It is well known that a vector space over an infinite field [cannot be a finite union of its proper subspaces](https://mathoverflow.net/questions/26/can-a-vector-space-over-an-infinite-field-be-a-finite-union-of-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?



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> Every $R$-module can not be equal to a union of finite number of its proper submodules.
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Modules over infinite rings which can not be a finite union of their proper submodules
The two questions can be answered in the positive.



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> > **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.
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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



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> > **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.
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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]:



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> > **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 $n > r$. 
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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.



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




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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 $\lambda - \mu \in R^{\times}$.



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> > **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.
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> > *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. 
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This rewording of the classical proof has an immediate consequence:



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> > **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.
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> > *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.
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The above corollary generalizes [Lemma 2.9 of Apoorva Khare's preprint](https://arxiv.org/pdf/0906.1023v2.pdf) to semilocal rings.


