M algebraic-coding-theory-notes.pdf → algebraic-coding-theory-notes.pdf

M algebraic-coding-theory-notes.tex → algebraic-coding-theory-notes.tex

@@ -34,6 +34,11 @@ @misc{vanddetproof2,
  title = {Abstract Algebra, Theory and Applications. \emph{sec.22.2}},
  author = {Thomas W. Judson},
 }
+
+@misc{commutativealgebranotes,
+ title = {Commutative Algebra Notes},
+ howpublished = "\url{https://github.com/arnaucube/math/blob/master/commutative-algebra-notes.pdf}",
+}
 \end{filecontents}
 \nocite{*}
 
@@ -788,7 +793,7 @@
   $\longrightarrow$ for small $n$, it's fast to locate the received vector $y$ in the array, but for large $n$: use syndrome to find which coset (ie. which row of the array) contains $y$.
 \end{cor}
 
-\begin{eg}{7.9}
+\begin{eg}{7.10}
   (from example \ref{h.6.5})
   $$G= \begin{bmatrix}
     1 & 0 & 1 & 1\\
@@ -1427,7 +1432,7 @@ \begin{defn}{}[Packing problem]
   The \emph{packing problem} for $V(r,q)$ is that of determining the values of $max_s(r, q)$ and the optimal $(n,s)$-sets.
 \end{defn}
 
-\begin{thm}{14.1}
+\begin{thm}{14.1} \label{14.1}
 There exists an $[n,n-r,d]$-code over $GF(q)$ iff $\exists$ an $(n, d-1)$-set in $V(r, q)$.
 \end{thm}
 \begin{proof}
@@ -1691,6 +1696,604 @@
 $$
 = \prod_{i=2}^n (x_i - x_1) ~~\cdot \prod_{2 \leq j < i \leq n} (x_i - x_j) ~~= \prod_{1 \leq j < i \leq n} (x_i - x_j)
 $$
+\end{proof}
+
+
+\vspace{1.0cm}
+
+\section{Cyclic Codes}
+This section contains notes both from Hill \cite{hill} and Lint \cite{lint} books. Hill's notes are enumerated as $X.Y$ while Lint's notes as $X.Y.Z$.
+
+\subsection{Recap on ideals}
+More about ideals at the \emph{Commutative Algebra notes} \cite{commutativealgebranotes}.
+\vspace{0.6cm}
+
+In the cyclic codes context, we're interested in ideals because
+\begin{itemize}
+  \item cyclic codes are ideals of $R_n = \mathbb{F}_q[x]/(x^n-1)$.
+  \item $R_n$ is a PID: every ideal in $R_n$ is principal.
+\end{itemize}
+
+\vspace{0.6cm}
+
+\begin{defn}{}[ideal]
+  $I \subset R$ ($R$ ring) such that $0 \in I$ and $\forall x \in I,~ r \in R,~ xr, rx \in I$.\\
+  \hspace*{2em} ie. $I$ absorbs products in $R$.
+\end{defn}
+
+\begin{defn}{}[prime ideal]
+  if $a, b \in R$ with $ab \in P$ and $P \neq R$ ($P$ a prime ideal), implies $a \in P$ or $b \in P$.
+\end{defn}
+
+\begin{defn}{}[principal ideal]
+  generated by a single element, $(a)$.
+
+  $(a)$: principal ideal, the set of all multiples $xa$ with $x \in R$.
+\end{defn}
+
+\begin{defn}{}[prime spectrum - $Spec(A)$]
+  set of prime ideals of $A$. ie.
+
+  $$Spec(A) = \{ P ~|~ P \subset A~ \text{is a prime ideal} \}$$
+\end{defn}
+
+\begin{defn}{}[integral domain]
+  Ring in which the product of any two nonzero elements is nonzero.
+
+  ie. no zerodivisors.
+
+  ie. $\forall~ 0 \neq a,~ 0 \neq b \in A,~ ab \neq 0 \in A$.
+
+  Every field is an integral domain, not the converse.
+\end{defn}
+
+\begin{defn}{}[principal ideal domain - PID]
+  integral domain in which every ideal is principal. ie.
+  ie. $\forall I \subset R,~ \exists~ a \in I$ such that $I = (a) = \{ ra ~|~ r \in R \}$.
+\end{defn}
+
+
+\vspace{1.0cm}
+
+\subsection{Cyclic Codes intro}
+
+\begin{defn}{}[cyclic code]
+  a code $C$ is cyclic if
+  \begin{enumerate}[i.]
+    \item $C$ is a linear code
+    \item any cyclic shift of a codeword is also a codeword\\
+      ie. when $a_0 a_1 \ldots a_{n-1} \in C$,\\
+      then so $a_{n-1} a_0 a_1 \ldots a_{n-2} \in C$.
