# Confusion about how the tensor of a ring and module is constructed.

I'm reading Dummit and Foote chapter 10 section 4 on tensors. They write

It is ... natural to consider the free $$\mathbb Z$$-module ... on the set $$S\times N$$, i.e. the collection of all finite commuting sums of elements of the form $$(s_i,n_i)$$ where $$s_i\in S$$ and $$n_i\in N$$. This is an abelian group where there are no relations between any distinct pairs $$(s,n)$$ and $$(s',n')$$, i.e. no relationship between the formal products'' $$sn$$, and in this abelian group the original module $$N$$ has been thoroughly distinguished from the new "coefficients" from $$S$$. To satisfy the relations necessary for an $$S$$-module structure imposed [above] and the compatibility relation with the action of $$R$$ on $$N$$ [above], we must take the quotient of this abelian group by the subgroup $$H$$ generated by all elements of the form

$$(s_1+s_2,n) - (s_1,n)-(s_2,n)$$ $$(s,n_1+n_2) - (s,n_1)-(s,n_2)$$ $$(sr,n) - (s,rn)$$

Now my question is basically: isn't $$(s_1+s_2,n) - (s_1,n) - (s_2,n)$$ the same as $$(0,-n)$$? That just doesn't seem possible though, since if it were true we would then just be taking all elements of the form $$(0,n)$$ and $$(s,0)$$ as well as elements $$(sr-s,n-rn)$$. This wouldn't generally even be closed under addition.

I'm sure I must be misunderstanding the nature of the + operation on two pairs $$(s_1,n_1)+(s_2,n_2)$$. But in this section they do say that we take the product and that it's an abelian group. Well ... don't we define the plus operation on the product, coordinatewise? What has indicated so far that we don't use coordinate-wise addition, and what is the alternate notion of addition that we are instead using? The semi-direct plus from group theory?

• Since you're taking a free abelian group on those elements, there's no a-priori way to add different pairs together to obtain a new one. You're confusing the sum on the free group with the one on the direct sum. This construction can be confusing but remember the idea is to create a "bilinear" product of the modules. Apr 3, 2021 at 1:11

$$(s,n)$$ and $$(s',n')$$ are independent in $$F_{ab}(S\times N)$$ (free abelian group with basis $$S\times N$$), unless $$s=s'$$ and $$n=n'$$. The first two relations are distribution (over addition in the ring and module respectively) and the third says that the copy of $$R\subseteq S$$ acts the same as the already existing $$R$$-module structure.

Given any set $$A$$, the free $$\mathbb Z$$-module on $$A$$ is a $$\mathbb Z$$-module $$F^{\mathbb Z}(A)$$ together with a function $$j : A \to F^{\mathbb Z}(A)$$ with the following property:

For any function $$f$$ from $$A$$ to a $$\mathbb Z$$-module $$G$$ there exists a unique $$\mathbb Z$$-module homomorphism $$\tilde{f\,} : F^{\mathbb Z}(A) \to G$$ satisfying $$\tilde{f\,} \circ j = f$$.

Taking a $$\mathbb Z$$-module $$G$$ with two or more elements, it is easy to show that $$j$$ is injective, and then we can identify $$A$$ with a subset of $$F^{\mathbb Z}(A)$$. That is to say, $$F^{\mathbb Z}(A)$$ is a $$\mathbb Z$$-module that contains $$A$$ such that for any function $$f$$ from $$A$$ to a $$\mathbb Z$$-module $$G$$ there exists a unique $$\mathbb Z$$-module homomorphism $$\tilde{f\,} : F^{\mathbb Z}(A) \to G$$ that extends $$f$$.

Such a $$\mathbb Z$$-module $$F^{\mathbb Z}(A)$$ exists?

Well, if $$\mathbb Z^{(A)}$$ denotes the set of all functions $$A \to \mathbb Z$$ with finite support, $$\mathbb Z^{(A)}$$ is a $$\mathbb Z$$-module with the pointwise operations; and note that for all $$a \in A$$, the characteristic function of $$\{a\}$$, $$j(a) := \chi_{\{a\}}$$, is in $$\mathbb Z^{(A)}$$. Moreover, if $$\alpha \in \mathbb Z^{(A)}$$, then $$\alpha = \sum_{a \in A} \alpha(a)j(a)$$ (the sum is finite!). Thus, writting $$j(a)$$ as $$a$$, every element of $$\mathbb Z^{(A)}$$ can be written as a finite sum $$\sum_{a \in A} n_aa$$, with $$n_a \in \mathbb Z$$, and then $$F^{\mathbb Z}(A) := \mathbb Z^{(A)}$$ works:

If $$G$$ is a $$\mathbb Z$$-module and $$f : A \to G$$ is a function, consider the function $$\tilde{f\,} : \mathbb Z^{(A)} \to G$$ that sends $$\sum_{a \in A} n_aa$$ to $$\sum_{a \in A} n_af(a)$$. Then $$\tilde{f\,}$$ is the unique group homomorphism that extends $$f$$.

In conclusion, $$(s_1+s_2,n) - (s_1,n) - (s_2,n)$$ is not the same as $$(0,-n)$$, it is indeed $$j(s_1+s_2,n) - j(s_1,n) - j(s_2,n)$$.

Oh, I think I'm actually starting to get the idea. I think the idea is that the + operation is defined by the definition of being a "free abelian group" which I guess I didn't really appreciate as being different from the group just viewed as if it were a $$\mathbb Z$$-module. But I guess in the most general setting, a free abelian group merely has formal sums. Therefore even though we know that these pairs could be endowed with the usual coordinate-wise group operation, we're not going to be committed to that. Instead we just understand $$(s_1,n_1)+(s_2,n_2)$$ to be some element of $$S\times N$$ but we don't say what it is, any more than that. But no matter what the + operation is, we can always define this formal set of elements and it will always be a subgroup. And therefore we can quotient by it, and yada yada yada with the rest of the construction.

But I only kind of feel confident in this answer. Please correct me if I have anything wrong.

• Yes, this is exactly how it works. Apr 3, 2021 at 0:14
• Yes, that's basically it. Note that if I had been writing the exposition, I would have said something along the lines of: consider some set in bijection with $S \times N$, where we will write the elements formally as $e_{(s,n)}$ so the set in bijection with $S \times N$ is $\{ e_{(s,n)} \mid s \in S, n \in N \}$; and then form the free abelian group on that set. That will help to distinguish the elements and avoid your confusion with $S \times N$ which already has a preexisting abelian group structure. Apr 3, 2021 at 0:19