Example 4.0.2.

1.

Let \(R = \mathbb {Z}\). Then if \(M\) is any abelian group. We may define a \(\mathbb {Z}-\)module structure on \(M\) as follows \(\, (n,x) \longmapsto nx\)
Check that for \(n_1,\, n_2\in \mathbb {Z},\,\, x,\, y\in M \)

(i)

\(n_1\cdot (n_2\cdot x) = n_1 n_2 x\)

(ii)

\((n_1 + n_2)\cdot x = n_1\cdot x + n_2\cdot x\)

(iii)

\(n_1\cdot (x+ y) = n_1\cdot x + n_1 \cdot y\)

(iv)

\(1_R\cdot x = 1\cdot x = x\)

2.

Let \(R = K\) field. Then the module over \(K\) is a vector space. Many facts about vector spaces remain true for modules. e.g consider \(a\neq 0\) in a field \(K\), then \(\, ax = 0\, \implies \, x = 0\). But this is not true in general for any module.

3.

Let \(R\) be a ring. Consider then multiplication \(\, \bullet \, R\times R \longmapsto R\). This defines a module structure on the ring \(R\). This is both a right and left \(R-\)module.

4.

Let \(K\) be a field and let \(R = M_n(K)\) be the ring of \(n\times n\) matrices over the field \(K\). Let \(M = K^n\) visualized as a column vector of \(n\) entries. \[R\times M \longmapsto M\] \[(a,x) \longmapsto ax\]

\[a = \begin {pmatrix} a_{11} & \cdots \cdots & a_{1n}\\ a_{21} & \cdots \cdots & a_{2n}\\ \vdots & & \vdots \\ \vdots & & \vdots \\ a_{n1} & \cdots \cdots & a_{nn}\\ \end {pmatrix} \hspace {0.5cm},\hspace {0.5cm} x = \begin {pmatrix} x_1\\ \vdots \\ \vdots \\ x_n\\ \end {pmatrix}\]

\[(a,x) = \begin {pmatrix} \begin {pmatrix} a_{11} & \cdots \cdots & a_{1n}\\ a_{21} & \cdots \cdots & a_{2n}\\ \vdots & & \vdots \\ \vdots & & \vdots \\ a_{n1} & \cdots \cdots & a_{nn}\\ \end {pmatrix} \hspace {0.2cm},\hspace {0.2cm} \begin {pmatrix} x_1\\ \vdots \\ \vdots \\ x_n\\ \end {pmatrix} \end {pmatrix}\]

\[ ax = \begin {pmatrix} a_{11} x_1 + a_{12} x_2 + \, \cdots \cdots \, + a_{1n} x_n\\ \vdots \\ \vdots \\ a_{n1} x_1 + a_{n2} x_2 + \, \cdots \cdots \, + a_{nn} x_n\\ \end {pmatrix}\]

5.

Let \(R\) be a ring and \(n\) a rational number, the Cartesian product \(R^n\) is both a left and Right \(R-\)module (If we use component-wise operation).

6.

Let \(S\) be a non-empty set, \(M\) a left \(R-\)module, and \(M^s = \{ f: s\longrightarrow M\}\). Define addition and multiplication as follows: \[(f + g) (s) = f(s) + g(s)\,\hspace {0.4cm} ,\hspace {0.4cm} \, (rf) (s) = r\big (f(s)\big )\] The \(M^2\) is a left \(R-\)module.

Check
Let \(f,\, g\in S\, ,\hspace {0.2cm} s,\, t\in M\,\) then \(M^s\) satisfies:

(i)

\(f\cdot (g\cdot s) = f\cdot (g(s)) = (fg) (s) = (fg)\cdot s\)

(ii)

\((f + g)\cdot s = (f + g)\cdot s = f\cdot s + g\cdot s = f(s) + g(s)\)

(iii)

\(f\cdot (s + t) = f\cdot s + f\cdot t = f(s) + f(t)\)

(iv)

\(1_R\cdot s = 1\cdot s = 1(s) = s\)

7.

Let \(X\) be a \(\underbrace {\text {Smooth}}_{\text {(inifitely differentiable)}}\) \(\,\) manifold\((\mathbb {R}^2 = (x,y) \) a surface in \(2-D)\)

Sphere: \(\begin {vmatrix} z\\ \end {vmatrix} = R\hspace {0.3cm}, \hspace {0.3cm} \sqrt {x^2 + y^2 + z^2} \, = r\hspace {0.3cm} (x, y , r)\longmapsto r\)
Then consider the smooth functions on \(X\) to real numbers. This set of smooth functions, form a ring \(C^{\infty } (X)\). The set of all smooth vectors fields defined on \(X\) is module over \(C^{\infty }(X)\).

Assignment
Check that indeed the vector field \(\overrightarrow {\nabla }\, f\) forms a module over the ring of functions
\(\, f: x \longrightarrow \mathbb {R}\). \((\) Take \(X=\mathbb {R}^3)\), where \[\overrightarrow {\nabla } f = \frac {\partial f}{\partial x} \, \textbf {i} + \frac {\partial f}{\partial y}\, \textbf {j} + \frac {\partial f}{\partial z}\, \textbf {k}\]

8.

If \(R\) is a ring, define \(R^{op}\) with elements as in \(R\), but multiplication defined as follows: If \(ab = c\), then \(ba = c \in R^{op}\). Any left \(R-\)module \(M\) can be seen as the right module over \(R^{op}\).