Abstract Nonsense

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 $M_{\alpha+1}/M_\alpha$is 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, 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 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 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. 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. 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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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.

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