@ -324,47 +324,66 @@ $A[\psi] \subset End M$ is the subring geneerated by $A$ and the action of $\psi
\end{pmatrix}
\end{pmatrix}
$$
$$
Kronecker delta:
$\delta_{ij}=
\begin{cases}
1 &\text{if } i = j,\\
0 &\text{otherwise}
\end{cases}$
With the Kronecker delta, $\psi(x_i)$ can be expressed as
$$\psi(x_i)=\sum_{j=1}^n \delta_{ij}\psi(x_j)$$
so the previous matrix can be characterized as
$$\sum_{j=1}^n (\delta_{ij}\psi- a_{ij}) x_j =0$$
The entries of the matrix are \emph{endomorphisms} (elements of the ring $A[\psi]$)
\begin{itemize}
\item the term $(\psi- a_{11})$ is an operator that acts on $x_1$; as $(\psi(x_1)-a_{11}\cdot x_1)$
\item the term $(-a_{12})$ is an operator that acts on $x_2$; as multiplication by it, ie. $(-a_{12}\cdot x_2)$
\end{itemize}
Since $A$ is a commutative ring, and $\psi$ commutes with any $a \in A$,
the ring of operators $A[\psi]$ is a commutative ring.
Denote the previous matrix by $\Phi$. Let $m$ denote the vector $(x_1, x_2, \ldots, x_n)^T$ (ie. the vector of generators of the $A$-module $M$).\\
\hspace*{4em}Then we can write the previous equality as
\begin{equation}
\Phi\cdot m = 0
\label{eq:2.4.1}
\end{equation}
We know that
\begin{equation}
adj(\Phi) \Phi = det(\Phi) I
\label{eq:2.4.2}
\end{equation}
(aka. fundamental identity for the adjugate matrix).
So if at \eqref{eq:2.4.1} we multiply both sides by $adj(\Phi)$,
\begin{align*}
adj(\Phi) \cdot\Phi\cdot&m = 0\\
(\text{recall from \eqref{eq:2.4.2}:}~ &det(\Phi)\cdot I ~)\\
=det(\Phi) \cdot I \cdot&m = 0
\end{align*}
Thus,
\begin{align*}
det(\Phi) \cdot I \cdot&m = 0\\
\begin{pmatrix}
det(\Phi) & 0 &\ldots& 0\\
0 & det(\Phi) &\ldots& 0\\
\vdots\\
0 & 0 &\ldots& det(\Phi)
\end{pmatrix}
\cdot
&\begin{pmatrix}
x_1\\ x_2\\\vdots\\ x_n
\end{pmatrix}
=
\begin{pmatrix}
0\\ 0\\\vdots\\ 0
\end{pmatrix}
\end{align*}
$\Longrightarrow$
\begin{equation}
det(\Phi) \cdot x_i = 0 ~~\forall i \in [n]
\label{eq:2.4.3}
\end{equation}
ie. $det(\Phi)$ is an \emph{annihilator} of the generators $x_i$ of $M$, thus of the entire module $M$.
$\Longrightarrow~$ so we can treat the matrix as a matrix of real numbers and calculate its determinant.
We're interested in the determinant because it is the only way to turn a system of multiple equations in a single scalar-like equation that describes the endomorphism $\psi$.\\
$\rightarrow$ Because in module theory, we lack of "division", so can not "solve for $\psi$" the system of equations.\\
$\rightarrow$ The determinant provides a way to find a polynomial that \emph{annihilates} the module; the \emph{characteristic polynomial}, which related $\psi$ to the ideal $\aA$
So, we're interested into calculating the $det(\Phi)$.
