Modules

It suffices to show that $$\begin{align*} r\in R, ~t_1, t_2\in \operatorname{Tor}(M) \implies rt_1 + t_2 \in \operatorname{Tor}(M) .\end{align*}$$

We have $$\begin{align*} t_1 \in \operatorname{Tor}(M) &\implies \exists s_1 \neq 0 \text{ such that } s_1 t_1 = 0 \\ t_2 \in \operatorname{Tor}(M) &\implies \exists s_2 \neq 0 \text{ such that } s_2 t_2 = 0 .\end{align*}$$

Since $R$ is an integral domain, $s_1 s_2 \neq 0$. Then $$\begin{align*} s_1 s_2(rt_1 + t_2) &= s_1 s_2 r t_1 + s_1 s_2t_2 \\ &= s_2 r (s_1 t_1) + s_1 (s_2 t_2) \quad\text{since $R$ is commutative} \\ &= s_2 r(0) + s_1(0) \\ &= 0 .\end{align*}$$

Let $R = {\mathbf{Z}}/6{\mathbf{Z}}$ as a ${\mathbf{Z}}/6{\mathbf{Z}}{\hbox{-}}$module, which is not an integral domain as a ring.

Then $[3]_6\curvearrowright[2]_6 = [0]_6$ and $[2]_6\curvearrowright[3]_6 = [0]_6$, but $[2]_6 + [3]_6 = [5]_6$, where 5 is coprime to 6, and thus $[n]_6\curvearrowright[5]_6 = [0] \implies [n]_6 = [0]_6$. So $[5]_6$ is not a torsion element.

So the set of torsion elements are not closed under addition, and thus not a submodule.

Suppose $R$ has zero divisors $a,b \neq 0$ where $ab = 0$. Then for any $m\in M$, we have $b\curvearrowright m \coloneqq bm \in M$ as well, but then $$\begin{align*} a\curvearrowright bm = (ab)\curvearrowright m = 0\curvearrowright m = 0_M ,\end{align*}$$ so $m$ is a torsion element for any $m$.

• Suppose toward a contradiction $\operatorname{Tor}(M)$ has rank $n \geq 1$.
• Then $\operatorname{Tor}(M)$ has a linearly independent generating set $B = \left\{{\mathbf{r}_1, \cdots, \mathbf{r}_n}\right\}$, so in particular $$\begin{align*} \sum_{i=1}^n s_i \mathbf{r}_i = 0 \implies s_i = 0_R \,\forall i .\end{align*}$$
• Let $\mathbf{r}$ be any of of these generating elements.
• Since $\mathbf{r}\in \operatorname{Tor}(M)$, there exists an $s\in R\setminus 0_R$ such that $s\mathbf{r} = 0_M$.
• Then $s\mathbf{r} = 0$ with $s\neq 0$, so $\left\{{\mathbf{r}}\right\} \subseteq B$ is not a linearly independent set, a contradiction.

• Let $n = \operatorname{rank}M$, and let $\mathcal B = \left\{{\mathbf{r}_i}\right\}_{i=1}^n \subseteq R$ be a generating set.
• Let $\tilde M \coloneqq M/\operatorname{Tor}(M)$ and $\pi: M \to M'$ be the canonical quotient map.

Note that the proof follows immediately.

• Suppose that $$\begin{align*} \sum_{i=1}^n s_i (\mathbf{r}_i + \operatorname{Tor}(M)) = \mathbf{0}_{\tilde M} .\end{align*}$$

• Then using the definition of coset addition/multiplication, we can write this as $$\begin{align*} \sum_{i=1}^n \qty { s_i \mathbf{r}_i + \operatorname{Tor}(M)} = \qty{ \sum_{i=1}^n s_i \mathbf{r}_i} + \operatorname{Tor}(M) = 0_{\tilde M} .\end{align*}$$

• Since $\tilde{\mathbf{x}} = 0 \in \tilde M \iff \tilde{\mathbf{x}} = \mathbf{x} + \operatorname{Tor}(M)$ where $\mathbf{x} \in \operatorname{Tor}(M)$, this forces $\sum s_i \mathbf{r}_i \in \operatorname{Tor}(M)$.

