# Modules

## 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:

• 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:

• 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.

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

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

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

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

• 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

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

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.

$$\implies$$:

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.

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 $$X$$.

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