+  \end{enumerate}
+\end{defn}
+
+\begin{remark}{}[Notation]
+  From now on,
+  $$F_q = \mathbb{F}_q = GF(q)$$
+
+  $F_q[x]$ is a ring; and for $f(x)$ irreducible in $F_q[x]$, then $\frac{F_q[x]}{f(x)}$ is a field.
+
+  From now on,
+$$f(x) = x^n -1 ~~~\Longrightarrow~~ F_q[x]/(x^n-1) = R_n$$
+
+  \vspace{0.6cm}
+  Observe:\\
+  since $x^n \cong 1 \bmod{x^n -1}$, to reduce any polynomial by modulo $x^n-1$ can just replace $x^n$ by $1$, $x^{n+1}$ by $x$, $x^{n+2}$ by $x^2$, etc.
+
+  \vspace{0.6cm}
+  Correspondence:
+  $$a_0 a_1 \ldots a_{n-1} \in V(n,q) ~\Longleftrightarrow~ a(x)=a_0 + a_1 x + \ldots + a_{n-1} x^{n-1} \in R_n$$
+
+  In $R_n$,
+  \begin{align*}
+    x \cdot a(x) &= a_0 \cdot x + a_1 x^2 + \ldots + \underbrace{a_{n-1} x^n}_{=a_{n-1}\cdot 1}\\
+                 &=a_{n-1} + a_0 x + \ldots + a_{n-2} \cdot x^{n-1} \in R_n
+  \end{align*}
+
+  ie. $a_{n-1} a_0 a_1 \ldots a_{n-2} \in V(n, q)$
+
+  \vspace{0.6cm}
+  In $R_n$:
+  \begin{itemize}
+    \item[$\rightarrow$] multiply by $x$ corresponds to performing a single cyclic shift
+    \item[$\rightarrow$] multiply by $x^m$ corresponds to a cyclcic shift through $m$ positions
+  \end{itemize}
+\end{remark}
+
+
+\vspace{1.0cm}
+\begin{thm}{12.6}[Algebraic characterization of cyclic codes] \label{12.6}
+  a code $C \in R_n$ is a cyclic code iff
+  \begin{enumerate}[i.]
+    \item $a(x),b(x) \in C ~\Longrightarrow~ a(x)+b(x) \in C$
+    \item $a(x) \in C$ and $r(x) \in R_n ~\Longrightarrow~ r(x) a(x) \in C$\\
+      ie. $C$ closed under multiplication by any element of $R_n$
+  \end{enumerate}
+\end{thm}
+\begin{proof}
+  \begin{enumerate}[i.]
+    \item let $C \in R_n$ cyclic code, then $C$ is linear, so (i) holds.
+    \item let $a(x) \in C$ and $r(x) = r_0 + r_1 x + \ldots + r_{n-1} x^{n-1} \in R_n$.
+
+      Since mult. by $x$ corresponds to a cyclic shift, then $x \cdot a(x) \in C$, thus
+      $$x \cdot (x \cdot a(x)) = x^2 \cdot a(x) \in C$$
+      and so on.
+
+      Hence,
+      $$r(x) \cdot a(x) = r_0 \cdot a(x) + r_1 \cdot x \cdot a(x) + \ldots + r_{n-1} \cdot x^{n-1} \cdot a(x) \in C$$
+      since each summand is in $C$; thus (ii) holds.
+
+      \vspace{0.5cm}
+      Conversely, suppose (i) and (ii) hold;\\
+      taking $r(x)$ to be a scalar, the conditions imply that $C$ is linear.\\
+      Taking $r(x)=x$, in (ii) shows that $C$ is cyclic.
+  \end{enumerate}
+\end{proof}
+
+\vspace{0.8cm}
+\begin{remark}{}
+  \begin{enumerate}[i.]
+    \item Cyclic codes are the ideals of the ring $R_n$.
+    \item Every ideal in $R_n$ is a \emph{principal ideal}.\\
+      ie. $R_n$ is a PID.
+  \end{enumerate}
+\end{remark}
+
+\vspace{0.5cm}
+\begin{thm}{6.1.3}
+  a linear code $C$ in $\mathbb{F}_q^n$ is cyclic iff $C$ is an ideal in $\mathbb{F}_q[x]/(x^n -1)$.