• Then there exists a scalar $\alpha\in R^{\bullet}$ such that $\alpha \sum s_i \mathbf{r}_i = 0_M$.

• Since $R$ is an integral domain and $\alpha \neq 0$, we must have $\sum s_i \mathbf{r}_i = 0_M$.

• Since $\left\{{\mathbf{r}_i}\right\}$ was linearly independent in $M$, we must have $s_i = 0_R$ for all $i$.

• Write $\pi(\mathcal B) = \left\{{\mathbf{r}_i + \operatorname{Tor}(M)}\right\}_{i=1}^n$ as a set of cosets.

• Letting $\mathbf{x} \in M'$ be arbitrary, we can write $\mathbf{x} = \mathbf{m} + \operatorname{Tor}(M)$ for some $\mathbf{m} \in M$ where $\pi(\mathbf{m}) = \mathbf{x}$ by surjectivity of $\pi$.

• Since $\mathcal B$ is a basis for $M$, we have $\mathbf{m} = \sum_{i=1}^n s_i \mathbf{r}_i$, and so $$\begin{align*} \mathbf{x} &= \pi(\mathbf{m}) \\ &\coloneqq\pi\qty{ \sum_{i=1}^n s_i \mathbf{r}_i} \\ &= \sum_{i=1}^n s_i \pi(\mathbf{r}_i) \quad\text{since $\pi$ is an $R{\hbox{-}}$module morphism}\\ &\coloneqq\sum_{i=1}^n s_i \mathbf{(}\mathbf{r}_i + \operatorname{Tor}(M)) ,\end{align*}$$ which expresses $\mathbf{x}$ as a linear combination of elements in $\mathcal B'$.

Notation: Let $0_R$ denote $0\in R$ regarded as a ring element, and $\mathbf{0} \in R$ denoted $0_R$ regarded as a module element (where $R$ is regarded as an $R{\hbox{-}}$module over itself)

An $R{\hbox{-}}$module $M$ is free if any of the following conditions hold:

• $M$ admits an $R{\hbox{-}}$linearly independent spanning set $\left\{{\mathbf{b}_\alpha}\right\}$, so $$\begin{align*}m\in M \implies m = \sum_\alpha r_\alpha \mathbf{b}_\alpha\end{align*}$$ and $$\begin{align*}\sum_\alpha r_\alpha \mathbf{b}_\alpha = 0_M \implies r_\alpha = 0_R\end{align*}$$ for all $\alpha$.
• $M$ admits a decomposition $M \cong \bigoplus_{\alpha} R$ as a direct sum of $R{\hbox{-}}$submodules.
• There is a nonempty set $X$ an monomorphism $X\hookrightarrow M$ of sets such that for every $R{\hbox{-}}$module $N$, every set map $X\to N$ lifts to a unique $R{\hbox{-}}$module morphism $M\to N$, so the following diagram commutes:

Equivalently, $$\begin{align*} \mathop{\mathrm{Hom}}_{\mathsf{Set}}(X, \mathop{\mathrm{Forget}}(N)) \xrightarrow{\sim} \mathop{\mathrm{Hom}}_{ {}_{R}{\mathsf{Mod}}}(M, N) .\end{align*}$$

• Define the annihilator: $$\begin{align*} \operatorname{Ann}(m) \coloneqq\left\{{r\in R {~\mathrel{\Big\vert}~}r\cdot m = 0_M}\right\} {~\trianglelefteq~}R .\end{align*}$$
  • Note that $mR \cong R/\operatorname{Ann}(m)$.
• Define the torsion submodule: $$\begin{align*} M_t \coloneqq\left\{{m\in M {~\mathrel{\Big\vert}~}\operatorname{Ann}(m) \neq 0}\right\} \leq M \end{align*}$$
• $M$ is torsionfree iff $M_t = 0$ is the trivial submodule.

• Let the following be an SES where $F$ is a free $R{\hbox{-}}$module: $$\begin{align*} 0 \to N \to M \xrightarrow{\pi} F \to 0 .\end{align*}$$

• Since $F$ is free, there is a generating set $X = \left\{{x_\alpha}\right\}$ and a map $\iota:X\hookrightarrow F$ satisfying the 3rd property from (a).

