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In contrast to the possibility of taking an arbitrary sequence of elements of submodules in the definition of direct product, the definition for the direct sum of submodules of a module requires the indexed elements to vanish cofinitely(i.e. except finitely many times).

More precisely, Let $R$ be a ring, and $\{M_i : i ∈ I\}$ a family of left $R-$modules indexed by the set $I$. The direct sum of ${M_i}$ is then defined to be the set of all sequences $(α_i)$ where $\alpha_i \in M_i$ and $α_i = 0$ for cofinitely many indices $i$. (The direct product is analogous but the indices do not need to cofinitely vanish.)(Source Wikipedia: Direct sum of modules.)We have similar definition for the sum of submodules.

I have not yet understood what pathology would be incurred without assuming or what simplification (if any) would be gained in assuming cofiniteness. Can you please explain this in simple terms (possibly, with examples) ?


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I recently asked the same question on MathOverflow. For reasons I do not agree with mildly,it was closed as "not a real question", though "not a research-level question" would have been a better excuse. My question is on the motivation for the necessity of cofiniteness in direct sums. I just do not see what motivated the definition. –  Unknown Jan 25 '11 at 18:54
I would have voted up every answer had I a rep point of at least 15! –  Unknown Jan 25 '11 at 19:53
You should still be able to accept one (if you are satisfied with one of them); that will also give you some rep. –  Arturo Magidin Jan 25 '11 at 20:32
@ Arturo thanks. Being a somewhat frequent reader of MO, I know most of the privileges. It has become a kind of culture ,however, that OP's wait a little longer even if they have received a complete answer. No wonder that I will accept yours! –  Unknown Jan 25 '11 at 22:40
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4 Answers

up vote 5 down vote accepted

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).

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Superb answer! A little typo $dim()$. You can also add the pathology you mentioned about vector spaces here:little typo in your third paragraph: $\dim(V_1 \oplus V_2)$. You may also add the pathology you mentioned as a false belief here:mathoverflow.net/questions/23478/… –  Unknown Jan 25 '11 at 20:45
@Elohemahab: Thanks; fixed. –  Arturo Magidin Jan 25 '11 at 20:53
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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.

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Zhen, thank you very much. I have understood most of what you wrote except. Did you use $N$ to stand for a family of homomorphisms? –  Unknown Jan 25 '11 at 18:49
$N$ is just another module. –  Zhen Lin Jan 25 '11 at 18:56
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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.

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Thanks for your answer which I understood only the first two lines. –  Unknown Jan 25 '11 at 19:49
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The point is that the direct sum and the direct product both exist and occur widely in nature. Sometimes you want one, sometimes the other.

Here is a simple, basic and important example where you absolutely want the "cofiniteness condition" (i.e., the direct sum). Let $V$ be an infinite-dimensional vector space over a field $K$. Then a basis $B$ for $V$ induces an isomorphism of $V$ with the infinite direct sum $\bigoplus_{b \in B} K$. Linear combinations of infinite sets of vectors still must have only finitely many nonzero coefficients. (At least in pure algebra. If you have a topology, then one can make sense of infinite linear combinations...)

Here is a simple example where you want the direct product (taken from yesterday's lecture in my commutative algebra course). If you have an infinite indexed family $\{R_i\}_{i \in I}$ of rings, then the direct product $R = \prod_{i \in I} R_i$ is a ring. The infinite direct sum is not a ring according to the modern definition if infinitely many of the factors are nonzero rings, because the multiplicative identity $1 = (1,1,1,\ldots)$ is not present.

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much obliged. The last sentence has confused me. Isn't the multiplicative identity absent even if finitely many of the factors are nonzero rings? –  Unknown Jan 25 '11 at 19:24
@Elohemahab: no. The (only possible) multiplicative identity is the element which is $1$ in each of the coordinates. This always exists in the direct product. It exists in the direct sum iff for all but finitely many $i$, the unit $1$ in $R_i$ is equal to the zero element $0$ in $R_i$. –  Pete L. Clark Jan 25 '11 at 23:25
Ah, I see. Thank You. –  Unknown Jan 29 '11 at 11:50
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