Abstract Nonsense

Crushing one theorem at a time

Submodules of Free Modules Need Not be Free Unless Ring is a PID (pt. II)

Point of Post: This is a continuation of this post.

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\mathbf{(3)}\implies\mathbf{(1)}: Let F be a free module with basis \{x_\alpha\}_{\alpha\in\mathcal{A}} and M\leqslant F. By the well-ordering theorem we can make \mathcal{A} into a well-ordered set \left(\mathcal{A},\prec\right), we then define a new set \mathcal{A}_\infty=\mathcal{A}\cup\{\infty\} for some \infty\notin\mathcal{A} and extend \prec to \mathcal{A}_\infty by declaring that \infty is larger than every element of \mathcal{A}. Evidently then \left(\mathcal{A}_\infty,\prec\right) is a well-ordered set where every element \alpha\in\mathcal{A} has a successor \alpha+1\in\mathcal{A}_\infty. We note the obvious fact that F_\alpha\subsetneq F_\beta if and only if \alpha<\beta, and moreover that \displaystyle \bigcup_{\alpha\in\mathcal{A}_\infty}F_\alpha=F. Now, define f:F_{\alpha+1}\to Rx_{\alpha+1} by sending x_{\alpha+1} to itself and every other x_\alpha to 0 and extending by linearity (this map is well-defined since it’s well-defined on a basis). Clearly f is an epi and \ker f=F_\alpha and so by the first isomorphism theorem F_{\alpha+1}/F_\alpha\cong Rx_{\alpha+1}, but since R is an integral domain it’s trivial that Rx_{\alpha+1}\cong R and so F_{\alpha+1}/F_\alpha\cong R. From this point we set M_\alpha to be equal to M\cap F_\alpha. We make the similar observations that \alpha<\beta implies M_\alpha\subseteq M_\beta, \displaystyle M=\bigcup_{\alpha\in\mathcal{A}_\infty}M_\alpha, and M_\alpha=M_{\alpha+1}\cap F_\alpha.

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The basic idea from here is simple. Note that M_{\alpha+1}/M_\alpha=M_{\alpha+1}/(M_{\alpha+1}+F_\alpha) which, by the third isomorphism theorem is isomorphic to (M_{\alpha+1}+F_{\alpha})/F_{\alpha}, but note that since M_{\alpha+1}+F_{\alpha+1}\subseteq F_{\alpha+1} and since F_{\alpha+1}/F_\alpha\cong R we may conclude that M_{\alpha+1}/M_\alpha is isomorphic to a submodule of R. But, since R is a PID every submodule is of the form y_\alpha R. If y_\alpha\ne 0 then since R is an integral domain y_\alpha R\cong R and if y_\alpha=0 then y_\alpha R=0. From this we may conclude that M_{\alpha+1}/M_\alpha is a free R-module of rank 0 or 1. That said, we have an obvious short exact sequence

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0\to M_\alpha\to M_{\alpha+1}\to M_{\alpha+1}/M_\alpha\to 0

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but since M_{\alpha+1}/M_\alphais free, we know this sequence splits and so there exists some backmap g:M_{\alpha+1}\to M_\alpha and we know that M_{\alpha+1}=M_\alpha\oplus \text{im }g and since M_{\alpha+1}/M_\alpha is free we know that \text{im }g=\langle m_\alpha\rangle where m_\alpha=0 if \text{rank}(M_{\alpha+1}/M_\alpha)=0 and m_\alpha\ne0 if \text{rank}(M_{\alpha+1}/M_\alpha)=1. We then let \mathscr{B}=\left\{m_\alpha:m_\alpha\ne0\right\}.

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We claim that \mathscr{B} is a basis of M.  To see first that \mathscr{B} is linearly independent, suppose that \displaystyle \sum_{j}r_jm_j=0 with each m_j in \mathscr{B}. Suppose for a second that one of the coefficients in this sum was nonzero, we could then select (by the well-ordering) a largest one, say k so that \displaystyle 0=\sum_{j}r_j m_j=\sum_{t<k}r_t m_t+r_km_k\in M_k\oplus Rm_k=M_{k+1}, but this clearly implies that r_k=0 which is contradictory to assumption. Thus, there does not exist an index for which r_j\ne 0 and so r_j=0 for all j. The linear independence of \mathscr{B} follows.

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To see that \mathscr{B} spans M we note that if we can prove that the set B_k=\text{span}\{m_j:j<k\} spans M_k then we’ll be done since we’d clearly have then that

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\displaystyle \text{span}\mathscr{B}=\bigcup_{j}B_j=\bigcup_{j}M_j=M

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Since our set \mathcal{A}_\infty is well-ordered we can prove this by transfinite induction. Suppose then that B_j=M_j for all j<k and let m\in M_k. We split this into two cases, depending on whether k is a successor in \mathcal{A}_\infty or not. If k=j+1 for some j\in\mathcal{A}_\infty then we know that M_k=M_j\oplus Rm_{j+1}. We see then that \displaystyle m=x+rm_{j+1} for some x\in G_j, and by the induction hypothesis x is in B_j and so evidently m\in B_j+Rm_{j+1}=B_{j+1}=B_k. Suppose now that k is not the immediate successor of anything. Since m\in M\cap F_k we know that m can be written as \displaystyle \sum_{j<k}r_j x_k. Choose the largest index, t, such that r_t\ne0 then evidently m\in F_{t+1}. That said, by assumption we have that t+1\leqslant k and since t+1\ne k by assumption we have that t+1<k. But, note then that m\in M\cap F_{t+1}=M_{t+1}. But, by the induction M_{t+1}=B_{t+1}\subseteq B_k and so m\in B_k. Either way, we see that m\in B_k and so the induction is complete. It thus follows that \mathscr{B} is a basis. \blacksquare

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In fact, it’s clear from the above that:

Corollary: Let R be a PID and F a free module. Then, any submodules M\leqslant F is free and \text{rank}(M)\leqslant\text{rank}(F).

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[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.

[4]  Hungerford, Thomas W. Algebra. New York: Holt, Rinehart and Winston, 1974. Print.


November 21, 2011 - Posted by | Algebra, Module Theory, Ring Theory | , , , , , , , , , , ,

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