Definitions
A nonzero module is simple if its only submodules are \(0\) and itself, indecomposable if it is not a direct sum of two nonzero submodules, and cyclic if it is generated by one element; a module is Noetherian if every ascending chain of submodules stabilizes. Every simple module is indecomposable and cyclic. Over \(\ZZ\), the module \(\ZZ\) is indecomposable and not simple, and \(\ZZ/4\) is cyclic and indecomposable and not simple.
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 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.
One-step submodule test (Tag FF-UYMWY)
Let \(R\) be a ring, let \(M\) be a left \(R\)-module, and let \(N\subseteq M\) be a nonempty subset. If \(rm+n\in N\) for all \(r\in R\) and \(m,n\in N\), then \(N\) is a submodule of \(M\).
Choose \(n\in N\); then \(0=(-1)n+n\in N\). For \(r\in R\) and \(m\in N\), \(rm=rm+0\in N\); in particular \(-m=(-1)m\in N\). For \(m,n\in N\), \(m+n=1\cdot m+n\in N\). So \(N\) is an additive subgroup closed under scalar multiplication.
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 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 \[ \phi(r x + y) = r\,\phi(x) + \phi(y) \quad \text{for all } r \in R \text{ and } x, y \in M. \]
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 \(R\).
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.
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 \(M=M_1\oplus M_2\).
Noetherian module (Tag FD-GNRTS)
Let \(R\) be a ring. An \(R\)-module \(M\) is Noetherian if there is no infinite strictly increasing sequence \(M_1 \subsetneq M_2 \subsetneq \cdots\) of submodules of \(M\).
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 \(M\).
The structure theorem
Let \(M\) be a finitely generated module over a PID \(R\) with torsion submodule \(M_t\). Then \(M\cong R^n\oplus M_t\), and \(M_t\) is the direct sum of its \(\gens p\)-primary components \(M_{(p)}=\ts{m\in M\st p^km=0\text{ for some }k}\) over the primes \(p\) of \(R\) up to associates, each a finite direct sum of modules \(R/\gens{p^e}\). The invariant factor and elementary divisor forms are two groupings of this decomposition.
Isomorphism theorems for modules (Tag PR-DJZLY)
Let \(R\) be a ring and \(M\) an \(R\)-module.
For an \(R\)-module homomorphism \(\phi\colon M \to M'\), there is an isomorphism \(M/\ker\phi \cong \im\phi\).
For submodules \(A, B \subseteq M\), \[ \frac{A+B}{B} \cong \frac{A}{A \intersect B}. \]
For submodules \(A \subseteq B \subseteq M\), \[ \frac{M/A}{B/A} \cong \frac{M}{B}. \]
For a submodule \(N \subseteq M\), there is an inclusion-preserving bijection \[ \begin{aligned} \correspond{\text{Submodules of } M \\ \text{containing } N} &\mapstofrom \correspond{\text{Submodules of } M/N} \\ A &\mapstofrom A/N. \end{aligned} \]
The bijection in (d) preserves sums and intersections: for submodules \(A, A' \subseteq M\) containing \(N\), \((A+A')/N = A/N + A'/N\) and \((A \intersect A')/N = A/N \intersect A'/N\).
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 \(M_1\cap M_2=0\).
Structure theorem for finitely generated modules over a PID (Tag PR-UVUS6)
Let \(R\) be a principal ideal domain and \(M\) a finitely generated \(R\)-module. Then there are an integer \(r\geq0\), nonzero nonunits \(r_1 \divides r_2 \divides \cdots\divides r_m\) of \(R\), and primes \(p_1,\ldots,p_s\) of \(R\), not necessarily distinct, with exponents \(e_i\geq1\), such that \[ M \cong R^r\oplus \bigoplus_{i=1}^m R/(r_i) \qquad\text{and}\qquad M \cong R^r\oplus \bigoplus_{i=1}^s R/(p_i^{e_i}). \] The first is the invariant factor decomposition and the second the elementary divisor decomposition. The rank \(r\) is unique, the \(r_i\) are unique up to units, and the \(p_i^{e_i}\) are unique up to units and reordering.
$\gens{p}$-primary module (Tag FD-LHLDU)
Let \(R\) be a principal ideal domain, \(p\in R\) a prime, and \(M\) an \(R\)-module. \(M\) is \(\gens{p}\)-primary if every \(m\in M\) is annihilated by some power of \(p\): for each \(m\in M\) there is \(n\geq 1\) with \(p^n m = 0\).
For any \(R\)-module \(M\), the \(p\)-primary component \(M[p^\infty] \coloneqq \theset{ m \in M \st p^n m = 0 \text{ for some } n\geq 1}\) is a submodule, and \(M\) is \(\gens{p}\)-primary if and only if \(M[p^\infty]=M\).
(Free part and torsion part)
Let \(M\) be a finitely generated module over a PID \(R\) with torsion submodule \(M_t\).
The quotient \(M/M_t\) is torsion-free: if \(r\neq0\) and \(r(m + M_t) = M_t\), then \(rm\in M_t\), so \(r'rm = 0\) for some \(r'\neq0\); since \(R\) is a domain, \(r'r\neq0\), so \(m\in M_t\) and \(m + M_t\) is zero.
A finitely generated torsion-free module over a PID is free, so \(M/M_t\cong R^n\) for some \(n\). The exact sequence \(0\to M_t\to M \to M/M_t\to 0\) splits because \(M/M_t\) is free, hence projective, and \(M \cong R^n\oplus M_t\).
