UDC 519.48

MATHEMATICS

V. I. ARNAUTOV

NONDISCRETE TOPOLOGIZABILITY OF COUNTABLE RINGS

(Presented by Academician P. S. Aleksandrov on 2 IX 1969)

By a topological ring we shall mean a not necessarily associative ring in which a Hausdorff topology is defined, with the operations of the ring continuous. It is obvious that in any ring one can introduce, in a trivial way, the (discrete) topology so as to turn it into a topological ring. In finite rings this is the only possible topology. The question of the possibility of introducing a nondiscrete topology in any infinite ring has not yet been solved. In the present paper it is proved that in any countable ring one can define a nondiscrete topology.

Let \(R\) be some ring and \(x\) some variable. By a monomial in \(x\) over \(R\) we shall mean an arbitrary formal product of elements of \(R\) and \(x\), with any placement of parentheses. The degree of a monomial is the number of occurrences of \(x\) in the given monomial. By a polynomial in \(x\) over \(R\) we shall mean an arbitrary finite sum of monomials. The degree of a polynomial is the maximal degree of the monomials occurring as summands in the polynomial. An element \(a \in R\) is called a root of the polynomial \(P(x)\) if \(P(a)=0\).

Analogously to the proof of Theorem 3 in \((^1)\), one proves

Theorem 1. In order that a countable ring \(R\) admit a nondiscrete topologization in which an ideal \(I\) is an open ideal, it is necessary and sufficient that, for any finite number of polynomials \(P_1(x), P_2(x), \ldots, P_n(x)\) over \(R\) with nonzero free terms, there exist a nonzero element of \(I\) which is not a root of any of these polynomials.

Let \(M\) be an arbitrary set and \(s\) some natural number. By \(M^{(s)}\) we denote the set of all subsets of the set \(M\) containing exactly \(s\) elements.

Lemma 1. Let \(M_0\) be some countable set. For any natural numbers \(s\) and \(r\), and any partition of the set \(M_0^{(s)}\) into \(r\) classes \(N_1, N_2, \ldots, N_r\), there exists a countable subset \(M \subseteq M_0\) and a number \(l \le r\) such that \(M^{(s)} \subseteq N_l\).

Proof. We prove the assertion by induction on the number \(s\). If \(s=1\), then the assertion of the lemma is obvious. Suppose that the lemma is true for the number \(s\), and prove it for \(s+1\).

Let

\[ M_0^{(s+1)}=\bigcup_{i=1}^{r} N_i \]

and let \(a_0\) be some element of \(M_0\). If we take

\[ N_i'=\{A\in (M_0\setminus \{a_0\})^{(s)}\mid \{a_0\}\cup A\in N_i\}, \]

then we obtain a partition of the set \((M_0\setminus \{a_0\})^{(s)}\) into \(r\) classes \(N_1', N_2', \ldots, N_r'\). By the induction hypothesis there exists an infinite subset \(M_1 \subseteq M_0\setminus \{a_0\}\) and a natural number \(l_0 \le r\) such that \(M_1^{(s)}\subseteq N_{l_0}'\). Then \(\{a_0\}\cup A\in N_{l_0}\) for any \(A\in N_1^{(s)}\), and

\[ M_1^{(s+1)}=\bigcup_{i=1}^{r}\bigl(N_i\cap M_1^{(s+1)}\bigr). \]

We now proceed with the set \(M_1\) in the same way as we proceeded above with the set \(M_0\). We obtain an element \(a_1\in M_1\), a natural number \(l_1\le r\), and such a...

an infinite subset \(M_2 \subseteq M_1 \setminus \{a_1\}\) such that \(\{a_1\} \cup A \in N_{l_1}\) for any \(A \subseteq M_2^{(s)}\).

Continuing this process indefinitely, we obtain sequences of elements \(a_0, a_1, a_2,\ldots\) from \(M_0\) and of natural numbers \(l_0, l_1, l_2,\ldots\), not greater than \(r\), and such a decreasing sequence of subsets
\[ M_0 \supset M_1 \supset M_2 \supset \cdots, \]
that \(a_j \in M_j\) and \(\{a_j\} \cup A_j \in N_{l_j}\) for any \(A_j \subseteq M_{j+1}^{(s)}\).

