Definitions and Basics

Four properties:

  • \(r(x+y) = rx + ry\)
  • \((r+s)x = rs + sx\)
  • \((rs)x= r(s(x))\)
  • \(1_Rx = x\)

Note that \(M\) is additionally an \(R{\hbox{-}}\)algebra if the multiplication map is \(R{\hbox{-}}\)bilinear and so given by \(m: M^{\otimes_R 2}\to M\) satisfying \begin{align*} r. m(a\otimes b) = m(r.a \otimes b) = m(a\otimes r.b) && \forall r\in R, a,b \in M .\end{align*}

\(N\subseteq M\) is an \(R{\hbox{-}}\)submodule iff \(N\) is nonempty and for every \(r\in R\) and \(x, y \in N\), we have \(rx+y\in N\).

A map \(f: M\to N\) is a morphism of modules iff \(f(rm + n) = rf(m) + f(n)\).

A map \(\phi: M\to N\) is a morphism in \({}_{R}{\mathsf{Mod}}\) iff \begin{align*} \phi(r.x + y) = r.\phi(x) + \phi(y) \in N && \forall r\in R, x,y\in M .\end{align*}

Quotients of modules are easier to reason about additively, writing \(M/N = \left\{{x + N}\right\}\) as cosets. Then \((x + N) + (y + N) = (x+y) + N\) and \((x+N)(y+N) = (xy) + N\).

A module is simple iff it has no nontrivial proper submodules.

A module \(M\) is decomposable iff it admits a direct sum decomposition \(M \cong M_1 \oplus M_2\) with \(M_1, M_2 \neq 0\). An indecomposable module is defined in the obvious way.

A module \(M\) is cyclic if there exists a single generator \(m\in M\) such that \(M = mR \coloneqq\left\langle{ m }\right\rangle\).

Structure Theorems

\begin{align*} M / \ker \phi &\cong \operatorname{im}\phi \\ {A+B \over B} &\cong {A\over A \cap B} \\ {M/A \over B/A} &\cong {M\over B} \\ \left\{{\substack{ \text{Submodules of } M \\ \text{containing }N }}\right\} &\rightleftharpoons \left\{{\substack{ \text{Submodules of } M/N }}\right\} \\ A &\rightleftharpoons A/N .\end{align*}

Note that the lattice correspondence commutes with sums and intersections of submodules.

If \(M_1, M_2 \leq M\) are submodules, then \(M = M_1 \oplus M_2\) if the following conditions hold:

  • \(M_1 + M_2 = M\)
  • \(M_1 \cap M_2 = 0\)

Exact Sequences

A sequence of \(R{\hbox{-}}\)module morphisms \begin{align*} 0 \xrightarrow{d_1} A \xrightarrow{d_2} B \xrightarrow{d_3} C \to 0 \end{align*} is exact iff \(\operatorname{im}d_i = \ker d_{i+1}\).

Note that \(C\cong B/d_1(A)\) always, but \(B\) is not a direct sum of the outer terms unless the sequence splits.

A short exact sequence \begin{align*} \xi: 0 \to A \xrightarrow{d_1} B \xrightarrow{d_2} C \to 0 \end{align*} has a right-splitting iff there exists a map \(s: C\to B\) such that \(d_2 \circ s = \operatorname{id}_{C}\). \(\xi\) has a left-splitting iff there exists a map \(t:B\to A\) such that \(t \circ d_1 = \operatorname{id}_A\).

Let \(\xi: 0 \to A \xrightarrow{d_1} B \xrightarrow{d_2} C \to 0\) be a SES, then TFAE

  • \(\xi\) admits a right-splitting \(s: C\to B\).
  • \(C\) is projective.
  • \(\xi\) admits a left-splitting \(t: B\to A\).
  • \(A\) is injective.
  • \(\xi\) is isomorphic to a SES of the form \(0\to A \to A \oplus C \to C \to 0\).

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*}

Free and Projective Modules

A free module \(M\) is a module satisfying any of the following conditions:

  • A universal property: There is a set \(\mathcal{B}\) and a set map \(M \xrightarrow{\iota} \mathcal{B}\) such that every set map \(\mathcal{B} \xrightarrow{N}\) lifts:

Link to Diagram

  • Existence of a basis:

    There is linearly independent (so \(\sum r_i \beta_i = 0 \implies r_i = 0\)) spanning set (so \(m\in M \implies m = \sum r_i \beta_i\) ) of the form \(\mathcal{B} \coloneqq\left\{{ \beta_i }\right\}_{i\in I}\),

  • Direct sum decomposition:

    \(M\) decomposes as \(M \cong \bigoplus_{i\in I} \beta_i R\), a sum of cyclic submodules.

\({\mathbf{Z}}/6\) is a \({\mathbf{Z}}{\hbox{-}}\)module that is not free, since the element \([3]\) is a torsion element, where \(2[3] = [6] = [0]\). This uses the fact that free modules over a PID are torsionfree.

If a module \(M\) is free, the free rank of \(M\) is the cardinality of any basis.

Every free \(R{\hbox{-}}\)module admits a basis (spanning \(R{\hbox{-}}\)linearly independent set).

An element \(m\in M\) is a torsion element if there exists a nonzero \(r\in R\) such that \(rm = 0_M\). A module \(M\) is torsion-free if and only if for every \(x\in M\), \(mx = 0_M \implies m=0_M\), i.e. \(M\) has no nonzero torsion elements. Equivalently, defining \(M_t \coloneqq\left\{{ m\in M {~\mathrel{\Big\vert}~}\exists r\in R, rm = 0_M }\right\}\) as the set of all torsion elements, \(M\) is torsion free iff \(M_t = 0\). If \(M_t = M\), we say \(M\) is a torsion module.

