I'm actually trying to fully understand an old reply here Forgetful functor from R-modules to abelian groups? but I figure my doubts can also be formulated as a separate question.

So consider the forgetful functor from the category of modules over $R$ into the category of abelian groups. I would like to understand what is the left adjoint to this functor, but I have trouble imagining what could it be. My approach to understanding is that if $G$ is the forgetful functor, $F$ is the adjoint we are looking for, and $c$ is an abelian group whereas $d$ is a module, then we want to pick $F$ in a way that there will be a natural bijection between group homomorphisms from $Hom(c,G(d)$ and module homomorphisms in $Hom(d, F(c))$.

In the post Forgetful functor from R-modules to abelian groups? the person replying states that the left adjoint sends a group $A$ to $R \otimes_{\mathbb{Z}} A$. I'm not very familiar with tensor products and tried reading about them but this just made me more confused. So I guess my question comes down to what kind of object is $R \otimes_{\mathbb{Z}} A$ and why is it the left adjoint here? Does the answer change if $R$ is a commutative ring?

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    $\begingroup$ I don't know if this is just a typo, but the natural bijection should be between $Hom_\mathbb{Z}(c,G(d))$ and $Hom_R(F(c), d)$. $\endgroup$ – Arnaud D. Jan 25 '18 at 18:11
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    $\begingroup$ The left adjoint is the tensor product $A\mapsto A\otimes_{\Bbb Z}R$. I would urge you to grasp the nettle and start getting to grips with tensor products... $\endgroup$ – Angina Seng Jan 25 '18 at 18:58
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    $\begingroup$ @LordSharktheUnknown But what does it mean to take the tensor product of a group and a ring? Could you recommend a source for that? $\endgroup$ – TheMountainThatCodes Jan 25 '18 at 19:15
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    $\begingroup$ Both $A$ and $R$ are $\mathbb{Z}$-modules (hence the $\otimes_\mathbb{Z}$). The $\mathbb{Z}$-module $A\otimes_\mathbb{Z} R$ comes with a natural $R$-module structure : $r\cdot (a\otimes r') = a\otimes rr'$ for pure tensors, extended linearly $\endgroup$ – Maxime Ramzi Jan 25 '18 at 19:49
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    $\begingroup$ In particular, $\mathbf{Ab}\cong\mathbf{Mod}(\mathbb Z)$. $\endgroup$ – Derek Elkins left SE Jan 25 '18 at 22:12

The left adjoint to a forgetful functor is always some kind of free construction functor.

In this case we take the elements of the given Abelian group $A$ and make a formal possibility to multiply them (from left) by elements of our ring $R$, and add them. We can call it the $R$-module freely generated on an Abelian group.

Its elements are [formal sums of] pairs $\langle r,a\rangle$, so that we can multiply these by elements of $R$ from the left.
Pairs of the form are considered equal: $\langle rn,a\rangle = \langle r,na\rangle=n\cdot\langle r,a\rangle$ for any $n\in\Bbb Z$.
There are also similar constrainments about the formal sums: $$\langle r_1+r_2,a\rangle = \langle r_1,a\rangle + \langle r_2,a\rangle \quad\quad\quad \langle r,a_1+a_2\rangle=\langle r,a_1\rangle+\langle r,a_2\rangle$$ The usual notation for such a pair $\langle r,a\rangle$ in this context is $r\otimes a$.

This construction can be made exact using the free Abelian group generated by $R\times A$ (so that we have the formal sums of our pairs) and then define a normal subgroup based on the above relations, and quotient it out.

Note that if we skip addition from the above, this situation might be easier to comprehend first:
Then we would have $A$ as a pure set, and $R$ as a monoid (a set with associative multiplication with unity).
The nonadditive $R$-modules are called $R$-sets or $R$-actions, and the left adjoint of the forgetful functor from $R$-actions to sets is simply $A\mapsto R\times A$ with the $R$-action $r\cdot\langle r_1,a\rangle:=\langle rr_1,a\rangle$.

If we then have a function $f:A\to M$ with $M$ (the underlying set of) an $R$-action, it induces a morphism of $R$-actions $\varphi:R\times A\to M$: $$\langle r,a\rangle \,\mapsto\, r\cdot f(a)$$ and conversely, such an $f$ can be recovered by taking $a\mapsto \varphi(\langle 1,a\rangle)$.

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    $\begingroup$ In one category theory course, the mnemonic was: often a left adjoint is constructed as "what objects have to exist, and what relations have to exist between them"? (Whereas right adjoint is "what data can you extract and what restrictions are there on that data".) So here, if you have a morphism of abelian groups $\phi : A \to B$ where $B$ is an $R$-module, then what has to exist in $B$ is $r \phi(a)$ for $r \in R$ and $a \in A$; and necessarily relations include $r \phi(a+b) = (r \phi(a)) + (r \phi(b))$. And you know you've found "enough" necessary elements and relations when... $\endgroup$ – Daniel Schepler Jan 26 '18 at 1:01
  • $\begingroup$ you can construct an $R$-module formally from the forms of the elements, making use of the relations. So in this example, you find you can't express $r \phi(a) + r' \phi(a')$ as a single $r_0 \phi(a_0)$, so you pass to $\sum_i r_i \phi(a_i)$, etc. etc. $\endgroup$ – Daniel Schepler Jan 26 '18 at 1:03
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    $\begingroup$ @Berci Would you mind clarifying the way that $f$ induces a morphism of $R$-actions? Unless I misunderstood i think you meant $\langle r, a \rangle \rightarrow r \cdot f(a)$ ? $\endgroup$ – TheMountainThatCodes Jan 26 '18 at 17:12
  • $\begingroup$ Thanks for the great answer, the example with the monoid does make it a lot easier to grasp. $\endgroup$ – TheMountainThatCodes Jan 26 '18 at 18:24

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