Why is cofiniteness included in the definition of direct sum of submodules?

Solution 1:

To see an example of some pathology, while the direct sum of free modules is always free (with the obvious free basis), the direct product of free modules may fail to be free.

Take $R=\mathbb{Z}$; then $M = \mathop{\oplus}\limits_{n=1}^{\infty}\mathbb{Z}$ is free abelian, with basis given by the "obvious" elements $e_i$ (which have a $1$ in the $i$th coordinate and zeros elsewhere). However, $N = \prod\limits_{n=1}^{\infty} \mathbb{Z}$ is not free abelian. For example, Specker proved (Additive Gruppen von Folgen ganzer Zahlen, Portugaliae Math. 9 (1950) 131-140) that $N$ has only countably many homomorphisms onto $\mathbb{Z}$. But since $N$ is uncountable, if it were free it would be free in uncountably many generators, and hence would have uncountably many homomorphisms onto $\mathbb{Z}$ (at least the projections). In fact, if $X$ is any infinite set, then $\mathop{\oplus}_{x\in X}\mathbb{Z}$ is free abelian of rank $|X|$, but $\prod_{x\in X}\mathbb{Z}$ is never free abelian.

You have a related pathology with vector spaces (of course, every vector space is free, so that's not what the problem will be, but rather when you think about "free on what set?"). When working with finitely many vector spaces, you have that $\dim(V_1\times V_2) = \dim(V_1\oplus V_2) = \dim(V_1)+\dim(V_2)$ (in the sense of sum of cardinalities). However, once you have infinitely many vector spaces, the equality breaks down for the product, while it holds for the direct sum: $$\dim\left(\bigoplus_{i=1}^{\infty} V_i\right) = \sum_{i=1}^{\infty}\dim(V_i)$$ but for products it need not hold: for a counterexample, take $V_i = \mathbb{Q}$ as a vector space over itself; the sum of dimensions is $\aleph_0$, but the direct product of denumerably many copies of $\mathbb{Q}$ is uncountable, so the dimension is $2^{\aleph_0}$ (so you have a "jump" in the dimension once you get to infinitely many elements).

Solution 2:

I'm going to try to motivate the difference by defining the direct sum and product in terms of their universal properties.

The direct sum of a family of $R$-modules $\{ M_\alpha : \alpha \in A \}$ is defined as a module $M$ together with a family of homomorphisms $\iota_\alpha : M_\alpha \to M$ such that for every family of homomorphisms $\phi_\alpha : M_\alpha \to N$, there is a unique homomorphism $\phi : M \to N$ such that $\phi \circ \iota_\alpha = \phi_\alpha$.

On the other hand, the direct product is defined in essentially the same way except with all the arrows reversed: so we have a module $P$ together with a family of homomorphisms $\pi_\alpha : P \to M_\alpha$ such that for every family of homomorphisms $\psi_\alpha: N \to M_\alpha$ we have a unique homomorphism $\psi: N \to M$ such that $\pi_\alpha \circ \psi = \psi_\alpha$. With this definition, we can probe $P$ by means of homomorphisms $R \to P$, where $R$ is regarded as the free $R$-module on one generator. Because every homomorphism into $P$ is determined by its projections, and every homomorphism out of $R$ is determined by the image of $1$, it turns out $P$ must contain everything in the set-theoretic Cartesian product $\displaystyle \prod_{\alpha \in A} M_\alpha$, and nothing else.

Unfortunately, it is not so easy to probe the direct sum $M$. All we know is that it contains isomorphic copies of each $M_\alpha$, and the definition of module only gives us finite sums of elements, so we only know that $M$ contains finite linear combinations of elements from each $M_\alpha$. It turns out this is sufficient to define a module, and it's exactly the direct sum as usually defined.

Solution 3:

I like to think of this as being the difference between polynomials (direct sum) and power series (direct product). Sometimes one object is more useful than or shows up in context instead of the other. For instance the cohomology ring of a space is the direct sum

$$ H^*(X)=\bigoplus_{n\geq 0}H^n(X). $$

So the cohomology ring of $\mathbb{C}P^\infty$ is $\mathbb{Z}[x]$, a polynomial ring in one variable. We choose the direct sum due to some topology issues with the cup product.

Other times we may want the direct product.