Notes - Rings and Modules HT24, Basic definitions for modules

An abelian (“A”) group $(M, +)$ and a multiplication (“M”) action $a : R \times M \to M$ written $(r, m) \mapsto rm$ that satisfies:

1. $(r _ 1 + r _ 2) m = r _ 1 m + r _ 2 m$ for all $r _ 1, r _ 2 \in R$ and $m \in M$ (“D”)
2. $(r _ 1 r _ 2) m = r _ 1 (r _ 2 m)$, for all $r _ 1, r _ 2 \in R$ and $m \in M$ (“A”)
3. $1 _ R m = m$, for all $m \in M$ (“I”)
4. $r(m _ 1 + m _ 2) = r m _ 1 + r m _ 2$ for all $r \in R$ and $m _ 1, m _ 2 \in M$ (“D”)

(AM DAID, “M” multiplication map, “A” Abelian group, “D” for 2 distributive axioms, “A” associativity, “I” identity)

It is a $\mathbb Z$-module, given by

\[\begin{aligned} na &= a + \cdots + a \quad (n \text{ times}, n > 0) \\\\ na &= -(a + \cdots + a) \quad (n \text{ times}, n < 0) \end{aligned}\]

Giving $M$ the structure of an $R$-module corresponds to specifying a multiplication action $a : R \times M \to M$. Then note that definining

\[a(r, m) = \phi(r)\big(m\big)\]

(where $\phi(r)$ is an endomorphism $M \to M$, and hence the brackets represent represent the application of a function) specifies $a$ in terms of $\phi$, and that

\[\phi(r) \big(m\big) = a(r, m)\]

specifies $\phi$ in terms of $a$.

• $N$ is a subgroup of $M$
• $r \in R$ and $n \in N$, then $rn \in N$ (closed under arbitrary multiplication)

It has a basis.

(Like in the normal linear algebra case, it is entirely possible this basis is infinite).

• Finitely generated: it is generated by a finite subset of elements
• Free: it has a basis

Unlike vector spaces, modules can be finitely generated but there is no linearly independent subset that generates $M$.

• Addition well defined: Since $N$ is a submodule, it is in particular a subgroup of an Abelian group and hence a normal subgroup. Then $(M / N, +)$ is a well-defined Abelian group.
• Multiplication well defined: Suppose $r \in R$ and $m _ 1, m _ 2 \in M$ are such that $m _ 1 + N = m _ 2 + N$. Then $m _ 1 - m _ 2 \in N$, so $r(m _ 1 - m _ 2) \in N$. Then $r m _ 1 + N = r m _ 2 + N$ as required.

Suppose:

• $M$ is an $R$-module
• $N$ is a submodule
• $q : M \to M/N$ is the corresponding quotient map

Then:

• If $S$ is a submodule of $M$, $q(S)$ is a submodule of $M/N$
• If $T$ is a submodule of $M / N$, $q^{-1}(T)$ is a submodule of $M$ that contains $N$

In particular, the map $T \mapsto q^{-1}(T)$ gives an injective map from submodules of $M/N$ to submodules of $M$ which contain $N$, thus submodules of $M/N$ correspond bijectively to submodules of $M$ which contain $N$.

Suppose:

• $R$ is a ring
• $M, N$ are $R$-modules
• $\phi : M \to N$ any $R$-module homomorphism
• $S \subseteq \ker \phi$
• $S$ is a submodule of $M$
• $q : M \to M/S$ is the quotient homomorphism

Then:

• $\exists ! \bar\phi : M/S \to N$ such that $\phi = \bar \phi \circ q$
• $\ker \bar \phi = \ker \phi / S = \{m + S \mid m \in \ker \phi\}$

• $M / \ker\phi \cong \text{Im }\phi$
• $(N _ 1 + N _ 2)/N _ 2 \cong N _ 1/(N _ 1 \cap N _ 2)$
• $(M / N _ 1)/(N _ 2 / N _ 1) \cong M/N _ 2$

It is generated by a single element.

1. $\phi(m _ 1 + m _ 2) = \phi(m _ 1) + \phi(m _ 2)$ for all $m _ 1, m _ 2 \in M _ 1$
2. $\phi(r m) = r \phi(m)$ for all $r \in R$ and $m \in M _ 1$

If $\phi$ were instead a ring homomorphisms, then we would actually have the second condition $\phi(rm) = \phi(r)\phi(m)$, which is different.

Because $R$-modules have no multiplicative identity themselves, they are just a Abelian group $(M, +)$ with a corresponding action. (The fact $\phi(0 _ M) = 0 _ N$ follows from $R$-linearity).

Define the map $\Phi : M _ 1 / \ker \phi \to \text{Im } \phi$ by $\Phi(m + \ker \phi) = \phi(m)$. Note that

\[\begin{aligned} &\phi(m_1) = \phi(m_2) \\\\ \iff &\phi(m_1 - m_2) = 0 \\\\ \iff &m_1 - m_2 \in \ker \phi \\\\ \iff &m_1 + \ker \phi = m_2 + \ker \phi \end{aligned}\]

Hence $\Phi$ is $1-1$ and well-defined. $\Phi$ is clearly onto and is a homomorphism since

\[\begin{aligned} \Phi(r_1 (m_1 + \ker \phi) + r_2 (m_2 + \ker \phi)) &= \Phi((r_1 m_1 + r_2 m_2) + \ker \phi) \\\\ &= \phi(r_1 m_1 + r_2 m_2) \\\\ &= r_1 \phi(m_1) + r_2 \phi(m_2) \\\\ &= r_1 \Phi(m_1 + \ker \phi) + r_2 \Phi(m_2 + \ker \phi) \end{aligned}\]

\[\mathbb Z / 2\mathbb Z\]

viewed as an $\mathbb Z$-module.

Finitely generated, e.g. by $\{1\}$, but since it is torsion, any candidate basis will contain a linear combination making $0$, since e.g. $2 \cdot 1 = 0$.

\[\mathbb Z[x]\]

viewed as a $\mathbb Z$-module. It has a basis $\{1, x, x^2, \cdots\}$ but no finite set generates all of $\mathbb Z[x]$.

Consider $M = \mathbb Z$ as a $\mathbb Z$-module. Then $\{2, 3\}$ is a spanning set (since $-1\cdot 2 + 1 \cdot 3 = 1$), but it is not a basis (since $2 \cdot 3 - 3 \cdot 2 = 0$), and no subset of it is a basis.

Consider $M = \mathbb Z$ as a $\mathbb Z$-module. Then $\{2\}$ is a linearly independent set, but no basis can contain $2$ as the only bases for $\mathbb Z$ are $\{1\}$ and $\{-1\}$.