) a maximal \(m\)-system containing \(M\) and not intersecting the ideal \(Q\). By what was said above

complement \(C(M^*)=P\) will be a minimal prime ideal containing \(Q\). Since \(b\notin C(M^*)=P\), we have \(P\not\supset B\). Let us prove that \(P\) is a completely prime ideal. For this purpose consider the set \(M'\) of all elements of the form \(x_1^{r_1}x_2^{r_2}\ldots x_n^{r_n}\), where \(r_i\) are natural numbers and \(x_1x_2\ldots x_n\in M^*\). Clearly, \(M^*\subseteq M'\). Note that \(M'\) is an \(m\)-system. Indeed, if \(a,b\in M'\), then

\[ a=a_1^{r_1}a_2^{r_2}\ldots a_m^{r_m},\quad a_1a_2\ldots a_m\in M^*;\quad b=b_1^{s_1}b_2^{s_2}\ldots b_n^{s_n},\quad b_1b_2\ldots b_n\in M^*. \]

Since \(M^*\) is an \(m\)-system, there exists an element \(x\in K\) such that \(a_1a_2\ldots a_mxb_1b_2\ldots b_n\in M^*\). But then, by the definition of the set \(M'\),

\[ axb=a_1^{r_1}a_2^{r_2}\ldots a_m^{r_m}xb_1^{s_1}b_2^{s_2}\ldots b_n^{s_n}\in M'. \]

We shall now show that the \(m\)-system \(M'\) does not meet the ideal \(Q\). For this, first note that from \(xy\in Q\) it follows that \(yx\in Q\), and conversely. Indeed, if \(xy\in Q\), then \((yx)^2=y(xy)x\in Q\), and, since \(Q\) is a completely semiprime ideal, \(yx\in Q\). Similarly, from \(yx\in Q\) it follows that \(xy\in Q\). Suppose now that \(M'\cap Q\ne\varnothing\), and let \(c\in M'\cap Q\). Then \(c=c_1^{r_1}c_2^{r_2}\ldots c_n^{r_n}\in Q\), \(c_1c_2\ldots c_n\in M^*\). If \(r_1>1\), then, by the remark just made above, \(c_1^{r_1-1}c_2^{r_2}\ldots c_n^{r_n}c_1\in Q\), whence \((c_1^{r_1-1}c_2^{r_2}\ldots c_n^{r_n})^2\in Q\), and therefore \(c_1^{r_1-1}c_2^{r_2}\ldots c_n^{r_n}\in Q\). After a finite number of steps we obtain \(c_1c_2^{r_2}\ldots c_n^{r_n}\in Q\). But then \(c_2^{r_2}\ldots c_n^{r_n}c_1\in Q\). Repeating the same arguments for \(r_2,\ldots,r_n\), after a finite number of steps we obtain \(c_1c_2\ldots c_n\in Q\), and therefore \(c_1c_2\ldots c_n\in Q\cap M^*\). But, by the choice of \(M^*\), \(Q\cap M^*=\varnothing\). The contradiction obtained proves that \(Q\cap M'=\varnothing\). Hence, and from the maximality of the \(m\)-system \(M^*\), we obtain \(M'=M^*=C(P)\). Note now that from \(a\in M^*\) it follows that \(a^2\in M^*\). Indeed, by the definition, \(M'a^2\subseteq M'\). Finally, let us prove that \(P\) is a completely prime ideal. Let \(ab\in P\). Then \((ba)^2\in P\), whence \((ba)^2\notin C(P)=M^*\). Consequently, \(ba\notin M^*\), and therefore \(ba\in C(M^*)=P\). But then for any \(x\in K\), from \(ab\in P\) it follows that \(abx\in P\), \(bxa\in P\), i.e. \(bKa\subseteq P\). By the primeness of the ideal \(P\) we get \(b\in P\) or \(a\in P\). The lemma is proved.

Note that, since \(M^*=C(P)\) and \(P\) is a completely prime ideal, from the proof of the lemma itself we obtain

Corollary 1. Every maximal \(m\)-system \(M^*\) not meeting a given completely semiprime ideal \(Q\ne K\) is a multiplicative system, i.e. from \(a\in M^*\), \(b\in M^*\) it follows that \(ab\in M^*\).

Corollary 1. In every ring \(K\ne 0\) without nonzero nilpotent elements there exists a completely prime ideal not containing a given nonzero ideal.

Indeed, in the assumptions of the lemma it suffices to put \(Q=0\).

Theorem 1. Every completely semiprime ideal of a ring is the intersection of all completely prime ideals containing it.