If we take \(B_k=\{a_i\mid l_i=k\}\), then there exists a number \(l \le r\) such that \(B_l\) is an infinite set.

Take \(M=B_l\) and show that \(M\) is the desired set.

Let \(\{a_{i_1}, a_{i_2},\ldots,a_{i_{s+1}}\}\subseteq M^{(s+1)}\). Without loss of generality we may assume that \(i_1<i_2<\cdots<i_{s+1}\). Since \(a_{i_k}\in M_{i_k}\subseteq M_{i_1+1}\) for \(k\ge 2\), it follows that \(\{a_{i_2},\ldots,a_{i_{s+1}}\}\subseteq M_{i_1+1}^{(s)}\). Hence
\[ \{a_{i_1},a_{i_2},\ldots,a_{i_{s+1}}\}\subseteq \{\{a_{i_1}\}\cup A\mid A\subseteq M_{i_1+1}^{(s)}\}\subseteq N_l . \]
This proves the lemma completely.

The lemma easily implies

Theorem 2. If \(M\) is a countable set, then for any partition of the set of all finite subsets of \(M\) into a finite number of classes \(N_1,N_2,\ldots,N_r\) and any natural number \(s\), there exists an infinite subset \(M_s\subseteq M\) and numbers \(l_1,l_2,\ldots,l_s\) such that \(M_s^{(k)}\subseteq N_{l_k}\) for every \(k\le s\).

Theorem 3 (van der Waerden). For any numbers \(m\) and \(r\) there exists a number \(n(m,r)\) such that, under any partition of an interval of natural numbers of length \(n(m,r)\) into \(r\) subsets, at least one of them contains an arithmetic progression of length \(m\) (see, for example, (2)).

Lemma 2. For any natural numbers \(m,r\) there exists a number \(s(m,r)\) such that, under any partition of the interval of the natural series from \(1\) to \(s(m,r)\) into \(r\) subsets, at least one of them contains numbers
\[ l_1,l_2,\ldots,l_m, \]
such that any sum of the form
\[ \sum_{j=1}^{q} l_{i_j}, \]
where \(i_j\ne i_k\), also belongs to this same subset.

Proof. Define natural numbers \(s_i\) by induction, taking \(s_0=1\) and \(s_{i+1}=n(s_i+1,r)\). We shall show that the number \(s=s_{m\cdot r}\) is the desired one.

Let the interval of the natural series from \(1\) to \(s\) be divided into \(r\) subsets \(A_1,A_2,\ldots,A_r\). For any number \(0\le j\le m\cdot r\), define a partition of the interval of the natural series from \(1\) to \(s_{m\cdot r-j}\) into \(r\) subsets
\[ A_{1,j}, A_{2,j},\ldots,A_{r,j} \]
as follows.

Take \(A_{i,0}=A_i\) for \(i=1,2,\ldots,r\). Suppose that a partition of the interval of the natural series from \(1\) to \(s_{m\cdot r-j}\), for \(j<m\cdot r\), into \(r\) subsets
\[ A_{1,j}, A_{2,j},\ldots,A_{r,j} \]
has already been defined. Then at least one of these subsets contains an arithmetic progression of length \(1+s_{m\cdot r-j-1}\), i.e.
\[ \{\,n_{j+1}+ip_{j+1}\mid i=0,1,\ldots,s_{m\cdot r-j-1}\,\} \]
is contained in some subset \(A_{t,j}\). Put
\[ A_{k,j+1}=\{\,i\mid ip_{j+1}\in A_{k,j}\,\} \]
for \(k=1,2,\ldots,r\). Since
\[ ip_{j+1}\le n_{j+1}+ip_{j+1}\le s_{m\cdot r-j} \]
for \(i\le s_{m\cdot r-j-1}\), we have
\[ \bigcup_{k=1}^{r} A_{k,j}\supseteq \{\,ip_{j+1}\mid 1\le i\le s_{m\cdot r-j-1}\,\}. \]
Then
\[ \bigcup_{k=1}^{r} A_{k,j+1} = \{1,2,\ldots,s_{m\cdot r-j-1}\}, \]
i.e. we have obtained a partition of the interval of the natural series from \(1\) to \(s_{m\cdot r-j-1}\) into \(r\) subsets
\[ A_{1,j+1}, A_{2,j+1},\ldots,A_{r,j+1}. \]

