UDC 519.48

MATHEMATICS

Academician of the Academy of Sciences of the MSSR V. A. Andrunakievich, V. I. Arnautov

INVERTIBILITY IN TOPOLOGICAL RINGS

It is known (see, for example, (¹)) that any topological ring \(R\) can be embedded as an everywhere dense subring in a complete ring \(\hat R\), i.e., a ring in which every Cauchy filter has a limit (\(\hat R\) is determined up to isomorphism and is called the completion of the ring \(R\)).

In the present paper it is proved that, in order that the completion \(\hat R\) of a topological ring \(R\) contain an identity and that every element of \(R\) be invertible in \(\hat R\), it is necessary and sufficient that \(R\) contain no closed one-sided ideals* and no generalized zero divisors (for the definition of a generalized zero divisor see below). From this, as a consequence, it follows that a complete ring containing no closed one-sided ideals and no generalized zero divisors is a field. It is proved that a locally bicompact ring \(R\) with identity, containing no closed one-sided ideals, is a field.

Lemma 1. In order that a topological ring \(R\) with nonzero multiplication contain no closed left (right) ideals, it is necessary and sufficient that for any elements \(0 \ne a, b \in R\) and any neighborhood of zero \(V\) there exist an element \(x \in R\) such that \(xa + b \in V\) \((ax + b \in V)\).

Sufficiency. Let \(I\) be some left ideal. We shall show that \([I]_R = R\)***. Indeed, if \(b\) is some element of \(R\) and \(0 \ne a \in I\), then for any neighborhood of zero \(V\) there exists an element \(x \in R\) such that \(xa + b \in V\). Then \(xa \in -b + V\). Consequently, every neighborhood of the element \(-b\) has a nonempty intersection with \(I\). Hence \(-b \in [I]_R\). From the arbitrariness of \(b\) it follows that \([I]_R = R\).

Necessity. Let \(a \ne 0\) and \(b\) be some elements of \(R\), and let \(V\) be any neighborhood of zero. Consider the left ideal \(Ra\). If \(Ra = 0\), then \(I=\{c\mid Rc=0\}\) is a closed ideal. Since \(a \in I\), it follows that \(I=R\). Then \(R^2=0\). But this is impossible, since \(R\) is a ring with nonzero multiplication. Hence \(Ra \ne 0\). Since \([Ra]_R=R\), we have \(-b \in [Ra]_R\). Consequently, \(-b - V \cap R\cdot a = \varnothing\), and hence \(b+xa \in V\) for some \(x \in R\).

Recall that an element \(a\) of a topological ring \(R\) is called a left (right) generalized zero divisor if there exists a neighborhood of zero \(V\) such that for every neighborhood of zero \(U\) one can find an element \(x \notin V\) for which \(ax \in U\) \((xa \in U)\). An element \(a\) is called a generalized zero divisor if \(a\) is a left and right generalized zero divisor.

Lemma 2. If \(a\) is not a left (right) generalized zero divisor, then \(a^2\) is also not a left (right) generalized zero divisor.

Lemma 3. If the completion \(\hat R\) of a topological ring \(R\) contains an identity \(e\) and every element of \(R\) is left (right) invertible in \(\hat R\), then \(R\) contains no closed left (right) ideals.

* By an ideal we shall mean a proper ideal, i.e., a nonzero ideal distinct from the whole ring.

** By a neighborhood of a point \(x\) we shall mean any set containing an open set that contains the point \(x\).

*** If \(M\) is some set in a topological ring \(R\), then \([M]_R\) denotes the closure of \(M\) in \(R\).

Lemma 4. If the completion \(\hat R\) of a topological ring \(R\) contains an identity \(e\) and an element \(a\in R\) is left (right) invertible in \(\hat R\), then \(a\) is not a left (right) generalized zero divisor in \(R\).

Lemma 5. If a topological ring \(R\) contains no nonzero closed one-sided ideals and an element \(a\in R\) is not a right (left) generalized zero divisor, then the completion \(\hat R\) contains an identity \(e\), and \(a\) is left (right) invertible in \(\hat R\).

Proof. If \(\{V_\alpha\}\) is some base of neighborhoods of zero, then, since \(a\) is not a right generalized zero divisor, there exists a family \(\{U_\alpha\}\) of neighborhoods of zero in \(R\) such that from \(xa^2\in U_\alpha\) it follows that \(x\in V_\alpha\). There exists a base of neighborhoods of zero \(\{W_\alpha\}\) in \(R\) such that \(W_\alpha-W_\alpha\subseteq U_\alpha\). Since \(a^2\ne0\), \(R\) cannot be a ring with zero multiplication. By Lemma 1, for every neighborhood \(W_\alpha\) there exists \(x_\alpha\in R\) such that \(x_\alpha a^2-a\in W_\alpha\).

