diff --git a/commutative-algebra-notes.pdf b/commutative-algebra-notes.pdf index 36759e4..25f4ae9 100644 Binary files a/commutative-algebra-notes.pdf and b/commutative-algebra-notes.pdf differ diff --git a/commutative-algebra-notes.tex b/commutative-algebra-notes.tex index 5c88d5c..3b5e162 100644 --- a/commutative-algebra-notes.tex +++ b/commutative-algebra-notes.tex @@ -324,47 +324,66 @@ $A[\psi] \subset End M$ is the subring geneerated by $A$ and the action of $\psi \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)$. - $$det(M) \cdot x_i = 0~~ \forall i$$ - where $x_i$ are the generators of $M$. + By the Leibniz formula, + $$\det(A) = \sum_{\sigma \in S_n} sgn(\sigma) \prod_{i=1}^n a_{i, \sigma(i)}$$ - Since $det(M)$ kills every generator, it must kill every element in $M$\\ - $\Longrightarrow~~ det(M)$ is the zero map. + thus, + $$det(\Phi) = \underbrace{(\psi - a_{11}) (\psi - a_{22}) \ldots (\psi - a_{nn})}_{\text{diagonal of $\Phi$, leading term of the determinant}} - \ldots$$ - Leibniz formula of the determinant of an $n \times n$ matrix: - $$ - det(M) = \sum_{\sigma \in S_n} sign(\sigma) \prod_{i=1}^n M_{i, \sigma(i)} - $$ - - so, - $$(\psi - a_{11}) (\psi - a_{22}) \ldots (\psi - a_{nn})$$ - expanding it, + The \emph{determinant trick} is that the terms that go after the "leading term of the determinant", will belong to $\aA$ and their combinations with $\psi$ will not be bigger than $\psi^n$. Furthermore, when expanding it \begin{itemize} \item highest power is $1 \cdot \psi^n$ \item coefficient of $\psi^{n-1}$ is $-( \underbrace{ a_{11} + a_{22} + \ldots + a_{nn} }_{a_1})$,\\ @@ -373,15 +392,111 @@ $A[\psi] \subset End M$ is the subring geneerated by $A$ and the action of $\psi \end{itemize} So we have - $$p(\psi) = \psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n$$ + $$det(\Phi) = \psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n$$ with $a_i \in \aA$. - Since this determinant annihilates the generators (ie. $det(M)x_i=0$), the resulting enddomorphism $p(\psi)$ is the zero map on the entire module $M$, so: + \vspace{0.5cm} + + Now, notice that we had $det(\Phi) \cdot x_i = 0 ~\forall~ i\in [n]$. + + % next part might be removed + Since $M$ is a fingen $A$-module, any element $m \in M$ can be written as a linear combination of $M$'s generators $x_i$, ie. + $$m = r_1 x_1 + r_2 x_2 + \ldots r_n x_n \in M$$ + + If we multiply $m \in M$ by $d = det(\Phi)$, + \begin{align*} + d \cdot m &= d \cdot (r_1 x_1 + r_2 x_2 + \ldots r_n x_n)\\ + &= r_1(d \cdot x_1) + r_2 (d \cdot x_2) + \ldots + r_n (d \cdot x_n)\\ + (\text{every}~ &d \cdot x_i = det(\Phi)x_i = 0 ~\forall~ i)\\ + &= r_1 (0) + \ldots + r_n (0)\\ + &= 0 + \end{align*} + + Therefore, $det(\Phi) \cdot m = 0$. + % end of might be removed + + The matrix $\Phi$ is the \emph{characteristic matrix}, $xI-A$, viewed as an operator. Then, + $$det(\Phi) = det(xI-A) = p(x)$$ + where $p(x)$ is the \emph{characteristic polynomial}. + + If a linear transformation turns every basis vector ($x_i$) into zero, then that transformation is the zero transformation. So in our case, $det(\Phi)$ is the zero transformation, thus $det(\Phi)=0$. + Therefore, $$\psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n = 0$$ - with $a_i \in \aA$, as stated in the Cayley-Hamilton theorem. + + + %%%%%% OLD START + + % \vspace{3cm} + % + % Kronecker delta: + % $\delta_{ij} = + % \begin{cases} + % 1 & \text{if } i = j,\\ + % 0 & \text{otherwise} + % \end{cases}$ + % + % With the Kronecker delta, $\psi(x_i)$ can be expressed as + % $$\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$$ + % + % The entries of the matrix are \emph{endomorphisms} (elements