Direct sums of modules

#module_theory

As is always the case with category-theoretic constructions, there is a construction "dual" to that of the direct product of modules, called the direct sum of modules. It is characterized by the property "dual" to that of direct product. (It really should have been called the coproduct instead of the direct sum, but we'll save that discussion for later.)

The direct sum of two modules

First suppose $M_{1}$ and $M_{2}$ are two $R$ -modules. We can construct an $R$ -module called their direct sum, denoted $M_{1} ⨁ M_{2}$ , whose elements consist of all formal sums^[1] of the form $m_{1} + m_{2}$ , with "component-wise" addition and scaling.^[2] We also have two "inclusion" module morphisms $j_{i} : M_{i} \to M_{1} ⨁ M_{2}$ , which send an element $m_{i} \in M_{i}$ to the element $m_{1} + 0_{M_{2}}$ (for $i = 1)$ or $0_{M_{1}} + m_{2}$ (for $i = 2$ ).

These data satisfy the usual universal property for a coproduct, in that it is universal among all such modules equipped with morphisms from $M_{1}$ and $M_{2}$ . More precisely, for each module $P$ and pair of module morphisms $f_{1} : M_{1} \to P$ and $f_{2} : M_{2} \to P$ there is a unique module morphism $h : M_{1} ⨁ M_{2} \to P$ such that the diagram below commutates:

As with the direct product, we can actually deduce the unique map $h$ from the commutativity conditions. First suppose we take any element $m_{1} + m_{2}$ of $M_{1} \oplus M_{2}$ . We must have

\begin{aligned} h (m_{1} + m_{2}) & = h ((m_{1} + 0_{M_{2}}) + (0_{M_{1}} + m_{2})) \\ = h (m_{1} + 0_{M_{2}}) + h (0_{M_{1}} + m_{2}) \\ = h (j_{1} (m_{1})) + h (j_{2} (m_{2})) \\ = (h \circ j_{1}) (m_{1}) + (h \circ j_{2}) (m_{2}) \\ = f_{1} (m_{1}) + f_{2} (m_{2}) . \end{aligned}

So, the only possible map $h : M_{1} \oplus M_{2} \to P$ with the required factorization properties is defined by $h (m_{1} + m_{2}) = f_{1} (m_{1}) + f_{2} (m_{2})$ . All that remains is to verify that this map is indeed an $R$ -module morphism.

Coproducts in related categories

Observe that, as an abelian group, $M_{1} ⨁ M_{2}$ is the usual direct sum (i.e., coproduct) of abelian groups; however, as a set, $M_{1} ⨁ M_{2}$ is not (bijective to) the usual disjoint union (i.e., coproduct) of sets.

More precisely, let $U_{1} : R - Mod \to Ab$ and $U_{2} : R - Mod \to Set$ denote the usual forgetful functors. Then $U_{1} (M_{1} ⨁ M_{2}) ≃ U_{1} (M_{1}) ⨁ U_{2} (M_{2})$ but $U_{2} (M_{1} ⨁ M_{2}) ≄ U_{2} (M_{1}) ⊔ U_{2} (M_{2})$ . We might reasonably say that the forgetful functor $U_{1}$ preserves binary coproducts, but the forgetful functor $U_{2}$ does not.

What is a "formal sum"?

In the construction above, we wrote that the elements of $M_{1} \oplus M_{2}$ consist of "formal sums" of the form $m_{1} + m_{2}$ with $m_{i} \in M_{i}$ . But what is a "formal sum" and why is it not just called a "sum"? For an initial answer to that question, note that for elements $m_{1} \in M_{1}$ and $m_{2} \in M_{2}$ there is no ambient operation that allows us to add $m_{1}$ to $m_{2}$ . That's because we are not given any module $M$ in which we can simultaneously view $M_{1}$ and $M_{2}$ as submodules. In fact, you can view the construction of $M_{1} \oplus M_{2}$ as precisely the construction of such a module.

The question still remains, however, as to what we're precisely doing here. In which previously existing set is the formal sum $m_{1} + m_{2}$ supposed to be contained? We'll give an answer to that question at the bottom of this note.

The direct sum of an arbitrary family of modules

As with product of modules, let's now jump to the general case. Suppose ${M_{s} ∣ s \in S}$ is a family of $R$ -modules indexed by some set $S$ ; i.e., a functor $F : S \to R - Mod$ . Following the pattern we've established, the direct sum of this family is an $R$ -module, denoted $⨁_{s \in S} M_{s}$ , together with module morphisms $j_{t} : M_{t} \to ⨁_{s \in S} M_{s}$ for every $t \in S$ , universal among all such data. As an abelian group, this is the usual direct sum of the corresponding abelian groups.

How is the module $⨁_{s \in S} M_{s}$ constructed? Recall that with the infinite direct product, we formally defined the elements of $\prod_{s \in S} M_{s}$ as set maps $f : S \to Z$ such that $f (s) \in M_{s}$ for every $s \in S$ , where $Z$ was any set containing all of the elements of every $M_{s}$ . Here we do the same, with one small twist. The elements of $⨁_{s \in S} M_{s}$ are the set maps $f : S \to Z$ such that:

$f (s) \in M_{s}$ for every $s \in s$
$f (s) = 0_{M_{s}}$ for all but finitely many $s \in S$ .

With this description of the direct sum, the "inclusion" morphism $i_{t} : M_{t} \to ⨁_{s \in S} M_{s}$ is the morphism that sends each $m \in M_{t}$ to the map $f_{m} : S \to Z$ defined by

f_{m} (s) = {\begin{cases} m, & if s = t \\ 0_{M_{s}}, & else \end{cases}

You might feel that this formal description is clunky, but there is one immediate benefit. Written this way, it is clear that there is a "canonical" injective module morphism

⨁_{s \in S} M_{s} \to \prod_{s \in S} M_{s},

which is an isomorphism when $S$ is finite (but usually not an isomorphism when $S$ is finite).

But why the requirement that $f (s) = 0_{M_{s}}$ for all but finitely many $s \in S$ ? This new restriction turns out to be a consequence of the desired universal property, one that was not implied/required for the universal property of the direct product.

Indeed, suppose we did not require this. Now suppose we had a collection of $R$ -module morphisms $g_{s} : M_{s} \to P$ , one for every element . The desired universal property of $⨁_{s \in S} M_{s}$ would then require the existence of a unique module morphism $h : ⨁_{s \in S} M_{s} \to P$ through which every $g_{s}$ factored. This factorization property would force $h$ to satisfy

h (f) = \sum_{s \in S} g (f (s)) .

However, that sum occurs in the module $P$ , where arbitrary sums of elements are not necessarily defined. (As an $R$ -module, $P$ is endowed with a binary operation, which can be repeated to define an $n$ -ary operation, but not necessarily anything beyond that.) In other words, it's not clear such a map $h$ is well defined (or even possible).

The finiteness condition on our formal sums removes this issue, and indeed the desired universal property is now straightforward to verify.

Suggested next notes

Sums of submodules
Direct products vs. direct sums vs. sums

See below if you don't like this "formal sum" business. ↩︎
The choice of symbol is deliberate. The direct sum of modules is not the same as the (direct) sum of submodules of a given module, so the notation is use to distinguish the two constructions. However, both share the same universal property, only in different categories. Thus the desire to give both a "sum-like" notation. ↩︎