+\end{thm}
+\begin{proof}
+  If $C$ ideal in $\mathbb{F}[x]/(x^n -1)$ and $c(x)=c_0 + c_1 x + \ldots + c_{n-1}x^{n-1} \in C$, then
+  \begin{align*}
+    x \cdot c(x) &= c_0 x + c_1 x \cdot x + \ldots + \underbrace{c_{n-1} x^{n-1} \cdot x}_{=c_{n-1} \cdot 1 \pmod{x^n-1}}\\
+                 &= c_{n-1} + c_0 x + c_1 x^2 + \ldots + c_{n-2} x^{n-1} \in C\\
+    \Longrightarrow&~ (c_{n-1}, c_0, c_1, \ldots, c_{n-2}) \in C
+  \end{align*}
+
+  \vspace{0.5cm}
+  Conversely, if $C$ cyclic, then $\forall c(x) \in C,~ x \cdot c(x) \in C$.
+
+  Therefore $x^i \cdot c(x) \in C ~\forall i \in [n]$
+
+  and since $C$ is linear, $a(x) \cdot c(x) \in C$
+
+  ie.
+  $$a_0 \cdot c(x) + a_1 \cdot x \cdot c(x) + a_2 \cdot x^2 \cdot c(x) + \ldots + a_{n-1} \cdot x^{n-1} \cdot c(x) \in C$$
+  $\forall a(x) \in \mathbb{F}_q[x]/(x^n-1)$.
+\end{proof}
+
+\begin{cor}{}
+  $\mathbb{F}_q[x]/(x^n -1)$ is a PID (Principal Ideal Domain).
+
+  Thus $C$ consists of the multiples of $g(x) \in C$ monic and of lowest degree.
+
+  Therefore,
+  \begin{itemize}
+    \item[$\Longrightarrow$] $g(x)$ is the \emph{generator polynomial} (more at \ref{12.9}) of $C$: $C= \langle g(x) \rangle$.
+    \item[$\Longrightarrow$] $g(x)$ is divisor of $x^n -1$, ie. $g(x) | (x^n-1)$.\\
+      since otherwise $gcd(x^n-1, g(x))$ would be a polynomial in $C$ with degree $< deg(g(x))$, a contradiction.
+
+  \end{itemize}
+\end{cor}
+\vspace{0.5cm}
+
+\vspace{1.0cm}
+Let $f(x) \in R_n$, $\langle f(x) \rangle$ the subset of $R_n$ consisting of all multiples of $f(x)$ (reduced modulo $x^n-1$). ie.
+$$\langle f(x) \rangle = \{ r(x) \cdot f(x) ~|~ r(x) \in R_n \}$$
+
+
+\begin{thm}{12.7}
+  for any $f(x) \in R_n$, the set $\langle f(x) \rangle$ is a cyclic code; named the \emph{code generated by $f(x)$}.
+\end{thm}
+\begin{proof}
+  check conditions (i), (ii) of Theorem \ref{12.6}:
+  \begin{enumerate}[i.]
+    \item if $a(x)f(x),~ b(x)f(x) \in \langle f(x) \rangle$ then
+      $$a(x)f(x)+b(x)f(x)= (a(x)+ b(x)) f(x) \in \langle f(x) \rangle$$
+
+    \item if $a(x)f(x) \in \langle f(x) \rangle$ and $r(x) \in R_n$, then
+      $$r(x) (a(x)f(x))= (r(x) \cdot a(x)) f(x) \in \langle f(x) \rangle$$
+  \end{enumerate}
+\end{proof}
+
+\begin{eg}{12.8}
+  $C=\langle 1+x^2 \rangle \in R_3$, with $GF(2)$.
+
+  $R_3$ has $2^3=8$ elements.
+
+  Multiply $1+x^2$ by each of the 8 elements of $R_3$ (reducing modulo $x^3-1$), obtaining 4 distinct codewords:
+  $$0, 1+x, 1+x^2, x+x^2$$
+  Thus $C$ is the code $\{ 000, 110, 101, 011 \}$.
+\end{eg}
+
+
+\subsection{Generator polynomial and check polynomial}
+
+\begin{thm}{12.9} \label{12.9}
+  let $C$ a nonzero cyclic coded in $R_n$. Then
+  \begin{enumerate}[i.]
+    \item $\exists !~ g(x)$ monic, of smallest degree in $C$
+    \item $C=\langle g(x) \rangle$
+    \item $g(x)$ is a factor of $x^n -1$
+  \end{enumerate}
+\end{thm}
+\begin{proof}
+  \begin{enumerate}[i.]
+    \item $\exists !~ g(x)$ monic, of smallest degree in $C$:\\
+      let $g(x), h(x) \in C$ monic and of smallest degree. Then $g(x)-h(x) \in C$ with smaller degree, which is a contradiction if $g(x) \neq h(x)$
+
+    \item $C=<g(x)>$:\\
+      let $a(x) \in C$. By division algorithm for $F_q[x]$,
+      $$a(x) = q(x) g(x) + r(x)$$
+      with $deg(r(x)) < deg(g(x))$.
