Over a commutative ring \(R\), free modules are projective, and over an integral domain projective modules are torsion-free. Neither implication reverses in general: the ideal \((2,1+\sqrt{-5})\) of \(\ZZ[\sqrt{-5}]\) is a projective module that is not free, and \(\QQ\) is a torsion-free \(\ZZ\)-module that is not projective. A short exact sequence \(0\to A\to B\to C\to0\) of \(R\)-modules splits if \(C\) is projective. The tensor product \(M\otimes_RN\) represents \(R\)-bilinear maps out of \(M\times N\), and \(-\otimes_RN\) is right exact and in general not left exact: applying \(-\otimes_\ZZ\ZZ/2\ZZ\) to the injection \(\ZZ\xrightarrow{2}\ZZ\) gives the zero map \(\ZZ/2\ZZ\to\ZZ/2\ZZ\).
Modules over a ring (Tag D-NQZUY)
Let \(R\) be a ring. A left \(R\)-module is an abelian group \((M,+)\) together with a map \(R\times M\to M\), \((r,x)\mapsto rx\), such that for all \(r,s\in R\) and \(x,y\in M\):
- \(r(x+y) = rx + ry\);
- \((r+s)x = rx + sx\);
- \((rs)x= r(sx)\);
- \(1_Rx = x\).
Let \(R\) be a commutative ring. An \(R\)-algebra structure on an \(R\)-module \(M\) is an \(R\)-bilinear multiplication \(M\times M\to M\), equivalently an \(R\)-linear map \(m\colon M\otimes_R M\to M\); bilinearity means \[ r\, m(a\otimes b) = m(ra \otimes b) = m(a\otimes rb) \qquad \text{for all } r\in R \text{ and } a,b \in M. \]
One-step test for $R$-module homomorphisms (Tag PR-JBDUI)
Let \(R\) be a ring and \(M, N\) left \(R\)-modules. A map of sets \(\phi\colon M \to N\) is an \(R\)-module homomorphism if and only if \[ \phi(r x + y) = r\,\phi(x) + \phi(y) \quad \text{for all } r \in R \text{ and } x, y \in M. \]
Module morphisms (Tag D-JUYFQ)
Let \(R\) be a ring and \(M,N\) be $R$-modules. A map \(f\colon M\to N\) is a morphism of modules, or \(R\)-linear map, if \[ f(rm + m') = rf(m) + f(m')\qquad\text{for all } r\in R \text{ and } m,m'\in M. \]
One-step submodule test (Tag PR-TAPZF)
Let \(R\) be a ring and \(M\) an \(R\)-module. A subset \(N\subseteq M\) is an \(R\)-submodule if and only if \(N\) is nonempty and \(rx+y\in N\) for all \(r\in R\) and \(x, y \in N\).
A submodule contains \(0\) and is closed under addition and scalar multiplication, so it satisfies the condition. Conversely, choose \(x\in N\); taking \(r=-1\) and \(y=x\) gives \(0\in N\). Taking \(y=0\) gives \(rx\in N\) for all \(r\in R\) and \(x\in N\), in particular \(-x\in N\), and taking \(r=1\) gives \(x+y\in N\). Hence \(N\) is an additive subgroup of \(M\) closed under scalar multiplication.
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 \(m\in M\).
Cyclic modules (Tag D-HY7UU)
Let \(R\) be a ring. An $R$-module \(M\) is cyclic if there exists \(m\in M\) such that \[ M = Rm \coloneqq \theset{ rm \st r\in R }. \]
Cyclic module (Tag FD-S2W5E)
Let \(R\) be a ring and \(M\) an \(R\)-module. \(M\) is cyclic if there exists \(m\in M\) with \(M = Rm\coloneqq\theset{rm \st r\in R}\).
An \(R\)-module \(M\) is cyclic if and only if \(M \cong R/I\) for some left ideal \(I\) of \(R\).
If \(M=Rm\), the \(R\)-linear map \(\phi\colon R\to M\), \(r\mapsto rm\), is surjective, and its kernel \(I\coloneqq\theset{r\in R \st rm=0}\) is a left ideal, so \(M\cong R/I\) by the first isomorphism theorem. Conversely, \(R/I=R\cdot(1+I)\) is cyclic, and an isomorphism \(R/I\to M\) sends the generator \(1+I\) to a generator of \(M\).