Indeed, let \(B=\bigcap_\alpha P_\alpha\), where \(P_\alpha\) runs through all completely prime ideals containing the completely prime ideal \(Q\). Clearly, \(B\supseteq Q\). Suppose that \(B\supset Q\). Then, by the lemma, there exists a completely prime ideal \(P_{\alpha_0}\) such that \(P_{\alpha_0}\supseteq Q\) and \(P_{\alpha_0}\not\supset B\), which contradicts the definition of the ideal \(B\). Consequently,

\[ Q=B=\bigcap_\alpha P_\alpha. \]

It follows at once from Theorem 1 that the intersection \(D_1\) of all completely semiprime ideals containing a given ideal \(A\) coincides with the intersection \(D_2\) of all completely prime ideals containing \(A\). Indeed, it is clear that \(A\subseteq D_1\subseteq D_2\) and, since \(D_1\) is a completely semiprime ideal, by Theorem 1,

\[ D_1=D_2. \]

Theorem 2. In order that a ring \(K\ne 0\) be a ring without nilpotent elements \(\ne 0\), it is necessary and sufficient that \(K\) be isomorphic to a subdirect sum of rings without zero divisors (see also \((10)\)).

Indeed, suppose first that \(K\) is a subdirect sum of rings without zero divisors. Since any subring of a complete direct sum of rings without zero divisors is, obviously, a ring without nonzero nilpotent elements, \(K\) will also be a ring of the same kind. Suppose now that the ring \(K \ne 0\) has no nilpotent elements \(\ne 0\). Since \(0\) is a completely prime ideal, by Theorem 1,

\[ 0=\bigcap_\alpha P_\alpha, \]

where \(P_\alpha\) ranges over all completely prime ideals of the ring \(K\). Consequently, \(K\) is isomorphic to a subdirect sum of rings without zero divisors \(\overline K_\alpha=K/P_\alpha\). Let \(K\) be an arbitrary associative ring. We construct by induction a chain of ideals of the ring \(K\)

\[ 0=\mathfrak N_0(K)\subseteq \mathfrak N_1(K)\subseteq \ldots \subseteq \mathfrak N_i(K)\subseteq \mathfrak N_{i+1}(K)\subseteq \ldots, \]

where \(i\) ranges over the natural numbers, and \(\mathfrak N_{i+1}(K)\) is the ideal of the ring \(K\) generated by the set

\[ \mathfrak n_i(K)=\{a\mid a\in K,\ a^n\in \mathfrak N_i(K)\ \text{for some natural } n=n(a)\}, \]

that is, by the set of all elements of the ring \(K\) nilpotent modulo \(\mathfrak N_i(K)\). Put

\[ \mathfrak N(K)=\bigcup_0^\infty \mathfrak N_i(K). \]

Proposition 2. The ideal \(\mathfrak N(K)\) is the least completely semiprime ideal of the ring \(K\), i.e. the least among all such ideals \(Q\) for which the quotient ring \(K/Q\) has no nilpotent elements \(\ne 0\).

Indeed, let \(Q\) be a completely semiprime ideal. It is clear that \(\mathfrak N_0\subseteq Q\). If it has already been proved that \(\mathfrak N_i\subseteq Q\) for some \(i\), then \(\mathfrak n_i(K)\subseteq Q\). But then also \(\mathfrak N_{i+1}\subseteq Q\). Consequently, for all \(i\), \(\mathfrak N_i\subseteq Q\), and therefore \(\mathfrak N(K)\subseteq Q\). On the other hand, \(\mathfrak N(K)\) is a completely semiprime ideal, since, if \(b^n\in \mathfrak N(K)\), then \(b^n\in \mathfrak N_i(K)\) for some \(i\), and therefore \(b\in \mathfrak N_{i+1}(K)\subseteq \mathfrak N(K)\).

From Proposition 2 and Theorem 1 we obtain

Proposition 3. The ideal \(\mathfrak N(K)\) is the intersection of all completely prime ideals of the ring.

Since the generalized nil radical \(N_g\) of the ring \(K\) also coincides with the intersection of all its completely prime ideals (see \((^5,^6)\); \((^7)\), p. 153), it follows that

Corollary 2. The ideal \(\mathfrak N(K)\) coincides with the generalized nil radical \(N_g\) of the ring \(K\), or, what is the same, with the compressive radical (see \((^8)\)).

Corollary 3. The ideal \(\mathfrak N(K)\) is a special radical. The \(\mathfrak N\)-radical rings are precisely the rings that do not map onto nonzero rings without zero divisors, or, what is the same, the rings without completely prime ideals. The \(\mathfrak N\)-semiprime rings coincide with subdirect sums of rings without zero divisors, i.e. with rings without nonzero nilpotent elements. For any ideal \(A\) of the ring \(K\), the equality \(\mathfrak N(A)=A\cap \mathfrak N(K)\) holds.

From what was said above it follows that the generalized nil radical \(N_g\) is the least (see \((^9)\)) among all such radicals \(r\) that in any ring \(K\) all nilpotent elements belong to the \(r\)-radical \(r(K)\) of the ring \(K\), i.e. \(N_g(K)\subseteq r(K)\) for every such radical. Consider the property \(\nu\) of being the ring generated as an ideal by its nilpotent elements. Then from the preceding it follows that the generalized nil radical \(N_g\) coincides with the lower radical in the sense of Kurosh generated by the property \(\nu\) (see \((^9)\)). In conclusion we note that the majority of the results obtained carry over without change to semigroups.

Received
29 I 1967

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