For each \(\alpha\) define the set
\[ \mathfrak R_\alpha=\{x_\gamma\mid x_\gamma a^2-a\in W_\alpha\}. \]
Since
\[ \mathfrak R_\alpha\cap\mathfrak R_\beta =\{x_\gamma\mid x_\gamma a^2-a\in W_\alpha\cap W_\beta\} \supseteq \mathfrak R_\rho, \]
where \(\rho\) is such that \(W_\rho\subseteq W_\alpha\cap W_\beta\), the family of sets \(\mathfrak R_\alpha\) is a base of some filter \(F\).

We prove that the filter thus obtained is a Cauchy filter. For this it is enough to prove that \(\mathfrak R_\alpha-x_\alpha\subseteq V_\alpha\). If \(x_\gamma\in\mathfrak R_\alpha\), then \(x_\gamma a^2-a\in W_\alpha\) and \(x_\alpha a^2-a\in W_\alpha\). Hence
\[ (x_\gamma-x_\alpha)a^2\in W_\alpha-W_\alpha\subseteq U_\alpha. \]
From the definition of \(U_\alpha\) it follows that \(x_\gamma-x_\alpha\in V_\alpha\). Consequently, \(\mathfrak R_\alpha-x_\alpha\subseteq V_\alpha\).

Let \(\hat x\) be the limit of the filter \(F\) in \(\hat R\). Put \(e=\hat x a\) and show that \(e\) is an identity in \(\hat R\).

Since \(\mathfrak R_\alpha a^2-a\subseteq W_\alpha\) and \(\hat x\) is the limit of the filter \(F\), we have \(\hat x a^2-a=0\), i.e. \(\hat x a^2=a\), and therefore \(ea=a\). Since \(R\) contains no closed right ideals, \(aR\) is everywhere dense in \(R\), and hence \(aR\) is everywhere dense in \(\hat R\). Now let \(b\) be some element of \(\hat R\), and suppose that \(eb-b\ne0\). There exists a neighborhood of zero \(V\) in \(\hat R\) such that \(eb-b\notin V\), and a neighborhood of zero \(U\) in \(\hat R\) such that \(U+eU\subseteq V\). Since \(aR\) is everywhere dense in \(\hat R\), it follows that there exists an element \(c\in R\) for which \(ac-b\in U\). Then
\[ eb-b=e(b-ac)+eac-b\in e\cdot U+U\subseteq V. \]
We have obtained a contradiction to the assumption. Consequently, \(eb-b=0\), i.e. \(e\) is a left identity in \(\hat R\).

We now show that \(e\) is also a right identity in \(\hat R\). Suppose the contrary, i.e. that \(be-b\ne0\) for some \(b\in\hat R\). There exists a neighborhood of zero \(W\) in \(\hat R\) such that
\[ (W+be-b)\cap W=\varnothing. \]
Since \(R\) is everywhere dense in \(\hat R\), for any neighborhood of zero \(W'_\alpha\) from the base of neighborhoods of zero \(\{W'_\alpha\}\) in \(\hat R\),
\[ R\cap\bigl(be-b+(W\cap W'_\alpha)\bigr)\ne\varnothing. \]
Let
\[ y_\alpha\in R\cap\bigl(be-b+(W\cap W'_\alpha)\bigr), \]
and let \(M\) be the family of all such \(y_\alpha\). Then
\[ M\cap W\subseteq (be-b+W)\cap W=\varnothing. \]
If \(V\) is an arbitrary neighborhood of zero in \(\hat R\), then there exists a neighborhood of zero \(W'_\gamma\) such that \(W'_\gamma\cdot a\subseteq V\). Then
\[ y_\gamma\cdot a\in R\cap\bigl((be-b+W'_\gamma)a\bigr) =R\cap\bigl(bea-ba+W'_\gamma a\bigr) =R\cap(W'_\gamma a)\subseteq R\cap V. \]
From the fact that \(y_\gamma\notin W\) and the arbitrariness of \(V\) it follows that \(a\) is a right generalized zero divisor in \(R\). We have obtained a contradiction. Consequently, \(be=b\) for all \(b\in\hat R\), and hence \(e\) is a two-sided identity in \(\hat R\). Then \(\hat x\) is a left inverse for \(a\).

