Modules

Right-splitting implies direct sum:

• Use that \(B \subset \ker d_2 + \operatorname{im}s\), writing \(b = (b - sd_2(b) ) + sd_2(b)\) and noting \begin{align*} d_2(b - sd_2(b)) = d_2(b) - d_2sd_2(b) = d_2(b) - d_2(b) = 0 .\end{align*}
• Show \(\ker d_2 \cap\operatorname{im}s=0\), writing \(b\) with \(d_2(b) = 0\) and \(b = s(c)\) for some \(c\) yields \begin{align*} 0 = d_2(b) = d_2s(c) = \operatorname{id}_C(c) = c .\end{align*}

• If \(M\) is finitely generated, write \(M = \left\langle{X}\right\rangle\) with \(X\coloneqq\left\{{x_1, \cdots, x_m}\right\}\) and \({\sharp}X<\infty\) a finite generating set.
• Since \(M\) is free, there is some maximal subset of generators \({\mathcal{B}}\coloneqq\left\{{x_1, \cdots, x_n}\right\} \subseteq X\) where \(n\leq m\) that is linearly independent.
• Consider \(N\leq M\) defined by \(\left\langle{{\mathcal{B}}}\right\rangle\); this is a basis for \(N\) and makes \(N\) free. The claim is now that \(M\cong N\), so that any maximal linearly independent subset of generators is all of \(X\).
• If \(N \not\cong M\), set \({\mathcal{B}}^c \coloneqq X\setminus{\mathcal{B}}= \left\{{x_{n+1}, \cdots, x_m}\right\}\) to be all generators for \(M\) that the basis \({\mathcal{B}}\) misses.
• Then \({\mathcal{B}}^c \cup\left\{{x_{j}}\right\}\) for any \(n+1\leq j \leq m\) has a linear dependence, and \(r_j x_j + \sum_{k=1}^n r_n x_n = 0\) for some \(r_j\neq 0\) implies \(r_j x_j = - \sum_{k=1}^n r_n x_n\).
• Let \(r\) be the product of all of the scalars obtained this way, so \(r = \prod_{k=n+1}^m r_j\), and consider the submodule \(rX \leq N \leq M\). We get \(rM \leq N \leq M\) since \(X\) is a generating set for \(M\), so it now suffices to show \(rM \cong M\).
• Just define a map \(\phi_r: M\twoheadrightarrow rM\) where \(m\mapsto rm\), and note \(\ker \phi_r =\left\{{ m\in M {~\mathrel{\Big\vert}~}rm = 0}\right\} = 0\) since \(M\) is torsionfree. So \(M = M/\ker \phi_r \cong rM\).

• Let \(M\) be free, so that the universal property gives us this diagram:

Link to Diagram

• To show \(M\) is projective, we need to produce a lift in the following diagram, where \(B, C\) are arbitrary:

Link to Diagram

• It suffices to produce a map \(\mathcal{B}\to B\), since the universal property then provides \(M\to B\). Here’s the schematic:

Link to Diagram

• Here we write \({\mathcal{B}}\coloneqq\left\{{e_i}\right\}\), included into \(M\), and mapped by \(f\) to \(C\). Then use surjectivity to choose preimages in \(B\) under \(g\) arbitrarily, and this defines a morphism \({\mathcal{B}}\to B\).

\begin{align*}
M \cong F \bigoplus_{i=1}^m R/(r_i) \quad\text{where } r_1 \divides r_2 \divides \cdots
\end{align*}

\(\implies\):

Suppose \(I\) is free as an \(R{\hbox{-}}\)module, and let \(B = \left\{{\mathbf{m}_j}\right\}_{j\in J} \subseteq I\) be a basis so we can write \(M = \left\langle{B}\right\rangle\). Suppose that \({\left\lvert {B} \right\rvert} \geq 2\), so we can pick at least 2 basis elements \(\mathbf{m}_1\neq \mathbf{m}_2\), and consider \begin{align*} \mathbf{c} = \mathbf{m}_1 \mathbf{m}_2 - \mathbf{m}_2 \mathbf{m}_1, \end{align*}

which is also an element of \(M\) . Since \(R\) is an integral domain, \(R\) is commutative, and so \begin{align*} \mathbf{c} = \mathbf{m}_1 \mathbf{m}_2 - \mathbf{m}_2 \mathbf{m}_1 = \mathbf{m}_1 \mathbf{m}_2 - \mathbf{m}_1 \mathbf{m}_2 = \mathbf{0}_M \end{align*}

However, this exhibits a linear dependence between \(\mathbf{m}_1\) and \(\mathbf{m}_2\), namely that there exist \(\alpha_1, \alpha_2 \neq 0_R\) such that \(\alpha_1 \mathbf{m}_1 + \alpha_2 \mathbf{m}_2 = \mathbf{0}_M\); this follows because \(M \subset R\) means that we can take \(\alpha_1 = -m_2, \alpha_2 = m_1\). This contradicts the assumption that \(B\) was a basis, so we must have \({\left\lvert {B} \right\rvert} = 1\) and so \(B = \left\{{\mathbf{m}}\right\}\) for some \(\mathbf{m} \in I\). But then \(M = \left\langle{B}\right\rangle = \left\langle{\mathbf{m}}\right\rangle\) is generated by a single element, so \(M\) is principal.

\(\impliedby\): Suppose \(M{~\trianglelefteq~}R\) is principal, so \(M = \left\langle{\mathbf{m}}\right\rangle\) for some \(\mathbf{m} \neq \mathbf{0}_M \in M \subset R\).

Then \(x\in M \implies x = \alpha\mathbf{m}\) for some element \(\alpha\in R\) and we just need to show that \(\alpha\mathbf{m} = \mathbf{0}_M \implies \alpha = 0_R\) in order for \(\left\{{\mathbf{m}}\right\}\) to be a basis for \(M\), making \(M\) a free \(R{\hbox{-}}\)module. But since \(M \subset R\), we have \(\alpha, m \in R\) and \(\mathbf{0}_M = 0_R\), and since \(R\) is an integral domain, we have \(\alpha m = 0_R \implies \alpha = 0_R\) or \(m = 0_R\). Since \(m \neq 0_R\), this forces \(\alpha = 0_R\), which allows \(\left\{{m}\right\}\) to be a linearly independent set and thus a basis for \(M\) as an \(R{\hbox{-}}\)module.