Free modules, rank, and torsion
An \(R\)-module is free if it has a basis, equivalently if it is isomorphic to \(\bigoplus_{i\in I}R\) for some set \(I\); over a commutative ring \(R\neq0\) the cardinality of \(I\) is its rank. Over an integral domain, a free module is torsion-free; if \(r\neq0\) and \(x=\sum_i c_ie_i\neq0\) in a basis \((e_i)\), then \(rx=\sum_i rc_ie_i\neq0\).
Free module (Tag D-LIEMF)
Let \(R\) be a ring. An $R$-module \(M\) is free if it has a basis: a family \((\beta_i)_{i\in I}\) in \(M\) such that every \(m\in M\) can be written as \(m=\sum_{i\in I} r_i\beta_i\) with \(r_i\in R\), all but finitely many zero, and \(\sum_{i\in I} r_i\beta_i = 0\) implies \(r_i = 0\) for all \(i\).
Let \(R\) be a ring and \(M\) an \(R\)-module. The following are equivalent:
\(M\) is free, with basis \((\beta_i)_{i\in I}\);
there are a set \(\mathcal{B}\) and a map of sets \(\iota\colon\mathcal{B}\to M\) such that for every \(R\)-module \(N\) and every map of sets \(f\colon\mathcal{B}\to N\) there is a unique \(R\)-linear map \(\tilde f\colon M\to N\) with \(\tilde f\circ\iota = f\):
- there is a family \((\beta_i)_{i\in I}\) in \(M\) such that \(M = \bigoplus_{i\in I} R\beta_i\) is the internal direct sum of the cyclic submodules \(R\beta_i\), and \(r\mapsto r\beta_i\) is injective for each \(i\), so that \(R\beta_i\cong R\).
Given a basis \((\beta_i)_{i\in I}\), take \(\mathcal B=I\) and \(\iota(i)=\beta_i\); the unique extension of \(f\) is \(\tilde f(\sum_i r_i\beta_i)=\sum_i r_if(i)\), which is well defined because the coefficients \(r_i\) are unique. Conversely, if \((\mathcal B,\iota)\) has the universal property, compare \(M\) with the module \(R^{(\mathcal B)}\) of finitely supported functions \(\mathcal B\to R\) and its standard basis \((e_b)_{b\in\mathcal B}\): the linear maps \(M\to R^{(\mathcal B)}\) extending \(b\mapsto e_b\) and \(R^{(\mathcal B)}\to M\) extending \(e_b\mapsto\iota(b)\) compose in both orders to maps that restrict to the identity on the generators, so by uniqueness they are inverse isomorphisms, and \((\iota(b))_{b\in\mathcal B}\) is a basis of \(M\). Finally, the spanning condition says that \(M=\sum_i R\beta_i\), and linear independence says exactly that this sum is direct and each \(r\mapsto r\beta_i\) is injective.
Let \(R\) be a ring, \(M\) a left \(R\)-module, \(S\subseteq M\) a subset, and \(R^{(S)} \coloneqq \bigoplus_{s\in S} R\) with standard basis \((e_s)_{s\in S}\). There is a unique \(R\)-linear map \(\varphi\colon R^{(S)}\to M\) with \(\varphi(e_s) = s\) for all \(s\in S\).
\(\varphi\) is surjective if and only if \(S\) is a generating set of \(M\); in that case \(M\cong R^{(S)}/\ker\varphi\).
\(\varphi\) is an isomorphism if and only if \(S\) is a basis of \(M\). In particular, \(M\) is free if and only if \(M\cong R^{(T)}\) for some set \(T\) [@DF04].
Rank of a free module (Tag FD-CVEAI)
Let \(R\) be a nonzero commutative ring and \(M\) a free \(R\)-module. The rank of \(M\) is the cardinality of a basis of \(M\).
Let \(R\) be a nonzero commutative ring and \(M\) a free \(R\)-module. Any two bases of \(M\) have the same cardinality.
Since \(R\neq0\), \(R\) has a maximal ideal \(\mfm\), and \(k\coloneqq R/\mfm\) is a field. If \((e_i)_{i\in I}\) is a basis of \(M\), then \(M\cong R^{(I)}\), so \(M/\mfm M\cong (R/\mfm)^{(I)}\) and the images of the \(e_i\) form a basis of the \(k\)-vector space \(M/\mfm M\). Hence \(\abs{I}=\dim_k M/\mfm M\), which does not depend on the basis.
