Do free modules over ring without identity exist?

Question

If {$e_i$} is the generating set of a free R-module M, and there is no unity in R, how does, say $e_1$ exist in M anyway?

Context Edit (by jgon): The original author appears to have abandoned the question, as there are no new comment replies, and the question is now closed and has a delete vote, but I think this is a fairly worthwhile question to address and keep on the site, so I wanted to explain why I think this is so.

When $R$ is a ring with unity, we have the well known notion of free $R$-modules, which have many definitions. Per the definition suggested by the OP in the comments below, they are $R$ modules $M$ equipped with a basis $\{e_i\}$ such that every element $m$ of $M$ has a unique expression $$m = \sum_i a_i(m)e_i,$$ for coefficients $a_i(m)\in R$, only finitely many of which are nonzero.

Now if we thought about how to adapt this definition to define a free $R$-module over a ring $R$ which doesn't have unity, we run into problems immediately. For example, if we pick a basis $\{e_i\}$, it's not even clear that we can express any member of this basis, like $e_1$, in terms of the basis, since we can't write $e_1=1\cdot e_1$ any longer. Thus we'd like to know how to generalize and make sense of free modules in the context of not necessarily unital rings.

Answer 1

Let me start by saying I am sorry this answer is not definitive. I hope, though, to justify the content of the question and boost interest by pointing out the underlying problems.

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In your comment you elaborated that you want the basis-definition for the definition of a free module.

An $R$ module $M$ is free if there exists a basis $B=\{e_i\mid i\in I\}\subseteq M$. That is, the $R$-combinations of things in $B$ generate $M$, and the elements of $B$ are $R$-linearly independent.

You already recognized one deficiency: apparently $e_i\in \sum e_jR$, but its representation is no longer obvious (it used to be the linear combination $e_i\cdot 1$.)

That is a valid concern. It calls into question what a linear combination out to be.

Unfortunate victims of that definition

I can't immediately say that I know there are (nonzero) rngs which have no nonzero free modules of that sort, but what I can say is that things we'd want to be free modules don't work with this definition. Specifically I'm thinking of the module $R^n$.

If $R$ is a nonzero ring such that $R^2=\{0\}$, for example, then $R^n$ cannot have an $R$ basis like the one you describe above. Every product between something in $R^n$ and something in $R$ is zero, so there is no hope of generating $R^n$ at all that way.

A likely alternative

Perhaps, like in the definition of a principle ideal of a ring without identity, it is necessary to include $\mathbb Z$ multiples as well. A revised definition might be something like:

1. Everything in $M$ is of the form $\sum e_ir_i +\sum e_in_i$ where $r_i\in R$ and $e_in_i$ is the sum of $e_i$ with itself $n_i$ times, and

2. $\sum e_ir_i +\sum e_in_i=0$ implies $e_ir_i+e_in_i=0$ for all $i$.

I have never seen this proposed in writing as a solution to the issue of "what do free modules look like for rings without identity" but it would be one candidate to check out. Perhaps it has its own flaws.

Free modules are at least defined

As for the title question: in any category, free objects are at least defined in any category, and the category of modules over a rng is just another category. In any category of modules the zero module is going to be a free module on an empty basis, so the question is really whether or not nonzero free modules exist for rngs.

Answer 2

rschwieb gave an excellent answer, but I thought I'd expand on some of the details, as well as on Hurkyl's comments beneath that answer.

What are free objects?

rschwieb's answer has already pointed out some of the problems with trying to directly generalize the definition of free modules in terms of linear combinations of basis elements, as well as a rough solution. Thus, I want to define free objects more generally, so that we know what we're talking about.

Let $C$ be a concrete category, and let $U:C\to\newcommand\Set{\mathbf{Set}}\Set$ be the forgetful functor (if you're not too familiar with concrete categories, you can think of this category as having objects that are sets with extra structure and morphisms that are functions between these sets that respect this structure).

In our case, $C$ will be the category of modules over a certain rng, and $U$ will just return the underlying set of the module.

A free object $F$ with basis $B\subseteq UF$, is an object such that for any other object $G$, and any function $\phi_0 : B\to UG$, there exists a unique extension $\phi : F\to G$ which is a morphism in $C$. In other words, to specify a morphism from a free object we can freely pick any elements we like of $G$ to send $B$ to.

The free functor

It turns out that in many cases, if we are given a set $X$, we can construct a free object $FX$ with basis $X$, and this assignment $X\mapsto FX$ is functorial. This functor, $F$, is called the free functor, and it turns out to be left adjoint to the forgetful functor.

Brief aside

When discussing modules over rngs, it is important to recall that even if a rng $R$ has a unit, the modules over $R$ as a rng and as a ring are different, because ring modules are required to satisfy the additional axiom that $1\cdot m=m$, whereas we cannot require this for modules over rngs.

Understanding modules over rngs

Let $R$ be a rng. Let $C$ be the category of rng modules over $R$, and let $D$ be the category of ring modules over $\Bbb{Z}\oplus R$ (addition and multiplication given by $(n,r)+(m,s)=(n+m,r+s)$, $(n,r)\cdot (m,s)= (nm,mr+ns+rs)$). Then $C$ and $D$ are isomorphic as concrete categories.

Proof.

If $M$ is a module over $\Bbb{Z}\oplus R$, it is already a module over $R$, since $R$ is a subring of $\Bbb{Z}\oplus R$. Similarly, if $M$ is a rng module over $R$, it is also naturally a module over $\Bbb{Z}\oplus R$, since we can define $$ (n,r) \cdot m = nm + rm.$$

Thus we have a bijective correspondence between the objects of $C$ and $D$ that preserves the underlying set, and it is functorial as well, since $R$-linear maps are additive by definition, and thus $\Bbb{Z}$-linear as well, and $\Bbb{Z}\oplus R$-linear maps are already $R$-linear.

Conclusion

Thus, since the categories of rng modules over $R$ and ring modules over $\Bbb{Z}\oplus R$ are isomorphic as concrete categories, they have the same free objects as well. Therefore, as Hurkyl suggests, the free rng modules over $R$ are of the form $(\Bbb{Z}\oplus R)^n$.

Furthermore, one can check that the alternative definition of linear combination suggested by rschwieb will also produce the same free objects, since the definition suggested for linear combinations in a rng module is that one should take $\Bbb{Z}\oplus R$ linear combinations.