Free ideals
An ideal of an integral domain is free if and only if it is principal (Tag PR-ASW5L)
Let \(R\) be an integral domain and \(I \subseteq R\) an ideal. Then \(I\) is a free \(R\)-module if and only if \(I\) is a principal ideal.
Over a ring with zero divisors, a principal ideal need not be free. In \(\ZZ/6\ZZ\), the principal ideal \(\gens{2}\) is not a free \(\ZZ/6\ZZ\)-module, since \(3 \cdot 2 = 0\) with \(3 \neq 0\), whereas a nonzero element of a basis of a free module has zero annihilator.
Let \(R\) be an integral domain and \(I\neq0\) an ideal.
\(\implies\): Suppose \(I\) is free with basis \(B\), and suppose \(B\) contains distinct elements \(m_1,m_2\). Since \(R\) is commutative, \[ (-m_2)\, m_1 + m_1\, m_2 = 0, \] which is a relation with nonzero coefficients \(-m_2\) and \(m_1\). This contradicts \(B\) being a basis, so \(B = \ts m\) has one element and \(I = \gens{m}\) is principal.
\(\impliedby\): Suppose \(I = \gens{m}\) with \(m \neq 0\). Every \(x\in I\) has the form \(\alpha m\), and \(\alpha m = 0\) implies \(\alpha = 0\) because \(R\) is a domain. So \(\ts{m}\) is a basis of \(I\).
Tensor products and duals
For finite-dimensional vector spaces \(V\) and \(W\) over a field \(k\), \(\dualof{V}\tensor_k W\cong\Hom_k(V,W)\) and \(\dualof{V}\tensor_k \dualof{W}\dualof{\cong(V\tensor_k W)}\), and the trace pairing \((f,g)\mapsto\tr(g\circ f)\) identifies \(\Hom_k(V,W)\) with \(\dualof{\Hom_k(W,V)}\). For a module \(M\) over a commutative ring \(R\), the functor \(M\tensor_R-\) preserves injective maps if and only if \(M\) is flat.
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 \(r\cdot(m\tensor n) = m\tensor(rn)\).
$\dualof{V}\tensor_k W\cong\Hom_k(V,W)$ for finite-dimensional $V$ and $W$ (Tag PR-O5YUI)
Let \(k\) be a field and let \(V\) and \(W\) be finite-dimensional \(k\)-vector spaces. The \(k\)-linear map \[ \begin{aligned} \Phi\colon \dualof{V}\tensor_k W &\to \Hom_k(V,W),\\ \tilde v\tensor w &\mapsto \big(v\mapsto \tilde v(v)\,w\big), \end{aligned} \] is an isomorphism.
The map \(\dualof{V}\times W\to\Hom_k(V,W)\), \((\tilde v,w)\mapsto \tilde v(\wait)\,w\), is \(k\)-bilinear, so \(\Phi\) is well defined by the universal property of the tensor product. Let \(e_1,\ldots,e_n\) be a basis of \(V\) with dual basis \(e_1^*,\ldots,e_n^*\) of \(\dualof{V}\), and let \(f_1,\ldots,f_m\) be a basis of \(W\). The elements \(e_i^*\tensor f_j\) form a basis of \(\dualof{V}\tensor_k W\). The map \(\Phi(e_i^*\tensor f_j)\) sends \(e_i\) to \(f_j\) and \(e_l\) to \(0\) for \(l\neq i\); in these bases its matrix is the matrix unit \(E_{ji}\). The matrix units \(E_{ji}\) form a basis of \(\Hom_k(V,W)\), so \(\Phi\) sends a basis to a basis and is an isomorphism.
The trace pairing identifies $\Hom(V,W)$ with $\dualof{\Hom(W,V)}$ (Tag PR-TGFTL)
Let \(k\) be a field and let \(V\) and \(W\) be finite-dimensional \(k\)-vector spaces. The map \[ \begin{aligned} \Hom_k(V, W) &\to \dualof{\Hom_k(W, V)},\\ T &\mapsto \big(S\mapsto\Tr(T \circ S)\big), \end{aligned} \] is an isomorphism of \(k\)-vector spaces, where \(T\circ S\colon W\to W\).
The map is \(k\)-linear because the trace is, and both sides have dimension \(\dim_k V\cdot\dim_k W\). Choose bases, so that \(T\) is an \(m\times n\) matrix and \(S\) an \(n\times m\) matrix; then \(\Tr(TS)=\sum_{a,b}T_{ab}S_{ba}\). If \(\Tr(TS)=0\) for all \(S\), taking \(S\) to be the matrix unit \(E_{ba}\) gives \(T_{ab}=0\) for all \(a,b\), so \(T=0\). Thus the map is injective, hence an isomorphism.
$V^*\otimes W\cong\Hom(V,W)$ 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.
The trace pairing identifies $\Hom(V,W)$ with $\dualof{\Hom(W,V)}$ (Tag PR-GXII2)
Let \(k\) be a field and \(V, W\) finite-dimensional \(k\)-vector spaces. The map \[ \begin{aligned} \Hom_k(V, W) &\to \dualof{\Hom_k(W, V)} \\ T &\mapsto \big(S \mapsto \Tr(T \circ S)\big) \end{aligned} \] is an isomorphism of \(k\)-vector spaces, where \(T \circ S\colon W \to W\).
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.