Abstract Nonsense

Crushing one theorem at a time

Countable Coproduct of Free Modules are Free, but not Arbitrary Products


Point of Post: In this post we prove that the coproduct of an arbitrary number of free modules is free, yet the same isn’t true for arbitrary products.

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Motivation

It is fairly obvious that the arbitrary direct sum of free modules are free modules, but what about direct products? Is it clear, or even likely that every product can be expressed as a coproduct–that is the content of the question. The answer as it turns out, is no. Namely, we can take the arbitrary (even countable) product of free modules and get modules which aren’t free. This, once again, flies in the face of the case for vector spaces and modules over division rings since their arbitrary direct product is still free (still being a module over a division ring).

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Coproducts and Products of Free Modules

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What we first prove is the following obvious theorem which says, in essence, that the arbitrary coproduct of free modules are free, and moreover that what we want to be a basis is a basis. In particular:

Theorem: Let \left\{F_\alpha\right\}_{\alpha\in\mathcal{A}} be a set of free R-modules. Then, given any set of bases \{\mathscr{B}_\alpha\}_{\alpha\in\mathcal{A}} the set \displaystyle \mathscr{B}=\bigcup_{\alpha\in\mathcal{A}}\widehat{\mathscr{B}}_\alpha, where \widehat{\mathscr{B}}_\alpha is the image of \mathscr{B}_\alpha under the usual inclusion \displaystyle\iota_\alpha:F_\alpha\hookrightarrow \bigoplus_{\alpha\in\mathcal{A}}F_\alpha, is a basis for \displaystyle \bigoplus_{\alpha\in\mathcal{A}}F_\alpha. In particular, the arbitrary coproduct of free modules is free. Moreover, if R is an IBN-ring then \displaystyle \text{rank}\left(\bigoplus_{\alpha\in\mathcal{A}}F_\alpha\right)=\sum_{\alpha\in\mathcal{A}}\text{rank}(F_\alpha).

Proof: To see that \mathscr{B} is a basis we note first that if (m_\alpha) in the coproduct is arbitrary then it can be written as \displaystyle \sum_{\alpha\in\mathcal{A}}r_\alpha\iota_\alpha(n_\alpha) where each n_\alpha\in F_\alpha. But, since \mathscr{B}_\alpha is a basis for F_\alpha we see that n_\alpha is a linear combination of elements of \mathscr{B}_\alpha and since \iota_\alpha is an R-map it clearly follows that each \iota_\alpha(n_\alpha) is a linear combination of elements of \iota_\alpha(\mathscr{B}_\alpha)=\widehat{\mathscr{B}}_\alpha and so clearly (m_\alpha) is a linear combination of elements of \mathscr{B}. To see that \mathscr{B} is linearly indepdendent we merely note that if a linear combination of the elements of \mathscr{B} is zero, then each coordinate would have to zero, which would tell us that a linear combination of the \mathscr{B}_\alpha is zero, which says that all the coefficients are zero. Doing this for each \alpha tells us that every coefficient is zero, and so linear independence follows. Thus, \mathscr{B} is a basis as desired.

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The other two statements follow immediately from this. \blacksquare

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But, as stated, not everything works out nicely for arbitrary products (finite products are fine since they coincide with coproducts).  Indeed, we have the following, somewhat surprising theorem:

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Theorem: Let R be a PID that is not a field. Then, R^\mathbb{N} is not a free R-module.

Proof: Suppose for a contradiction that R^\mathbb{N} has a basis \mathscr{B}. Define, for (x_n)\in R^\mathbb{N},  the support \text{supp}(x_n) to be the finite set of elements of \mathscr{B} which have nonzero coefficients in the \mathscr{B}-expansion of (x_n).  The important thing to note is that \text{supp}(x_n)\cap\text{supp}(y_n)=\varnothing then \text{supp}(x_n+y_n)=\text{supp}(x_n)\cup\sup(y_n). Since R is not a field we can choose some r\in R which is nonzero and noninvertible. Define then the n-support \text{supp}_n(x_n) to be the set of basis vectors in the \mathscr{B}-expansion whose coefficient is not divisible by r^n.  Of course we have that \text{supp}_{n+1}(x_n)\supseteq\text{supp}_n(x_n).

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Choose a sequence k_n of elements of R^\mathbb{N} for which the first n-1 coordinates of k_n is zero and the support of any two k_n,k_m are disjoint (it’s clear how to do this). Divide k_n by the maximal power of r which is permissible, to obtain a new sequence \ell_n. Define \displaystyle s=\sum_{n=1}^{\infty}r^n\ell_n where this sum makes sense since coordinatewise this is a finite sum. So, we claim that the existence of s is a contradiction. Why? Since \mathscr{B} is a basis we should be able to express s as the linear combination of some finite subset of \mathscr{B}. That said, we shall now show that \text{supp}(s) can be made arbitrarily large. Indeed, let \displaystyle S_m=\sum_{n=1}^{m}r^n\ell_n. Note then that \displaystyle \text{supp}_{m+1}(S_m) is equal to, by construction, \text{supp}_{m}(S_{m-1})\cup\text{supp}_{m+1}(r^m\ell_m), but note that \text{supp}_{m+1}(r^m\ell_m) is non-empty by construction. Continuing this way down, recalling that since the \ell_n‘s have disjoint supports they distribue we can clearly see that \#(\text{supp}_{m+1}(S_m))\geqslant m. But, since \text{supp}_{m+1}(S_m)\subseteq\text{supp}(s) it follows that \text{supp}(s) can be made arbitrarily large, which is a contradiction. \blacksquare

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From this we the following we get the following corollary:

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Theorem: Let R be a commutative unital ring and F,K two free left R. Then, \text{Hom}_R(F,K) need not be free.

Proof: Clearly \mathbb{Z},\mathbb{Z}^{\oplus\mathbb{N}} are free \mathbb{Z}-modules, but because of what we know about Hom sets and coproducts we have that \text{Hom}_\mathbb{Z}(\mathbb{Z}^{\oplus\mathbb{N}},\mathbb{Z})\cong \text{Hom}_\mathbb{Z}(\mathbb{Z},\mathbb{Z})^\mathbb{N}\cong\mathbb{Z}^\mathbb{N} and by the previous theorem this is not free. \blacksquare

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References:

[1] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. Module Theory. Clarendon, 1990. Print.

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November 23, 2011 - Posted by | Algebra, Module Theory, Ring Theory | , , , , , , , , , ,

1 Comment »

  1. […] this would be equivalent to the statement that the product of free modules is free, which we have shown before this is not true in […]

    Pingback by Projective Modules (Pt. II) « Abstract Nonsense | February 7, 2012 | Reply


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