of the ring $A[\psi]$) + % \begin{itemize} + % \item the term $(\psi - a_{11})$ is an operator that acts on $x_1$; as $(\psi(x_1)-a_{11}\cdot x_1)$ + % \item the term $(-a_{12})$ is an operator that acts on $x_2$; as multiplication by it, ie. $(-a_{12} \cdot x_2)$ + % \end{itemize} + % + % We need a single element $x \in A$ that \emph{annihilates} every $m \in M$ simultaneously, ie. $xM=0$. We're going to use the determinant for getting $x$. + % + % Since $A$ is a commutative ring, and $\psi$ commutes with any $a \in A$, + % the ring of operators $A[\psi]$ is a commutative ring. + % + % $\Longrightarrow~$ so we can treat the matrix as a matrix of real numbers and calculate its determinant. + % + % + % \vspace{0.75cm} + % This is called \emph{"the determinant trick"}.\\ + % We're interested in the determinant because it is the only way to turn a system of multiple equations in a single scalar-like equation that describes the endomorphism $\psi$.\\ + % $\rightarrow$ Because in module theory, we lack of "division", so can not "solve for $\psi$" the system of equations.\\ + % $\rightarrow$ The determinant provides a way to find a polynomial that \emph{annihilates} the module; the \emph{characteristic polynomial}, which related $\psi$ to the ideal $\aA$ + % + % $$det(M) \cdot x_i = 0~~ \forall i$$ + % where $x_i$ are the generators of $M$. + % + % Use $\Phi$ to denote the previous matrix. The determinant is the only function that can take that matrix $\Phi$ and produce a single scalar $x=det(\Phi)$ such that the following identity holds: $adj(\Phi)\cdot \Phi=det(\Phi) \cdot \aA$. + % \vspace{0.5cm} + % + % Since $det(M)$ kills every generator, it must kill every element in $M$\\ + % $\Longrightarrow~~ det(M)$ is the zero map. + % + % Leibniz formula of the determinant of an $n \times n$ matrix: + % $$ + % det(M) = \sum_{\sigma \in S_n} sign(\sigma) \prod_{i=1}^n M_{i, \sigma(i)} + % $$ + % + % so, + % $$(\psi - a_{11}) (\psi - a_{22}) \ldots (\psi - a_{nn})$$ + % expanding it, + % \begin{itemize} + % \item highest power is $1 \cdot \psi^n$ + % \item coefficient of $\psi^{n-1}$ is $-( \underbrace{ a_{11} + a_{22} + \ldots + a_{nn} }_{a_1})$,\\ + % where, since each $a_{ii} \in \aA,~~ a_1 \in \aA$ + % \item the rest of coefficients of $\psi^k$ are also elements in $\aA$ + % \end{itemize} + % + % So we have + % $$p(\psi) = \psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n$$ + % with $a_i \in \aA$. + % + % Since this determinant annihilates the generators (ie. $det(M)x_i=0$), the resulting enddomorphism $p(\psi)$ is the zero map on the entire module $M$, so: + % $$\psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n = 0$$ + % with $a_i \in \aA$, as stated in the Cayley-Hamilton theorem. + + %%%%% OLD END \end{proof} +\vspace{0.5cm} \begin{cor}{AM.2.5} \label{2.5} Let $M$ a fingen $A$-module, let $\aA$ an ideal of $A$ such that $\aA M = M$. @@ -466,7 +581,7 @@ $A[\psi] \subset End M$ is the subring geneerated by $A$ and the action of $\psi $$\frac{\aA M + N}{N} = \left\{ y + N ~~|~~ y \in \aA M +N \right\} = \aA M + N$$ thus every $y \in \aA M +N$ can be written as - $$y=x+n,~~ \text{with} x \in \aA M,~ n\in N$$ + $$y=x+n,~~ \text{with}~ x \in \aA M,~ n\in N$$ which comes from \eqref{eq:2.7.1}. Thus, $y + N = (x+n)+N = x+N$, since $n \in N$ is zero in the quotient. @@ -524,7 +639,7 @@ $A[\psi] \subset End M$ is the subring geneerated by $A$ and the action of $\psi Split exact sequence. TODO \end{prop} -\section{Noetherean rings} +\section{Noetherian rings} \begin{defn}{}{Ascending Chain Condition} A partially orddered set $\Sigma$ has the \emph{ascending chain condition} (a.c.c.) if every chain