Thus, for any number \(0\le j<m\cdot r\), a partition of the interval of the natural series from \(1\) to \(s_{m\cdot r-j}\) into \(r\) subsets
\[ A_{1,j}, A_{2,j},\ldots,A_{r,j} \]
is defined, and at least one of these subsets contains an arithmetic progression
\[ n_{j+1}+ip_{j+1},\quad i=0,1,\ldots,1+s_{m\cdot r-j-1}. \]

Since \(j\) takes \(m\cdot r\) values, there exist a number \(t_0\) and numbers
\[ 0\le j_1<j_2<\cdots<j_m<m\cdot r \]
such that in each of the sets \(A_{t_0,j_k}\)

\(k=1,2,\ldots,m\), contains the corresponding arithmetic progression
\(n_{j_{k+1}}+i p_{j_{k+1}},\ i=0,1,\ldots,s_{m\cdot r-j_k-1}\).

For any \(1\le k\le m\) take \(l_k=(p_1p_2\ldots p_{j_k})n_{j_{k+1}}\) and show that the numbers \(l_1,l_2,\ldots,l_m\) are the desired ones.

Before completing the proof of the lemma, let us prove some properties of the sets \(A_{t,j}\).

(1) If \(i\in A_{t,j}\), then \((p_1p_2\ldots p_j)i\in A_{t,0}\) for any \(j\le m\cdot r\).

Indeed, from the definition of the sets \(A_{t,j}\) it follows that \(p_j i\in A_{t,j-1}\). Then \(p_{j-1}p_j i\in A_{t,j-2}\), and so on. Applying this process \(j\) times, we obtain \((p_1p_2\ldots p_j)i\in A_{t,0}\).

\[ \text{(2)}\quad \sum_{j=1}^{i+q} n_{j+1}(p_{i+1}p_{i+2}\ldots p_j)\le s_{m\cdot r-1} \]
for any \(i\).

Indeed, if \(q=0\), then \(n_{i+1}\le s_{m\cdot r-i}\). Suppose that
\[ \sum_{j=i}^{i+q} n_{j+1}(p_{i+1}p_{i+2}\ldots p_j)\le s_{m\cdot r-i} \]
for any \(i\). Then
\[ \sum_{j=1}^{i+q+1} n_{j+1}(p_{i+1}p_{i+2}\ldots p_j) = n_{i+1}+p_{i+1}\left[\sum_{j=i+1}^{i+1+q} n_{j+1}(p_{i+2}p_{i+3}\ldots p_j)\right]\le \]
\[ \le n_{i+1}+p_{i+1}s_{m\cdot r-i-1}\le s_{m\cdot r-i}. \]

Let us now proceed to complete the proof of the lemma. Since, by property (2),
\[ \sum_{i=2}^{q} (p_{j_{k_1}+2}p_{j_{k_1}+3}\ldots p_{j_{k_i}})\,n_{j_{k_i}+1}\le \]
\[ \le \sum_{i=j_{k_1}+1}^{j_{k_q}} n_{i+1}(p_{j_{k_1}+2}p_{j_{k_1}+3}\ldots p_i)\le s_{m\cdot r-j_{k_1}-1}, \]
we have
\[ n_{j_{k_1}+1}+p_{j_{k_1}+1} \left[\sum_{i=2}^{q} (p_{j_{k_1}+2}p_{j_{k_1}+3}\ldots p_{j_{k_i}})\,n_{j_{k_i}+1}\right]\in A_{t_0,j_{k_1}}. \]

Then, by property (1), we have
\[ \sum_{i=1}^{q} l_{k_i} = (p_1p_2\ldots p_{j_{k_1}}) \left[ n_{j_{k_1}+1} + \sum_{i=2}^{q}(p_{j_{k_1}+1}\ldots p_{j_{k_i}})n_{j_{k_i}+1} \right]\in A_{t_0,0}. \]

This proves the lemma completely.

Lemma 3. If \(P(x)\) is a polynomial of degree \(m\) without constant term over a ring \(R\), then for any element \(a\in R\)
\[ P(x+a)=P(x)+P(a)+Q(x), \]
where \(Q(x)\) is a polynomial of degree at most \(m-1\) without constant term.

