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I have a question very similar to that asked in Free $\mathbb{Z}$-modules, but is not answered in that thread. This thread cleared up some of my confusion, but still leaves me with the question:

Why are free abelian groups equivalent to free $\mathbb{Z}$-modules rather than having a more general definition where we can specify the ring? Or am I incorrect in this interpretation?

Some more info:

In my study of homology theory, I work with tons of free modules where the scalars come from $\mathbb{Z}/p$. However, when I was first learning homology theory, the fundamental object of study was referred to as a free abelian group by every text I could find (because we were working with free $\mathbb{Z}$-modules). This has led to a lot of confusion on my part about how to refer to free $\mathbb{Z}/p$-modules, since I've want to keep calling them free abelian groups generated by a set $S$, but now with coefficients in $\mathbb{Z}/p$. I interpret the discussion in the question I link above to suggest that this is wrong and I should not use the term "free abelian group" to refer to a free $\mathbb{Z}/p$-module. Is this the case? If so, is there an analogous way to refer to a free $\mathbb{Z}/p$-module that emphasizes "abelian group" in the same way that we can refer to a free $\mathbb{Z}$-module as a free abelian group? Or should I just stick with using "free $\mathbb{Z}/p$-module"?

rosterherik
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  • Any module over the $p$-element field (which you call $\mathbb{Z}/p$) is free. Abelian groups and $\mathbb{Z}$-modules are one and the same thing. – egreg Nov 16 '18 at 22:53

2 Answers2

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A free abelian group is a free $\mathbb{Z}$-module. That's because an abelian group is essentially the same thing as a $\mathbb{Z}$-module: every abelian group can be made into a $\mathbb{Z}$-module in exactly one way (and homomorphisms of abelian groups are the same as homomorphisms of $\mathbb{Z}$-modules). A module over some other ring $R$ always has an underlying abelian group structure, but if $R$ is different from $\mathbb{Z}$, then not every abelian group has a unique $R$-module structure, so $R$-modules are not the same as abelian groups in the same way $\mathbb{Z}$-modules are.

If you have a module $M$ over some ring $R$, then calling $M$ a free abelian group would mean that the underlying abelian group of $M$ (when you forget about the scalar multiplication of $R$) is a free $\mathbb{Z}$-module, not that $M$ is a free $R$-module. So, for instance, a free $\mathbb{Z}/p$-module should definitely not be called a free abelian group.

There isn't any special name like "free abelian group" for free modules over any ring other that $\mathbb{Z}$. If you want to talk about free $\mathbb{Z}/p$-modules, you should just call them free $\mathbb{Z}/p$-modules. (Or you could just call them $\mathbb{Z}/p$-modules, since every $\mathbb{Z}/p$-module is free, but it makes sense to say "free" if you are talking about a module which is free over some specific set of generators.)

Eric Wofsey
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I interpret the discussion in the question I link above to suggest that this is wrong and I should not use the term "free abelian group" to refer to a free Z/p-module.

Correct: free $\mathbb{Z}/p$ modules are not free as abelian groups.

When people use "abelian group" and "$\mathbb{Z}$-module specifically, they mean something very specific:

Consider the functor $F: \mathbf{Ab}\to \,_\mathbb{Z}\mathbf{Mod}$ that sends an abelian group $G$ to the module whose underlying abelian group is $G$, with scalar multiplication given by $n\cdot g = g + \ldots + g$, where there are $n$ copies of $g$ being summed. This functor is an isomorphism of categories (which means exactly what you'd expect it to mean). In particular, all categorical constructions in either $\mathbf{Ab}$ or $_\mathbb{Z}\mathbf{Mod}$ are identical, regardless of which category you do them in, and everything commutes with $F$. In particular, being free is a categorical thing, so the free $\mathbb{Z}$-modules are exactly those whose underlying abelian group is free. There is no functor with this property $\mathbf{Ab}\to\,_{\mathbb{Z}/p}\mathbf{Mod}$.

user3482749
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