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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 sum of its nontrivial cyclic $R$-submodules $S\oplus S = M_1+\cdots + M_n$, each of which must be a proper submodule. This is a representation of an $R$-module as a finite sum of proper submodules. \\\

[1 implies 2] First suppose that $R$ has a finite module $M$ that is a sum 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, $M=\sum_{i=1}^n M_i$. 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 $G = \cup_{i=1}^n a_iH_i$, 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 $M=\sum_{i=1}^n M_i$ forces some $M_i$ 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.

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.

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 sum of its nontrivial cyclic $R$-submodules $S\oplus S = M_1+\cdots + M_n$, each of which must be a proper submodule. This is a representation of an $R$-module as a finite sum of proper submodules. \\\

[1 implies 2] First suppose that $R$ has a finite module $M$ that is a sum 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, $M=\sum_{i=1}^n M_i$. 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 $G = \cup_{i=1}^n a_iH_i$, 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 $M=\sum_{i=1}^n M_i$ forces some $M_i$ 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.

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 sum of its nontrivial cyclic $R$-submodules $S\oplus S = M_1+\cdots + M_n$, each of which must be a proper submodule. This is a representation of an $R$-module as a finite sum of proper submodules. \\\

[1 implies 2] First suppose that $R$ has a finite module $M$ that is a sum 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, $M=\sum_{i=1}^n M_i$. 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 $G = \cup_{i=1}^n a_iH_i$, 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 $M=\sum_{i=1}^n M_i$ forces some $M_i$ 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.

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 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] If $I$ is a proper ideal of finite index in $R$, then $R/I$ is a nontrivial finite quotient ring. \\\

[4 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 sum of its nontrivial cyclic $R$-submodules $S\oplus S = M_1+\cdots + M_n$, each of which must be a proper submodule. This is a representation of an $R$-module as a finite sum of proper submodules. \\\

[1 implies 2] First suppose that $R$ has a finite module $M$ that is a sum 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, $M=\sum_{i=1}^n M_i$. 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 $G = \cup_{i=1}^n a_iH_i$, 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 $M=\sum_{i=1}^n M_i$ forces some $M_i$ 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.

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 sum of its nontrivial cyclic $R$-submodules $S\oplus S = M_1+\cdots + M_n$, each of which must be a proper submodule. This is a representation of an $R$-module as a finite sum of proper submodules. \\\

[1 implies 2] First suppose that $R$ has a finite module $M$ that is a sum 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, $M=\sum_{i=1}^n M_i$. 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 $G = \cup_{i=1}^n a_iH_i$, 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 $M=\sum_{i=1}^n M_i$ forces some $M_i$ 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.