The proof is obvious.

Lemma 4. If for a polynomial \(P(x)\) of degree \(m\) over a ring \(R\) there exist elements \(a_1,a_2,\ldots,a_{m+1}\) such that every sum of the form \(\sum_{i=1}^{q} a_{k_i}\), where \(k_i\ne k_j\), is a root of \(P(x)\), then the constant term of the polynomial \(P(x)\) is equal to zero.

Proof. The proof is easily carried out by induction on the degree of the polynomial \(P(x)\).

Theorem 4. For any finite number of polynomials over a ring \(R\) with nonzero constant terms, in any infinite subgroup \(I\) of the addi-

of the additive group of the ring \(R\) there exists a nonzero element \(a\) which is not a root of any of these polynomials.

Proof. Suppose the contrary, i.e., that there is a finite number of polynomials \(P_1(x), P_2(x), \ldots, P_r(x)\) over \(R\) with nonzero constant terms such that every nonzero element of \(I\) is a root of one of these polynomials. Choose a countable set \(M\) of elements of \(I\) such that any finite sum of pairwise distinct elements \(a_i \in M\) is nonzero.

We now define a partition of the set of all finite subsets of the set \(M\) into \(r\) classes \(N_1, N_2, \ldots, N_r\) as follows: assign the subset \(\{c_1, c_2, \ldots, c_n\}\) of the set \(M\) to the class \(N_j\), where

\[ j=\min\left\{k \mid P_k\left(\sum_{i=1}^{n} c_i\right)=0\right\}. \]

If \(m\) is the greatest degree of the polynomials \(P_1(x), P_2(x), \ldots, P_r(x)\) and \(s=s(m+1,r)\), then by Theorem 2 there exists an infinite subset \(M_s\) such that \(M_s^{(k)} \subseteq N_{l_k}\) for any \(k \le s\).

We now define a partition of the interval of the natural numbers from 1 to \(s\) into \(r\) subsets \(A_1, A_2, \ldots, A_r\) as follows: put the number \(i \le s\) into the subset \(A_j\) if and only if \(M_s^{(i)} \subseteq N_j\).

By Lemma 2 there exist natural numbers \(j_0 \le r\) and \(l_1 < l_2 < \cdots < l_{m+1} < s\) such that any sum \(l_{i_1}+l_{i_2}+\cdots+l_{i_q}\), where \(i_j \ne i_k\), belongs to the subset \(A_{j_0}\). Then

\[ M_s^{\left(\sum_{j=1}^{q} l_{i_j}\right)} \subseteq N_{j_0}. \]

If \(M_s=\{b_1,b_2,\ldots\}\), then define the elements \(a_1=b_1+b_2+\cdots+b_{l_1}\),

\[ a_2=b_{l_1+1}+b_{l_1+2}+\cdots+b_{l_1+l_2},\ldots,\quad a_{m+1}=b_{l_1+l_2+\cdots+l_m+1}+b_{l_1+l_2+\cdots+l_m+2}+\cdots \]

\[ \cdots+b_{l_1+l_2+\cdots+l_m+l_{m+1}}. \]

From the definition of the classes \(N_j\) it follows that any sum \(\sum_{j=1}^{q} a_{i_j}\) is a root of the polynomial \(P_{j_0}(x)\). Since the degree of the polynomial \(P_{j_0}(x)\) is no greater than \(m\), by Lemma 4 the constant term of \(P_{j_0}(x)\) is equal to zero. We have obtained a contradiction with the condition of the theorem. Consequently, in \(I\) there is a nonzero element which is not a root of any of the polynomials \(P_j(x)\).

From Theorems 1 and 4 it follows that

Theorem 5. For any infinite ideal \(I\) of a countable ring \(R\), there exists a nondiscrete topologization of the ring \(R\) in which \(I\) is an open ideal.

Corollary. Every countable ring admits a nondiscrete topologization.

Institute of Mathematics with Computing Center
Academy of Sciences of the MSSR
Kishinev

Received
22 VIII 1969

References

1. V. I. Arnautov, Sibirsk. Mat. Zh., 9, No. 6 (1968).
2. A. Ya. Khinchin, Three Pearls of Number Theory, Moscow, 1948.