  • If we construct any map $f: X\to M$, the universal property modules will give a lift $\tilde f: F\to M$

• Identify $X$ with $\iota(X) \subseteq F$.

• For every $x\in X$, the preimage $\pi^{-1}(x)$ is nonempty by surjectivity. So arbitrarily pick any preimage.

• $\left\{{\iota(x_\alpha)}\right\} \subseteq F$ and $\pi$ is surjective, so choose fibers $\left\{{y_\alpha}\right\} \subseteq M$ such that $\pi(y_\alpha) = \iota(x_\alpha)$ and define $$\begin{align*} f: X&\to M \\ x_\alpha &\mapsto y_\alpha .\end{align*}$$

• The universal property yields $h: F\to M$:

• It remains to check that it’s a section.
  • Write $f= \sum r_i x_i$, then since both maps are $R{\hbox{-}}$module morphism, by $R{\hbox{-}}$linearity we can write $$\begin{align*} (\pi \circ h)(f) &= (\pi \circ h)\qty{ \sum r_i x_i } \\ &= \sum r_i (\pi \circ h)(x_i) ,\end{align*}$$ but since $h(x_i) \in \pi^{-1}(x_i)$, we have $(\pi \circ h)(x_i) = x_i$. So this recovers $f$.

• Free implies projective

  • Universal property of projective objects: for every epimorphism $\pi:M\twoheadrightarrow N$ and every $f:P\to N$ there exists a unique lift $\tilde f: P\to M$:

  

  • Construct $\phi$ in the following diagram using the same method as above (surjectivity to pick elements in preimage):

Link to Diagram

• Now take the identity map, then commutativity is equivalent to being a section.

There is a SES

• Let $m+M_t \in M/M_t$ be arbitrary. Suppose this is a torsion element, the claim is that it must be the trivial coset. This will follow if $m\in M_t$
• Since this is torsion, there exists $r\in R$ such that $$\begin{align*} M_t = r(m + M_t) \coloneqq(rm) + M_t \implies rm\in M_t .\end{align*}$$
• Then $rm$ is torsion in $M$, so there exists some $s\in R$ such $s(rm) = 0_M$.
• Then $(sr)m = 0_M$ which forces $m\in M_t$

By the correspondence theorem, submodules of $M/N$ biject with submodules $A$ of $M$ containing $N$.

So

• $M$ is maximal:

• $\iff$ no such (proper, nontrivial) submodule $A$ exists

• $\iff$ there are no (proper, nontrivial) submodules of $M/N$

• $\iff M/N$ is simple.

Identify ${\mathbf{Z}}{\hbox{-}}$modules with abelian groups, then by (a), $N$ is maximal $\iff$ $M/N$ is simple $\iff$ $M/N$ has no nontrivial proper subgroups.\

By Cauchy’s theorem, if ${\left\lvert {M/N} \right\rvert} = ab$ is a composite number, then $a\divides ab \implies$ there is an element (and thus a subgroup) of order $a$. In this case, $M/N$ contains a nontrivial proper cyclic subgroup, so $M/N$ is not simple. So ${\left\lvert {M/N} \right\rvert}$ can not be composite, and therefore must be prime.

• Let $G = \left\{{x \in {\mathbf{C}}{~\mathrel{\Big\vert}~}x^n=1 \text{ for some }n\in {\mathbb{N}}}\right\}$, and suppose $H < G$ is a proper submodule.

• Since $H\neq G$, there is some $p$ and some $k$ such that $\zeta_{p^k}\not\in H$.

  • Otherwise, if $H$ contains every $\zeta_{p^k}$ it contains every $\zeta_n$

Then there must be a prime $p$ such that the $\zeta_{p^k} \not \in H$ for all $k$ greater than some constant $m$ – otherwise, we can use the fact that if $\zeta_{p^k} \in H$ then $\zeta_{p^\ell} \in H$ for all $\ell \leq k$, and if $\zeta_{p^k} \in H$ for all $p$ and all $k$ then $H = G$.

But this means there are infinitely many elements in $G\setminus H$, and so $\infty = [G: H] = {\left\lvert {G/H} \right\rvert}$ is not a prime. Thus by (b), $H$ can not be maximal, a contradiction.