Section A0.III.6

Products, coproducts, etc., in $\RMod$

1. Products and coproducts

The situation is very similar to $\mathbf{Ab}$. You have a direct sum $M \oplus N$ which is just an $R$-module structure on the cartesian product $M \times N$. The projections take $(m,n)$ apart into $m$ and $n$, while the injections map $m$ to $(m,0)$ and $n$ to $(0,n)$.

In the infinite case, direct sums have finite support while direct products don’t have this restriction.

The pairing $(f, g)$ is simply $p \mapsto (f(p), g(p))$. The copairing $[f,g]$ is $(m,n) \mapsto f(m) + g(n).$

See 6.10 and 6.11 for fibered products and coproducts.

The exercise 6.07 shows that in the infinite case products and coproducts no longer agree in $\RMod.$

2. Kernels and cokernels

Let $\varphi : M \to N$ be an $R$-module homomorphism. Then for any $\alpha : P \to M$ such that $\varphi \circ \alpha = 0$ factors uniquely through $\ker \varphi.$

Kernels exist in $\RMod$

See Proposition (K1)

And similarly, any $\beta : N \to P$ such that $\beta \circ \varphi = 0$ factors uniquely through ${\rm coker} \varphi.$

Cokernels exist in $\RMod$

See Proposition (C1)

For kernels, the situation is already familiar. For cokernels, note that

$${\rm coker} \varphi = \frac{N}{{\rm im} \varphi}.$$

It’s easy to see why this holds: whatever $\beta$ does, it must at least kill the image of $\varphi$ in $N.$ This means that it factors through the quotient $N/{\rm im}\varphi$ in a unique way.

The content of proposition 6.2 is proven in Exercise A0.III.6.12.

3. Free modules and free algebras

For free modules, the situation is essentially analogous to that of $\mathbf{Ab}.$

The universal property of a free object lets us uniquely complete any set function from the basis set to an $R$-module homomorphism.

The exercise 6.1 proves claim 6.3.

In the case of free commutative $R$-algebras, the situation is a bit different. In proposition 6.4 Aluffi proves that $R[A],$ the polynomial ring on intermediates indexed by a (finite) set $A$, is a free commutative $R$-algebra generated by $A.$

The general case of free objects in $\mathbf{R\text{-}Alg}$ is also not too complicated: a free $R$-algebra on $A$ is isomorphic to the monoid ring $R\langle F^{\mathbf{Mon}}(A)\rangle$ on the free monoid on $A.$

4. Submodule generated by a subset; Noetherian modules

By mapping from the free $R$-module on $A \subseteq M$ into $M$ we get to a submodule of $M$ generated by $A.$ This is pretty much exactly how generating sets of groups work, so it’s not a huge surprise.

Exercise 6.14 gives an example of a submodule of a finitely generated module that fails to be finitely generated.

5. Finitely generated vs. finite type

WIP

TODO: Exercise 6.18