Internal direct sum of two submodules (Tag PR-RIJSC)
Let \(R\) be a ring, \(M\) an \(R\)-module, and \(M_1, M_2 \leq M\) submodules. The \(R\)-linear map \(M_1\oplus M_2\to M\), \((x,y)\mapsto x+y\), is an isomorphism, so that \(M = M_1 \oplus M_2\) internally, if and only if
\(M_1 + M_2 = M\), and
\(M_1 \cap M_2 = 0\).
The image of the map is \(M_1+M_2\), so it is surjective if and only if \(M_1+M_2=M\). Its kernel is \(\theset{(x,-x)\suchthat x\in M_1\cap M_2}\), so it is injective if and only if \(M_1\cap M_2=0\).
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.
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 \(g\circ\tilde f=f\).
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 \(g\circ\phi=f\).
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 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: \(\tor(M) = \theset{0}\).
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 \(\Ann_R(m)\).
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.
Exact sequences (Tag D-BJYH3)
Let \(R\) be a ring. A sequence of \(R\)-modules and \(R\)-linear maps \[ M_0 \mapsvia{f_1} M_1 \mapsvia{f_2} \cdots \mapsvia{f_n} M_n \] is exact at \(M_i\), for \(0 < i < n\), if \(\im f_i = \ker f_{i+1}\); it is exact if it is exact at \(M_i\) for every \(0 < i < n\).
A sequence of \(R\)-module maps \[ 0 \mapsvia{d_1} A \mapsvia{d_2} B \mapsvia{d_3} C \mapsvia{d_4} 0 \] is exact if and only if \(\im d_i = \ker d_{i+1}\) for \(i = 1, 2, 3\), that is, if and only if \(d_2\) is injective, \(\im d_2 = \ker d_3\), and \(d_3\) is surjective.
Left and right splittings of a short exact sequence (Tag D-3JJJN)
Let \(R\) be a ring, and let \[ \xi\colon 0 \to A \mapsvia{d_1} B \mapsvia{d_2} C \to 0 \] be a short exact sequence of \(R\)-modules. A right splitting of \(\xi\) is an \(R\)-linear map \(s\colon C\to B\) with \(d_2 \circ s = \id_{C}\). A left splitting of \(\xi\) is an \(R\)-linear map \(t\colon B\to A\) with \(t \circ d_1 = \id_A\).
Splitting lemma for short exact sequences (Tag PR-OODAV)
Let \(R\) be a ring and let \(\xi\colon 0 \to A \mapsvia{d_1} B \mapsvia{d_2} C \to 0\) be a short exact sequence of \(R\)-modules. The following are equivalent:
\(\xi\) admits a right-splitting: an \(R\)-linear \(s\colon C\to B\) with \(d_2\circ s=\id_C\).
\(\xi\) admits a left-splitting: an \(R\)-linear \(t\colon B\to A\) with \(t\circ d_1=\id_A\).
There is an isomorphism \(\varphi\colon B\to A\oplus C\) with \(\varphi\circ d_1=\iota_A\) and \(\pi_C\circ\varphi=d_2\), where \(\iota_A\colon A\to A\oplus C\) is the inclusion and \(\pi_C\colon A\oplus C\to C\) is the projection.
A projective or injective module characterizes splitting for all sequences at once: \(C\) is projective if and only if every short exact sequence of \(R\)-modules \(0\to A'\to B'\to C\to 0\) splits, and \(A\) is injective if and only if every short exact sequence \(0\to A\to B'\to C'\to 0\) splits.
A single short exact sequence can split with \(C\) not projective. Over \(\ZZ\), the sequence \(0\to \ZZ/2\ZZ \to \ZZ/2\ZZ \oplus \ZZ/2\ZZ \to \ZZ/2\ZZ\to 0\) with the inclusion of the first summand and the projection onto the second splits, but \(\ZZ/2\ZZ\) is not a projective \(\ZZ\)-module, since the surjection \(\ZZ\to\ZZ/2\ZZ\) has no section.