Theorem 1. In order that the completion \(\hat R\) of a topological ring \(R\) contain an identity \(e\) and that every element of \(R\) be invertible in \(\hat R\), it is necessary and sufficient that \(R\) contain no closed one-sided ideals and no generalized zero divisors.

Proof. The necessity follows easily from Lemmas 3 and 4,

Sufficiency. Let \(a\) be an arbitrary element of \(R\). Since \(R\) contains no generalized zero divisors, \(a\) is not either a right or a left generalized zero divisor. Suppose, for definiteness, that \(a\) is not a right generalized zero divisor. Then, by Lemma 5, \(\hat R\) has an identity and \(a\) is left-invertible in \(\hat R\). By Lemma 4, \(a\) is not a left generalized zero divisor. Applying Lemma 5 once more, we obtain that \(a\) is right-invertible in \(\hat R\). Hence \(a\) is invertible in \(\hat R\).

Corollary. A complete topological ring \(R\) containing no generalized zero divisors and no closed one-sided ideals is a field.

Indeed, the completion \(\hat R\) has an identity and every element of \(R\) is invertible in \(\hat R\). Since \(\hat R\) is a complete topological ring, \(R=\hat R\). Consequently, \(R\) itself has an identity and every element of \(R\) is invertible in \(R\).

As usual, a set \(M\) of a topological ring \(R\) will be called topologically nilpotent if for every neighborhood \(V\) of zero there exists a number \(n\) such that
\[ M^k=\{a_1a_2\ldots a_k\mid a_i\in M\}\subseteq V \]
for all \(k\ge n\).

Theorem 2. If a topological ring \(R\) with identity \(e\) has a topologically nilpotent neighborhood of zero \(V\) and contains no closed one-sided ideals, then every element of \(R\) is invertible in the completion \(\hat R\).

Proof. Let \(n\) be such a number that
\[ V^k+V^{k+1}\subseteq V \]
for all \(k\ge n\). For any element \(a\in R\), by Lemma 1, there exists an element \(x\in R\) such that
\[ ax+e\in V. \]
Then
\[ (ax+e)^n=ax\sum_{i=1} C_n^i(ax)^{i-1}+e\in V^n. \]
Consequently, there exists an element
\[ y=x\sum_i C_n^i(ax)^{i-1}, \]
such that
\[ ay+e\in V^n. \]

For any number \(p\), put
\[ s_p=\sum_{k=0}^{p} y(ay+e)^k. \]
We shall show that the sequence of elements \(s_1,s_2,\ldots\) is a Cauchy sequence, i.e. that for every neighborhood of zero \(U\) there exists a number \(m\) such that
\[ s_p-s_q\in U \]
for all \(p,q\ge m\).

Let \(W\) be a neighborhood of zero in \(R\) such that
\[ y\cdot W\subseteq U, \]
and let \(m\) be such a number that
\[ V^k\subseteq W \]
for \(k\ge m\). Then
\[ s_p-s_q=y\sum_{k=q+1}^{p}(ay+e)^k = y(ay+e)^q\sum_{k=1}^{p-q}(ay+e)^k. \]
We shall show that
\[ \sum_{k=1}^{r}(ay+e)^k\in V \]
for every \(r\). Indeed, for \(r=1\) we have
\[ (ay+e)\in V^n\subseteq V. \]
Suppose that
\[ \sum_{k=1}^{r}(ay+e)^k\in V. \]
Then
\[ \sum_{k=1}^{r+1}(ay+e)^k =(ay+e)+(ay+e)\sum_{k=1}^{r}(ay+e)^k\in V^n+V^n\cdot V =V^n+V^{n+1}\subseteq V. \]
Consequently,
\[ s_p-s_q\in y(ay+e)^q\cdot V\subseteq y\cdot V^{q+1}\subseteq y\cdot W\subseteq U \]
for \(p,q\ge m\). From the arbitrariness of \(U\) it follows that the sequence \(s_1,s_2,\ldots\) is a Cauchy sequence.

There exists an element \(d\in\hat R\) such that
\[ d=\lim_p s_p \]
(i.e. every neighborhood of the element \(d\) contains all elements \(s_p\), starting from some one). But
\[ \begin{aligned} ad &=ad-as_p+as_p\\ &=a(d-s_p)+as_p+e-e\\ &=a(d-s_p)+a\sum_{k=0}^{p}y(ay+e)^k+e-e\\ &=a(d-s_p)+\left(\sum_{k=0}^{p}ay(ay+e)^k+e\right)-e\\ &=a(d-s_p)+(ay+e)^{p+1}-e\\ &=a(d-s_p)+s_{p+1}-s_p-e \end{aligned} \]
for all \(p\).