For \(R\) an integral domain, any finitely generated free \(R{\hbox{-}}\)module \(M\) is torsionfree.

  • 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 \(M = M/\ker \phi_r \cong rM\).

\({\mathbf{Q}}\in {}_{{\mathbf{Z}}}{\mathsf{Mod}}\) is torsionfree, but not free as a \({\mathbf{Z}}{\hbox{-}}\)module. This follows because any two elements \(a/b, p/q\) are in a single ideal, since taking \(d\coloneqq\gcd(b, q)\) we have \(1/a = 1/d + \cdots 1/d\) and similarly \(p/q = 1/a + \cdots + 1/a\), so these are in \(\left\langle{ 1/d }\right\rangle\). So any basis has size one, which would mean \({\mathbf{Q}}= \left\{{ \pm 1/d, \pm 2/d, \cdots }\right\}\) which in particular doesn’t include the average of the first two terms.

A module \(P\) is projective iff it satisfies any of the following conditions:

  • A universal property: for every surjective \(N \xrightarrow{g} M\) and \(P \xrightarrow{f} M\), the following lift exists:

Link to Diagram

  • Direct summand:

    \(P\) is a direct summand of a free module \(F\), so \(F = P \oplus T\) for some module \(T\leq F\).

  • Splitting:

    For every SES \(0\to A\to B\to P\to 0\), there is a right section \(P\to B\) such that \(P\to B\to P = \operatorname{id}_P\).

    Note that this implies \(B\cong \operatorname{im}(P\to B) \oplus \ker(B\to P)\).

  • Exactness:

    The (always left-exact) covariant hom functor \(\mathop{\mathrm{Hom}}(P, {-})\) is right-exact.

There is a nice way to remember the right diagrams for injective and projective modules. The slogan is that morphisms out of a projective module can be pulled back through epimorphisms/surjections, and morphisms into an injective module can be pushed forward through monomorphisms/injections.

Link to Diagram

Any free \(M\in {}_{R}{\mathsf{Mod}}\) is projective.

  • 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 \({\mathcal{B}}\to B\).

Let \(R_1, R_2\) be two nontrivial rings and set \(R \coloneqq R_1 \oplus R_2\). Then \(R_1, R_2\) are projective \(R{\hbox{-}}\)modules by construction, but each factor contains \(R{\hbox{-}}\)torsion: setting \(e \coloneqq(0, 1) \in R\) we have \(e \curvearrowright R_1 = 0_{R_1}\). Since free implies torsionfree, \(R_1\) can not be a free \(R{\hbox{-}}\)module.

Classification of Modules over a PID

Let \(M\) be a finitely generated modules over a PID \(R\). Then there is an invariant factor decomposition

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

and similarly an elementary divisor decomposition: \begin{align*} M \cong F \bigoplus_{i=1}^n R/ \left\langle{p_i^{e_i}}\right\rangle \end{align*} where \(F\) is free of finite rank and the \(p_i\) are not necessarily distinct primes in \(R\).

If \(I {~\trianglelefteq~}R\) is an ideal of \(R\), then \(I\) is a free \(R{\hbox{-}}\)module iff \(I\) is a principal ideal.


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.

This says every module \(M\) decomposes as \(M \cong F_M \oplus M_t\) where \(F_M\) is free (and thus torsionfree) and \(M_t\) is torsion, and moreover \(F_M \cong M/M_t\).

That \(M/M_t\) is torsionfree: suppose \(r(m+ M_t) = M_t\), so \(rm\in M_t\) is torsion. Then \(r'(rm)=0\) for some \(r'\), making \(m\) torsion, and \(m+ M_t = M_t\) is the zero coset.

That \(F_M \cong M/M_t\): take the SES \(0\to M_t\to M \to F\to 0\) to get \(F\cong M/M_t\). This splits since \(F\) is free and thus projective, so \(F\cong M \oplus M_t\).

Algebraic Properties

\begin{align*} r\curvearrowright(m\otimes n) \coloneqq(r\curvearrowright m)\otimes n .\end{align*}

If \(\dim_k V, \dim_k W < \infty\) then there is an isomorphism \begin{align*} V {}^{ \vee }\otimes_k W &\xrightarrow{\sim} \mathop{\mathrm{Hom}}_{ {}_{k}{\mathsf{Mod}}}(V, W) \\ \tilde v \otimes w &\mapsto \tilde v({-}) w .\end{align*}

If either of \(\dim_k V, \dim_k W\) is finite, then \begin{align*} V {}^{ \vee }\otimes_k W {}^{ \vee }&\xrightarrow{\sim} (V\otimes W) {}^{ \vee }\\ v\otimes w &\mapsto (x \otimes y \mapsto v(x) w(y)) .\end{align*}

\begin{align*} \mathop{\mathrm{Hom}}_{ {}_{k}{\mathsf{Mod}}}(V, W) &\xrightarrow{\sim} \mathop{\mathrm{Hom}}_{ {}_{k}{\mathsf{Mod}}}(W, V) {}^{ \vee }\\ T &\mapsto \operatorname{Tr}(T \circ {-}) .\end{align*}

If \(T:V \hookrightarrow W\) is injective, then \(T\otimes\one_X: V\otimes X \hookrightarrow W\otimes X\) is also injective for any \(X\). Thus \(F({-}) = ({-}\otimes X)\) is right-exact for any \(X\).

\({\mathbf{Z}}/2 \otimes_{\mathbf{Z}}{\mathbf{Z}}/3 = 0\):

Link to Diagram