If \(R\) is an integral domain, the rank of \(M\) is the supremum of the cardinalities of \(R\)-linearly independent families in \(M\). With \(K\) the field of fractions of \(R\) and \((e_i)_{i\in I}\) a basis, \(M\cong R^{(I)}\subseteq K^{(I)}\); a family that is \(R\)-linearly independent in \(R^{(I)}\) is \(K\)-linearly independent in \(K^{(I)}\), after clearing denominators in a \(K\)-linear relation, so it has at most \(\abs{I}\) elements, and the basis itself is an independent family of cardinality \(\abs{I}\).
Every free $R$-module admits a basis (Tag PR-DLPTR)
Let \(R\) be a ring. Every free \(R\)-module admits a basis, that is, an \(R\)-linearly independent spanning set.
Torsion elements, torsion modules, and torsion-free modules (Tag D-ZJJ7G)
Let \(R\) be a commutative ring and \(M\) an \(R\)-module.
An element \(m\in M\) is a torsion element if there exists a nonzero \(r\in R\) with \(rm=0\).
Write \(M_t\coloneqq\theset{m\in M \st rm=0 \text{ for some nonzero } r\in R}\) for the set of torsion elements of \(M\).
\(M\) is torsion-free if no nonzero element of \(M\) is a torsion element, that is, \(M_t\subseteq\theset{0}\).
\(M\) is a torsion module if \(M_t=M\).
Torsion elements of a module (Tag FF-CY5EA)
Let \(R\) be a commutative ring and let \(M\) be an \(R\)-module. An element \(m\in M\) is a torsion element if and only if its annihilator \(\Ann_R(m)\coloneqq\theset{r\in R\suchthat rm=0}\) is a nonzero ideal of \(R\).
By definition, \(m\) is a torsion element if and only if there is a nonzero \(r\in R\) with \(rm=0\), that is, a nonzero element of \(\Ann_R(m)\).
Torsion submodule (Tag FD-BPUNZ)
Let \(R\) be an integral domain and \(M\) an \(R\)-module. The torsion submodule of \(M\) is \[ \tor(M) \coloneqq \theset{m \in M \st rm = 0 \text{ for some nonzero } r \in R}. \]
\(\tor(M)\) is a submodule: if \(rm=0\) and \(sn=0\) with \(r,s\neq0\), then \(rs\neq0\) because \(R\) is a domain, \(rs(m+n)=0\), and \(r(am)=a(rm)=0\) for every \(a\in R\).
Torsion-free module (Tag FD-U6KUJ)
Let \(R\) be an integral domain and \(M\) an \(R\)-module. \(M\) is torsion-free if its torsion submodule is zero: \(\tor(M) = \theset{0}\).
Annihilator of a module (Tag FD-SK4ON)
Let \(R\) be a ring and \(M\) an \(R\)-module. The annihilator of \(M\) is \[ \Ann_R(M) \coloneqq \theset{r\in R\st rm=0 \text{ for every } m\in M}. \]
\(\Ann_R(M)\) is a two-sided ideal of \(R\): it is an additive subgroup, and if \(r\in\Ann_R(M)\) and \(s\in R\), then \((sr)m=s(rm)=0\) and \((rs)m=r(sm)=0\) for every \(m\in M\).
Free modules over an integral domain are torsionfree (Tag PR-4K4XZ)
Let \(R\) be an integral domain. Every finitely generated free \(R\)-module \(M\) is torsionfree.
Projective modules
An \(R\)-module is projective if and only if it is a direct summand of a free module. Every free module is projective, and every short exact sequence \(0\to A\to B\to P\to0\) with \(P\) projective splits. Over a PID, a finitely generated torsion-free module is free, so finitely generated projective and free modules coincide.