The \emph{determinant trick} is that the terms that go after the "leading term of the determinant", will belong to $\aA$ and their combinations with $\psi$ will not be bigger than $\psi^n$. Furthermore, when expanding it
\begin{itemize}
\begin{itemize}
\item highest power is $1\cdot\psi^n$
\item highest power is $1\cdot\psi^n$
\item coefficient of $\psi^{n-1}$ is $-(\underbrace{ a_{11}+ a_{22}+\ldots+ a_{nn}}_{a_1})$,\\
\item coefficient of $\psi^{n-1}$ is $-(\underbrace{ a_{11}+ a_{22}+\ldots+ a_{nn}}_{a_1})$,\\
@ -373,15 +392,111 @@ $A[\psi] \subset End M$ is the subring geneerated by $A$ and the action of $\psi
Since this determinant annihilates the generators (ie. $det(M)x_i=0$), the resulting enddomorphism $p(\psi)$ is the zero map on the entire module $M$, so:
\vspace{0.5cm}
Now, notice that we had $det(\Phi)\cdot x_i =0 ~\forall~ i\in[n]$.
% next part might be removed
Since $M$ is a fingen $A$-module, any element $m \in M$ can be written as a linear combination of $M$'s generators $x_i$, ie.
$$m = r_1 x_1+ r_2 x_2+\ldots r_n x_n \in M$$
If we multiply $m \in M$ by $d = det(\Phi)$,
\begin{align*}
d \cdot m &= d \cdot (r_1 x_1 + r_2 x_2 + \ldots r_n x_n)\\
The matrix $\Phi$ is the \emph{characteristic matrix}, $xI-A$, viewed as an operator. Then,
$$det(\Phi)= det(xI-A)= p(x)$$
where $p(x)$ is the \emph{characteristic polynomial}.
If a linear transformation turns every basis vector ($x_i$) into zero, then that transformation is the zero transformation. So in our case, $det(\Phi)$ is the zero transformation, thus $det(\Phi)=0$.
% The entries of the matrix are \emph{endomorphisms} (elements of the ring $A[\psi]$)
%\begin{itemize}
%\item the term $(\psi- a_{11})$ is an operator that acts on $x_1$; as $(\psi(x_1)-a_{11}\cdot x_1)$
%\item the term $(-a_{12})$ is an operator that acts on $x_2$; as multiplication by it, ie. $(-a_{12}\cdot x_2)$
%\end{itemize}
%
% We need a single element $x \in A$ that \emph{annihilates} every $m \in M$ simultaneously, ie. $xM=0$. We're going to use the determinant for getting $x$.
%
% Since $A$ is a commutative ring, and $\psi$ commutes with any $a \in A$,
% the ring of operators $A[\psi]$ is a commutative ring.
%
%$\Longrightarrow~$ so we can treat the matrix as a matrix of real numbers and calculate its determinant.
%
%
%\vspace{0.75cm}
% This is called \emph{"the determinant trick"}.\\
% We're interested in the determinant because it is the only way to turn a system of multiple equations in a single scalar-like equation that describes the endomorphism $\psi$.\\
%$\rightarrow$ Because in module theory, we lack of "division", so can not "solve for $\psi$" the system of equations.\\
%$\rightarrow$ The determinant provides a way to find a polynomial that \emph{annihilates} the module; the \emph{characteristic polynomial}, which related $\psi$ to the ideal $\aA$
%
%$$det(M)\cdot x_i =0~~ \forall i$$
% where $x_i$ are the generators of $M$.
%
% Use $\Phi$ to denote the previous matrix. The determinant is the only function that can take that matrix $\Phi$ and produce a single scalar $x=det(\Phi)$ such that the following identity holds: $adj(\Phi)\cdot\Phi=det(\Phi)\cdot\aA$.
%\vspace{0.5cm}
%
% Since $det(M)$ kills every generator, it must kill every element in $M$\\
%$\Longrightarrow~~ det(M)$ is the zero map.
%
% Leibniz formula of the determinant of an $n \times n$ matrix:
% Since this determinant annihilates the generators (ie. $det(M)x_i=0$), the resulting enddomorphism $p(\psi)$ is the zero map on the entire module $M$, so:
