Let $R$ be a unital ring and $M$ be an $R$-module. I have some questions about relation between the ring $\operatorname{End}_R M$ of endomorphisms and the notion of “dimension” of a module. I’m not really a ring theory expert, and because my intuition comes from linear algebra over a field, I want to invent a definition of 1-dimensional (more precisely, of dimension no more than 1) module. This came partially from my exercises on K-theory and partially from thinking on this math.SE question.

Definition: an $R$-module is called unidimensional iff any its $R$-endomorphism can be represented as a scalar multiplication by certain $r\in R$ (not necessarily unique). In other words, $R ↠ \operatorname{End}_R M$, or $\operatorname{End}_R M$ is a quotient ring of $R$ (by some ideal, possibly {0}).

Update from August 31: I realized that there is no canonical homomorphism $R → \operatorname{End}_R M$ for a general (not necessarily commutative) ring $R$. See details in “Problems with non-commutativity” section.

Let’s see what we obtained for ${\mathbb Z}$-modules, that are the same thing as abelian groups. Unidimensionality is, obviously, not the same thing as rank ≤ 1 (if rank is defined as the maximal cardinality of linearly independent subset): any torsion module has rank 0, but ${\mathbb Z_2}\oplus{\mathbb Z_2}$ isn’t a unidimensional ${\mathbb Z}$-module.

On the other hand, unidimensionality is logically weaker than “can be generated by one element”. One can directly check that any one-element-generated module is unidimensional, but the converse is not true. Let $M$ be the set of all rational numbers with square-free denominators:

$$ M = \{ \frac{n}{p_1p_2\cdots p_k}\ |\ n\in {\mathbb Z},\ p_1,\ldots p_k{\scriptstyle\text{ are prime and distinct }}\}.$$

It is a group under addition because the least common denominator of any two its elements (least common multiple of two square-free numbers) is square-free. But all elements of this $M$ don’t have a common divisor since the least common multiple of all square-free naturals is undefined. In spite of being not a finitely-generated ${\mathbb Z}$-module, this $M$ is indeed unidimensional: it hasn’t any endomorphisms but multiplications by integers because is not divisible by any natural number > 1.

Problems with non-commutativity

(added August 31, 2014)

The definition presented yesterday was, in fact, unclear. If $R$ is not commutative and $M$ is a left $R$-module, then we are sure that ${\mathbf m}\mapsto c{\mathbf m}$ is an $R$-endomorphism if and only if $c$ belongs to the centre of the ring (because an endomorphism must commute will scalar multiplication: $cλ{\mathbf m} = λc{\mathbf m}$). Generally, to make endomorphisms from arbitrary “scalars” one must apply multiplication on the side opposite to the module’s multiplication, i.e. to have a bimodule structure on $M$. If $M$ is initially a right $R$-module but has some auxiliary left multiplication structure that makes it an $(M,M)$-bimodule, then yes, $c\quad\mapsto\quad({\mathbf m}\mapsto c{\mathbf m})$ will be really a ring homomorphism $R → \operatorname{End}_R M$. So, there are two “versions” of an unidimensional module:

  • (weak, for any unital ring): An epimorphism $R ↠ \operatorname{End}_R M$ must exist. In other words, for some left multplication structure any endomorphism of $M$ as a right $R$-module is a left multiplication.
  • (strong, useful only for commutative rings): For the same multiplication (a.k.a. symmetric bimodule) any endomorphism of $M$ is a scalar multiplication.

For $R = {\mathbb Z}$ there is certainly no difference because on an abelian group the only ${\mathbb Z}$-multiplication is possible. But for commutative rings with non-trivial automorphisms there may be a gap between two versions of unidimensionality.


Does a name for my “unidimensional modules” exist already?

Does it admit an equivalent reformulation? At least for an important partial case, such as for commutative $R$.

(added August 31, 2014): Is there such example of a commutative $R$ and an $R$-module $M$ that $\operatorname{End}_R M ≃ R$, but $M$ is not unidimensional in the strong sense? (see the section above)

How can I prove that any rank > 1 module is not unidimensional? It is obvious for free modules and can be proven for projective ones, but is not obvious in general (although might be true, according to my intuition).

Is it true that, for commutative $R$, the tensor product of an unidimensional and an arbitrary modules consists only of elementary tensors? And is the tensor product of two unidimensional modules unidimensional?

How wide is the gap between rank 1 and unidimensionality? Starting to think about ${\mathbb Z}$, rational numbers ${\mathbb Q}$ form a divisible rank 1 ${\mathbb Z}$-module that is neither unidimensional nor finitely generated, and also doesn’t have torsion. (Must a projective rank 1 module be unidimensional? no, counterexamples are provided in comments.)

And what about the gap between unidimensionality and “can be generated by one element” (or, in other words, quotients by ideals in $R$)?

  • $\begingroup$ A projective rank 1 module is not necessarily unidimensional: a counterexample can be constructed using the fact that ℤ₂ ⊕ ℤ₃ = ℤ₆ (both as rings and as ℤ₆-modules; the statement generally known as the Chinese remainder theorem). ℤ₂ ⊕ ℤ₆ is a projective ℤ₆-module (its complement to a free module ℤ₆² is ℤ₃), and it obviously has the rank 1. But it isn’t unidimensional since has non-trivial projections among ℤ₆-endomorphisms. $\endgroup$ – Incnis Mrsi Aug 30 '14 at 16:06
  • $\begingroup$ More generally, example of a projective rank 1 module that isn’t unidimensional can be written as AR over R, where A is a non-trivial direct summand of R. Note that this type of example requires the ground ring R to be not simple. $\endgroup$ – Incnis Mrsi Aug 30 '14 at 16:43

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