# Modules over infinite rings which can not be a finite union of their proper submodules

It is well known that a vector space over an infinite field cannot be a finite union of its proper subspaces.

Does this fact have an immediate and obvious generalization to modules over infinite division rings (skew fields)?

Furthermore:

To which extent all infinite commutative unital rings $R$ with the following property have been classified?

Every $R$-module can not be equal to a union of finite number of its proper submodules.

• You might find something helpful in this preprint by A. Khare. – Fred Rohrer Mar 14 '18 at 18:56
• "Does this fact have an immediate and obvious generalization": why don't you try? – YCor Mar 14 '18 at 21:05
• I precisely rather tend to believe that the argument for fields carries over skew-fields, accounting as "immediate generalization". – YCor Mar 14 '18 at 22:09
• The title says "infinite rings", the body says "infinite division rings" and then "infinite commutative unital rings", so it's not clear what you want. The integers form an infinite commutative unital ring, the modules are just abelian groups, and certainly there are abelian groups that are finite unions of proper subgroups. – Gerry Myerson Mar 14 '18 at 22:22
• Lemma 2.9 of Khare's preprint shows that no module over a local ring can be covered by a finite union of proper submodules, provided the residue field is infinite. (Of course, any ring with a finite residue field is out of the game.) – Luc Guyot Mar 14 '18 at 23:07

The two questions can be answered in the positive.

Claim 1. Let $R$ be an infinite division ring. Then no left $R$-module can be the union of finitely many of its proper left $R$-submodules.

The proof is a straightforward generalization of the classical pigeonhole argument for finite covers of vector spaces, see the Corollary below.

If $R$ is a commutative ring with identity which has a finite residue field $k$, then any vector space over $k$ with finite dimension $> 1$ is an $R$-module that can be covered by finitely many (proper) lines. OP's second question is therefore addressed by

Claim 2. Let $R$ be a commutative ring with identity and whose residue fields are all infinite. Then no $R$-module can be the union of finitely many of its proper $R$-submodules.

The class of rings which satisfy the hypotheses of Claim 2 is wide: fix finitely many fields $k_1, \dots, k_n$ and let $\{K_{\alpha}\}_{\alpha}$ be a family of fields such for every $\alpha$, the field $K_{\alpha}$ contains (up to isomorphism) one of the fields $k_i$. Then there is a zero-dimensional ring $R$ whose residue fields are the fields $K_{\alpha}$ [1].

The proof of Claim 2 relies on [3, Lemma 10]:

Gottlieb's Lemma. Let $R$ be a commutative ring with identity. Let $M = M_1 \cup M_2 \cup \cdots \cup M_n$ be an $R$-module written as an union of proper submodules and suppose there is an element $x \in M_1 \setminus \bigcup_{i = 2}^n M_i$ and elements $a_1, a_2, \dots, a_r \in R$ which are pairwise distinct modulo every ideal $(M_k: x), k \ge 2$. Then $n > r$.

If $M$ is a module over $R$, $x \in M$, and $N$ is a submodule of $M$, we denote by $(N:x)$ the ideal of $R$ consisting of the elements $r$ such that $rx \in N$.

Now we can prove the second claim.

Proof of Claim 2. Let $M = M_1 \cup M_2 \cup \cdots \cup M_n$ be an $R$-module written as an union of proper submodules. We can assume that $n > 2$ and $M_1 \not\subset M'$ with $M' \Doteq M_2 \cup \cdots \cup M_n$. Let $x \in M_1 \setminus M'$ and pick some $r \ge n$. Each ideal $(M_k: x)$ for $k \ge 2$ is contained in some maximal ideal $\mathfrak{m_i}$ for $i = 1, \dots, N$ and $N \le k - 1$. Since $R/\mathfrak{m}_i$ is infinite, we can find for every $i$ some elements $a_{i1}, \dots, a_{ir} \in R$ which are pairwise distinct modulo $\mathfrak{m}_i$. By the Chinese Remainder Theorem, we can find $a_1, \dots, a_r \in R$ such that $a_j \equiv a_{ij} \text{ mod } \mathfrak{m}_i$ for every $i, j$. Gottlieb's Lemma yields $n > r$, a contradiction.

Addendum. I realized that Claim 2 is also an immediate consequence of [2, Lemma 3].

[1] R. Gilmer, W. Heinzer, "The family of residue fields of a zero-dimensional commutative ring", 1992.
[2] M. Zandt, "A note on unions of ideals and cosets of ideals", 1995.
[3] C. Gottlieb, "Modules covered by finite unions of submodules", 1998.

Below lie the remains of my initial (awkward) answer.

This is only a long comment.

Let $R$ a be unital ring which is not necessarily commutative and let $R^{\times}$ denote the set of left-invertible elements of $R$. We say that $R$ has property (U) if for every finite cover $R^{\times} \subset \bigcup_{i = 1}^n A_i$ of $R^{\times}$ by subsets $A_i \subset R$, there is at least one index $j$ such that $A_j$ contains two left-invertible elements $\lambda, \mu$ satisfying $\lambda - \mu \in R^{\times}$.

Claim 3. Let $R$ a be unital ring with property (U). Then no left $R$-module can be the union of finitely many of its proper left $R$-submodules.