Module structure on tensor products (Tag D-B5X33)
Let \(R\) be a commutative ring and let \(M\) and \(N\) be \(R\)-modules. The \(R\)-module structure on \(M\tensor_R N\) is the scalar multiplication determined on pure tensors by \[ r\cdot (m\tensor n) \coloneqq (rm)\tensor n, \qquad r\in R,\ m\in M,\ n\in N. \]
The scalar multiplication is well defined: for fixed \(r\in R\), the map \(M\times N\to M\tensor_R N\), \((m,n)\mapsto (rm)\tensor n\), is \(R\)-bilinear, so it induces an additive map \(M\tensor_R N\to M\tensor_R N\). Since \((rm)\tensor n = m\tensor (rn)\), the same structure is determined by \(r\cdot(m\tensor n) = m\tensor(rn)\).
$V^*\otimes W\cong\Hom(V,W)$ in finite dimensions (Tag PR-LPJLD)
Let \(k\) be a field and \(V, W\) finite-dimensional \(k\)-vector spaces. The \(k\)-linear map \[ \begin{aligned} \dualof{V} \tensor_k W &\to \Hom_k(V, W) \\ \tilde v \tensor w &\mapsto \big(x \mapsto \tilde v(x)\, w\big) \end{aligned} \] is an isomorphism.
$V^*\otimes W^*\cong(V\otimes W)^*$ when one factor is finite-dimensional (Tag PR-5PDNQ)
Let \(k\) be a field and let \(V, W\) be \(k\)-vector spaces, at least one of which is finite-dimensional. The \(k\)-linear map \[ \begin{aligned} \dualof{V} \tensor_k \dualof{W} &\to \dualof{(V \tensor_k W)} \\ v \tensor w &\mapsto \big(x \tensor y \mapsto v(x)\, w(y)\big) \end{aligned} \] is an isomorphism.
Tensoring preserves injections exactly when the factor is flat (Tag PR-KX7L7)
Let \(R\) be a commutative ring and \(X\) an \(R\)-module, and let \(F \coloneqq (\wait) \tensor_R X\) on \(R\)-modules.
\(F\) is right exact.
\(F\) preserves injections, that is, \(T \tensor \id_X\colon V \tensor_R X \to W \tensor_R X\) is injective for every injective \(R\)-linear map \(T\colon V \injects W\), if and only if \(X\) is flat, and in that case \(F\) is exact.
If \(R = k\) is a field, every \(k\)-module is flat, so \(T \tensor \id_X\colon V \tensor_k X \injects W \tensor_k X\) is injective for every injective \(k\)-linear map \(T\colon V \injects W\) and every \(k\)-vector space \(X\).
Over \(\ZZ\), tensoring does not preserve injections in general: \(\ZZ/2\ZZ\) is not a flat \(\ZZ\)-module. The map \(\ZZ \mapsvia{2} \ZZ\) is injective, and tensoring it with \(\ZZ/2\ZZ\) gives multiplication by \(2\) on \(\ZZ/2\ZZ\), which is the zero map and not injective.
Simple modules (Tag D-4KM4P)
Let \(R\) be a ring. An \(R\)-module \(M\) is simple if \(M\neq 0\) and every submodule \(M' \leq M\) is either \(0\) or \(M\). A commutative ring \(R\) is simple if it is simple as an \(R\)-module, that is, if \(R\neq 0\) and its only ideals are \(0\) and \(R\).
Semisimple modules (Tag D-CYAJI)
Let \(R\) be a ring. An \(R\)-module \(M\) is semisimple if there is a family \((M_j)_{j\in J}\) of simple submodules of \(M\) with \[ M = \bigoplus_{j\in J} M_j. \]
Decomposable and indecomposable modules (Tag D-4HEUB)
Let \(R\) be a ring and let \(M\) be an \(R\)-module. \(M\) is decomposable if there exist nonzero \(R\)-modules \(M_1, M_2\) with \(M \cong M_1 \oplus M_2\). \(M\) is indecomposable if \(M\neq 0\) and \(M\) is not decomposable: whenever \(M \cong M_1 \oplus M_2\), either \(M_1 = 0\) or \(M_2 = 0\).
Indecomposable module (Tag FD-2EUYX)
Let \(R\) be a ring. An \(R\)-module \(M\) is indecomposable if there are no nonzero submodules \(M_1,M_2\subseteq M\) with \(M=M_1\oplus M_2\).
Irreducible module (Tag FD-BVM37)
Let \(R\) be a ring. An \(R\)-module \(M\) is irreducible if its only submodules are \(0\) and \(M\); irreducible modules are also called simple modules.