Projective modules (Tag D-RHJMK)
Let \(R\) be a ring. An $R$-module \(P\) is projective if for every surjective \(R\)-linear map \(g\colon N \to M\) and every \(R\)-linear map \(f\colon P\to M\) there exists an \(R\)-linear map \(\tilde f\colon P\to N\) with \(g\circ\tilde f = f\):
Let \(R\) be a ring and \(P\) an \(R\)-module. The following are equivalent:
\(P\) is projective;
\(P\) is a direct summand of a free module: there are a free \(R\)-module \(F\) and a submodule \(T\leq F\) with \(F = P \oplus T\), up to isomorphism;
every short exact sequence \(0\to A\to B\xrightarrow{p} P\to 0\) of \(R\)-modules splits: there is an \(R\)-linear map \(s\colon P\to B\) with \(p\circ s = \id_P\);
the functor \(\Hom_R(P, \wait)\) from \(R\)-modules to abelian groups is exact.
(1)\(\Rightarrow\)(3): lift \(\id_P\) along the surjection \(p\). (3)\(\Rightarrow\)(2): choose a free module \(F\) with a surjection \(p\colon F\to P\), for example on a generating set of \(P\); a section \(s\) of \(p\) gives \(F=s(P)\oplus\ker p\) with \(s(P)\cong P\). (2)\(\Rightarrow\)(1): a free module \(F\) with basis \((e_i)\) is projective, since \(\tilde f(e_i)\) may be chosen as any preimage under \(g\) of \(f(e_i)\); if \(F=P\oplus T\) with inclusion \(\iota\colon P\to F\) and projection \(\pi\colon F\to P\), a lift \(h\) of \(f\circ\pi\) gives the lift \(\tilde f=h\circ\iota\) of \(f\). (1)\(\Leftrightarrow\)(4): \(\Hom_R(P,\wait)\) is always left exact, and it preserves surjections exactly when every \(f\colon P\to M\) lifts along every surjection \(g\colon N\to M\).
In (3), the section \(s\) gives \(B = s(P) \oplus \ker p\), and \(s(P)\cong P\).
Projective module (Tag FD-6XJ7D)
Let \(R\) be a ring. An \(R\)-module \(P\) is projective if for every surjective \(R\)-linear map \(g\colon M\to N\) and every \(R\)-linear map \(f\colon P\to N\) there exists an \(R\)-linear map \(\phi\colon P\to M\) with \(g\circ\phi=f\):
An \(R\)-module \(P\) is projective if and only if there are an \(R\)-module \(Q\) and a free \(R\)-module \(F\) with \(F\cong P\oplus Q\).
Suppose \(P\) is projective. Let \(F\) be the free \(R\)-module on the set \(P\) and \(\pi\colon F\to P\) the surjection sending each basis element to the corresponding element of \(P\). Applying the definition to \(g=\pi\) and \(f=\id_P\) gives \(s\colon P\to F\) with \(\pi\circ s=\id_P\), and then \(F\cong P\oplus\ker\pi\) via \(x\mapsto(\pi(x),x-s\pi(x))\).
Conversely, a free module \(F\) with basis \((e_i)_{i\in I}\) is projective: given \(g\colon M\to N\) surjective and \(f\colon F\to N\), choose \(m_i\in M\) with \(g(m_i)=f(e_i)\) and let \(\phi\) be the \(R\)-linear map with \(\phi(e_i)=m_i\). If \(F=P\oplus Q\) is free and \(f\colon P\to N\) is given, lift \(f\oplus 0\colon F\to N\) to \(\psi\colon F\to M\); the restriction \(\phi=\psi|_P\) satisfies \(g\circ\phi=f\).
Free modules are projective (Tag PR-RPL4Q)
Let \(R\) be a ring. Every free \(R\)-module is projective.
Let \(F\) be free with basis \((e_i)_{i\in I}\), let \(g\colon N\to M\) be a surjective \(R\)-linear map, and let \(f\colon F\to M\) be \(R\)-linear. For each \(i\in I\) choose \(n_i\in N\) with \(g(n_i)=f(e_i)\). There is a unique \(R\)-linear map \(\tilde f\colon F\to N\) with \(\tilde f(e_i)=n_i\) for all \(i\), and \(g\circ\tilde f\) agrees with \(f\) on the basis, so \(g\circ\tilde f=f\).
(Free, projective, flat, torsion-free)
For modules over an integral domain, \[ \text{free} \implies \text{projective} \implies \text{flat} \implies \text{torsion-free}, \] no implication reverses in general, and for finitely generated modules over a PID all four conditions coincide. Examples separating them are on Classify this module.