Proof. We reason by contradiction, considering a left $R$-module $M$ that is covered by a finitely many proper left $R$-submodules $M_1, \dots, M_n$. Without loss of generality, we can assume that $n > 1$, $M' \Doteq\bigcup_{i = 1}^{n -1} M_i \not \subset M_n$ and $M_n \not\subset M'$. Consider $z(u) = x + uy$ with $x \in M_n \setminus M'$ and $y \in M' \setminus M_n$ and $u \in R^{\times}$. We have $z(u) \notin M_n$ since otherwise $y$ would lie in $M_n$ too. Therefore $z(u) \in M'$ for every $u \in R^{\times}$. Set $A_i = \{ u \in R^{\times}\, \vert\, z(u) \in M_i\}$ for $i = 1, \dots, n - 1$. As $R$ has property (U), we can find $\lambda, \mu \in A_j$ for some $j \le n - 1$ such that $\lambda - \mu \in R^{\times}$. As $z(\lambda) - z(\mu) \in M_j$ we deduce that $y \in M_j$, a contradiction.

This rewording of the classical proof has an immediate consequence:

Corollary. Assume that $R$ is an infinite division ring or a commutative semilocal ring with identity and whose residue fields are infinite. Then no left $R$-module can be the union of finitely many of its proper left $R$-submodules.

Proof. If $R$ is an infinite division ring, then $R^{\times} = R \setminus \{0\}$ so that $R$ has clearly property (U) and the result follows from Claim 3. Assume now that $R$ is a unital commutative ring with finitely many maximal ideals $\mathfrak{m}_1, \dots, \mathfrak{m}_n$. Then $R^{\times} = R \setminus \bigcup_{i = 1}^{n} \mathfrak{m}_i$. Assume that $R^{\times} \subset \bigcup_{j = 1}^N A_j$ for some subsets $A_j \subset R$. We reason by contradiction, supposing further that the difference of any pair of units in $A_j$ is not invertible for every $j$. Then $R^{\times} \subset \bigcup_{1 \le i \le n, 1 \le j \le N} (x_{ij} + \mathfrak{m}_i)$ for some elements $x_{ij} \in R$. Since $R/\mathfrak{m}_i$ is infinite for every $i$, we can find $x_i \in R$ such that $x_i \not\equiv 0, x_i \not\equiv x_{ij} \mod \mathfrak{m_i}$ for every $j$. By the Chinese Remainder Theorem, there is $x \in R^{\times}$, such that $x \equiv x_i \mod \mathfrak{m_i}$ for every $i$, which contradicts the above inclusion. As result, $R$ has property (U) and Claim 3 yields the conclusion.

The above corollary generalizes Lemma 2.9 of Apoorva Khare's preprint to semilocal rings.

To which extent all infinite commutative unital rings $R$ with the following property have been classified?

Every $R$-module cannot be equal to a union of finite number of its proper submodules.

I will classify these rings without assuming commutativity. I claim

Thm. The following are equivalent for a ring $R$.

1. There is an $R$-module $M$ that is the union of finitely many proper submodules.

2. $R$ has a nontrivial finite module.

3. $R$ has a proper ideal of finite index.

4. $R$ has a maximal ideal of finite index.

5. $R$ has a nontrivial finite quotient ring.

[2 implies 3] If $M$ is a nontrivial finite left $R$-module, then its left annihilator is a proper ideal of finite index. \\\

[3 implies 4] Any proper ideal of finite index can be extended to a maximal ideal of finite index. \\\

[(3 or 4) implies 5] If $I$ is a proper/maximal ideal of finite index in $R$, then $R/I$ is a nontrivial finite quotient ring. \\\

[5 implies 1] Suppose that the ring $S=R/I$ is a nontrivial finite quotient of $R$. The left $S$-module $S\oplus S$ is a left $R$-module, by restriction of scalars. Since $S\oplus S$ is free of rank 2 as an $S$-module, and it is finite, it cannot be cyclic as an $S$-module. Hence it cannot be cyclic as an $R$-module. Write it as a union of its nontrivial cyclic $R$-submodules $S\oplus S = M_1\cup\cdots \cup M_n$, each of which must be a proper submodule. This is a representation of an $R$-module as a finite union of proper submodules. \\\

[1 implies 2] First suppose that $R$ has a finite module $M$ that is a union of finitely many proper submodules. Then $M$ is surely a nontrivial finite $R$-module.

Next suppose that $R$ has an infinite module $M$ that is a finite union of proper submodules, $M=\bigcup_{i=1}^n M_i$. I now employ an old and famous result of B. H. Neumann which asserts that if a group $G$ is a union of finitely many cosets of subgroups, say $G = \bigcup_{i=1}^n a_iH_i$, then one can discard all cosets of subgroups of infinite index and the remaining union still covers $G$. In particular, since modules have underlying group structure, the representation $M=\bigcup_{i=1}^n M_i$ forces some proper submodule $M_i$ to have finite index in $M$. That is, $M/M_i$ is a nontrivial finite $R$-module. \\\

Neumann, B. H. Groups covered by permutable subsets. J. London Math. Soc. 29, (1954). 236–248.

So, $R$ has the property that NO $R$-module is a finite union of proper submodules iff all maximal ideals of $R$ have infinite index.

• When you say and write sum in the proofs of the last two implications you mean union. The sum symbols should be union. – Benjamin Steinberg Apr 1 '18 at 20:56
• @BenjaminSteinberg: Fixed. – Keith Kearnes Apr 1 '18 at 20:59