Definition 9.5.2.
Let \(R\) be a ring, and let \(f\colon M\to N\) be an \(R\)-module homomorphism. The kernel of \(f\) is
\begin{equation*}
\ker(f)=\{m\in M:f(m)=0\}
\end{equation*}
and the image of \(f\) is
\begin{equation*}
{\rm Im}(f)=\{f(m)\in N:m\in M\}.
\end{equation*}
Let \(A\) be a commutative ring, and let \(M\) be a left \(A\)-module. We fix \(a\in A\) and consider the map \(\ell_a\colon M\to M\) defined by
\begin{equation*}
m\mapsto a\cdot m
\end{equation*}
The map \(\ell_a\) is an \(A\)-module homomorphism.
Let \(V\) be a finite-dimensional vector space over a field \(F\text{.}\) We fix an \(F\)-linear endomorphism of \(V\text{,}\) say \(T\text{.}\) We consider \(V\) as a left \(F[X]\)-module via the following scalar multiplication (refer to Example 9.2.4):
\begin{equation*}
F[X]\times V\to V;\quad f(X)\cdot v=f(T)(v)
\end{equation*}
Consider the following map.
\begin{equation*}
\varphi\colon V\to V\quad\text{defined by}\quad v\mapsto T(v).
\end{equation*}
It is left to the reader to verify that \(\varphi\) is an \(F[X]\)-module.
Let \(R\) be a ring and let \(M\) be a left \(R\)-module. Fix an element \(m\in M\text{.}\) The map \(f_m\colon R\to M\) define by
\begin{equation*}
r\mapsto r\cdot m
\end{equation*}
is an \(R\)-module homomorphism. The kernel of this map is called the annihilator of \(m\) and it is denoted by \({\rm Ann}_R(m)\text{.}\) Thus
\begin{equation*}
{\rm Ann}_R(m)=\{r\in R:r\cdot m=0\}
\end{equation*}
is a left ideal of \(R\text{.}\)
Consider the matrix ring over a field \(F\text{,}\) \(M_2(F)\text{.}\) Then
\begin{equation*}
M=\left\{\begin{pmatrix}a\amp 0\\b\amp 0\end{pmatrix}:a,b\in F\right\}\quad\text{and}\quad N=\left\{\begin{pmatrix}0\amp a\\0\amp b\end{pmatrix}:a,b\in F\right\}
\end{equation*}
are left ideals of \(M_2(F)\text{,}\) and hence left \(M_2(F)\)-submodules of \(M_2(F)\text{.}\) We define a map
\begin{equation*}
f\colon M\to N\quad\text{by}\quad \begin{pmatrix}a\amp 0\\b\amp 0\end{pmatrix}\mapsto\begin{pmatrix}0\amp a\\0\amp b\end{pmatrix}.
\end{equation*}
It is left to the reader to check that \(f\) is an \(M_2(F)\)-module homomorphism. Furthermore, the map
\begin{equation*}
g\colon N\to M\quad\text{defined by}\quad\begin{pmatrix}0\amp a\\0\amp b\end{pmatrix}\mapsto\begin{pmatrix}a\amp 0\\b\amp 0\end{pmatrix}
\end{equation*}
is also an \(M_2(F)\)-module homomorphism with
\begin{equation*}
f\circ g=\unit_N\quad\text{and}\quad g\circ f=\unit_M.
\end{equation*}
Thus \(M_2(F)\)-modules \(M,N\) are isomorphic. Furthermore we show that \(M\) is a simple left \(M_2(F)\)-module. By a similar reasoning \(N\) can be shown as a simple left \(M_2(F)\)-mdoule. Suppose that \(m= \begin{pmatrix}a\amp 0\\b\amp 0\end{pmatrix}\in M\) is a nonzero element and \(m^\prime=\begin{pmatrix}c\amp 0\\d\amp 0\end{pmatrix}\in M\) be an arbitrary element. We show that there exists \(X=\begin{pmatrix}x\amp y\\z\amp w\end{pmatrix}\in M_2(F)\) such that \(Xm=m^\prime\text{.}\) Indeed, we get the following equations from \(Xm=m^\prime\text{.}\)
\begin{equation*}
ax+by=c\quad\text{and}\quad az+bw=d.
\end{equation*}
This can be solved to obtain \(x,y,z,\) and \(w\text{.}\) Hence, \(M\) is a simple left \(M_2(F)\)-module. In other words, \(M\) is a minimal left ideal of \(M_2(F)\text{.}\)