For any neighborhood of zero \(U'\) in \(\hat R\) there exists a neighborhood of zero \(W'\) in \(R\) such that \(aW' + W' \subseteq U'\), and such a number \(r\) that \(d - s_p \in W'\) and \(s_{p+1}-s_p \in W'\) for all \(p \ge r\). Then \(ad = a(d-s_p)+s_{p+1}-s_p-e \in a\cdot W' + W' - e \subseteq -e+U'\). From the arbitrariness of \(U'\) it follows that \(ad=-e\). Hence \(a(-d)=e\).

The invertibility of the element \(a\) on the left is proved analogously.

Theorem 3. If the completion \(\hat R\) of a topological ring \(R\) contains an identity and every element of \(R\) is invertible in \(R\), then \([aV]_R\) is a neighborhood of zero in \(R\) for every \(a \in R\) and every neighborhood of zero \(V\) in \(R\).

Proof. Indeed, \([V]_{\hat R}\) is a neighborhood of zero in \(\hat R\). Since \(a\) is invertible in \(\hat R\), \(a[V]_{\hat R}\) is a neighborhood of zero in \(\hat R\). Then \([aV]_{\hat R}=a[V]_{\hat R}\) will also be a neighborhood of zero in \(\hat R\). Thus,
\([aV]_R=R\cap [aV]_{\hat R}\) will be a neighborhood of zero in \(R\).

Corollary. If a topological ring \(R\) has a topologically nilpotent neighborhood of zero \(V\) and contains no closed one-sided ideals, then \(R\) has a countable base of neighborhoods of zero.

Indeed, let \(\{V_\alpha\}\) be some base of neighborhoods of zero in \(R\). Without loss of generality, we may assume that all \(V_\alpha\) are closed sets. Since for any neighborhood of zero \(V_\alpha\) there exists such an \(n\) that
\([a^nV]_R \subseteq [V^{n+1}]_R \subseteq [V_\alpha]_R=V_\alpha\) for \(a \in V\), the collection \(\{[a^nV]_R\}\) is a base of neighborhoods of zero in \(R\).

Theorem 4. If a locally bicompact ring \(R\) with identity \(e\) contains no closed one-sided ideals, then \(R\) is a field.

Proof. If \(C\) is the component of zero in \(R\), then \(C\) is a closed ideal, and hence either \(C=0\), or \(C=R\).

Since the left annihilator of any element \(a \in R\) is a closed left ideal and \(e\cdot a\ne 0\), \(R\) contains no nonzero zero divisors.

If \(C=R\), then, by (4) (Theorem 2), \(R\) is a finite-dimensional algebra over the field of real numbers. By Frobenius’ theorem, \(R\) is either the field of real numbers, or the field of complex numbers, or the field of quaternions.

If, however, \(C=0\), then, by (4) (Lemma 4), \(R\) has a base of neighborhoods of zero that are subrings.

Let \(V\) be a bicompact neighborhood of zero, not containing \(e\) and being a ring. Since \(V\) is a ring without zero divisors, by (5) (Theorem 19), \(V\) is either radical in the sense of the Jacobson radical, or completely primary. If \(V\) were completely primary, then \(V\) would have an identity \(f\ne e\). Then \((e-f)f=0\). But this is impossible, since \(R\) contains no zero divisors. Thus, \(V\) is a radical ring in the sense of the Jacobson radical. From the fact that \(V\) is totally disconnected it follows that \(V\) is a topologically nilpotent ring. Hence \(V\) will be a topologically nilpotent set in \(R\). By Theorem 2, every element of \(R\) is invertible in the completion \(\hat R\). But since a locally bicompact ring is complete, \(R=\hat R\), and therefore \(R\) is a field.

Institute of Mathematics with Computing Center
Academy of Sciences of the Moldavian SSR

Received
4 V 1966

CITED LITERATURE

\(^{1}\) N. Bourbaki, General Topology, Moscow, 1958.
\(^{2}\) I. M. Gelfand, D. A. Raikov, G. E. Shilov, Commutative Normed Rings, Moscow, 1960.
\(^{3}\) A. G. Kurosh, Lectures on General Algebra, Moscow, 1962.
\(^{4}\) I. Kaplansky, Am. J. Math., 70, No. 4, 447 (1948).
\(^{5}\) I. Kaplansky, Am. J. Math., 69, No. 2, 153 (1947).
\(^{6}\) S. Warner, Math. Ann., 145, No. 1, 52 (1962).