+
+      But $r(x) = a(x) - q(x)g(x) \in C$ by the properties of cyclic codes (Theorem \ref{12.6}, $a,g \in C$ and $C$ an ideal)
+
+      By minimality of $deg(g(x))$, must have $r(x) = 0$, therefore
+      $$a(x) \in \langle g(x) \rangle \forall a(x) \in C$$
+      $$\Longrightarrow~ C=\langle g(x) \rangle$$
+
+    \item $g(x)$ is a factor of $x^n -1$:\\
+      by division algorithm,
+      $$x^n -1 = q(x) g(x) + r(x)$$
+      with $deg(r(x)) < deg(g(x))$.
+
+      But then $r(x) = -q(x) g(x) \bmod{x^n -1}$, ie. $r(x)$ is a multiple of $g(x)$, and so $r(x) \in \langle g(X) \rangle$.
+
+      By the minimality of $deg(g(x))$, we must have $r(X)=0$, which implies
+      $$x^n -1 = q(x) \cdot g(x) + 0$$
+      $\Longrightarrow~ g(x)$ is a factor of $(x^n-1)$.
+  \end{enumerate}
+\end{proof}
+
+
+\begin{defn}{}[Generator polynomial]
+  let $C$ cyclic code, monic polynomial of least ddegree (given by Theorem \ref{12.9}) is called the \emph{generator polynomial} of $C$.
+
+  Remark $C$ may contain other polynomials than the generator polynomial which also generate $C$.
+\end{defn}
+
+
+\begin{eg}{12.10}
+  All cyclic codes of length $3$:\\
+  ie. ideals in $\mathbb{F}_2[x]/(x^3-1) ~\Longrightarrow$ factor $x^3-1$ in $\mathbb{F}_2$, every cyclic code corresponds to a divisor of $x^3-1$.
+
+  $$x^3-1 = (x+1) \cdot (x^2+x+1)$$
+  so by Theorem \ref{12.9}(ii):
+
+  \begin{center}
+    \begin{tabular}{ c c c }
+      generator poly & code in $R_3$ & Corresponding code in $V(3,2)$\\
+      \hline
+      1 & all of $R_3$ & all of $V(3,2)$ \\ 
+      $x+1$ & $\{0, 1+x, x+x^2,1+x^2\}$ & $\{000, 110, 011, 101\}$ \\  
+      $x^2+x+1$ & $\{0, 1+x+x^2\}$ & $\{000, 111\}$ \\  
+      $x^3-1$ & $\{0\}$ & $\{000\}$
+    \end{tabular}
+  \end{center}
+
+  Notice that the 2nd column contains the multiples $\pmod{x^3-1}$ of the polynomials at the 1st column.
+\end{eg}
+
+\begin{lemma}{12.11}
+  let $g(x)=g_0 + g_1 + \ldots + g_r x^r$ be the generator polynomial. Then $g_0$ is nonzero.
+\end{lemma}
+\begin{proof}
+  by contradiction, let $g_0=0$. Then $x^{n-1} \cdot g(x) = x^{-1} \cdot g(x) \in C$ with degree $r-1$.
+
+  This contradicts minimality of $deg(g(x))$; thus $g_0 \neq 0$.
+\end{proof}
+
+\begin{thm}{12.12} \label{12.12}
+  let $C$ cyclic code with generator polynomial $g(x) = g_0 + g_1 x + \ldots + g_r x^r$. Then $dim(C)=n-r$, and a GM for $C$ is
+  $$
+  G= \begin{pmatrix}
+    g_0 & g_1 & g_2 & \cdots & g_r & 0 & 0 & \cdots & 0\\
+    0 & g_0 & g_1 & g_2 & \cdots & g_r & 0 & \cdots & 0\\
+    0 & 0 & g_0 & g_1 & g_2 & \cdots & g_r & \cdots & 0\\
+      & \vdots & & & & & & \vdots\\
+    0 & 0 & \cdots & 0 & g_0 & g_1 & g_2 & \cdots & g_r\\
+  \end{pmatrix}
+  $$
+\end{thm}
+\begin{proof}
+  the $n-r$ rows of $G$ are linear independent because the echelon of nonzero $g_0$'s with $0$'s below.
+
+  These $n-r$ represent the codewords $g(x), x \cdot g(x), x^2 g(x), \ldots, x^{n-r-1}g(x)$.
+
+  It remains to show that every codeword in $C$ can be expressed as a linear combination of them.
+
+  From Theorem \ref{12.9}.(ii), if $a(x) \in C$ then $a(x)=q(x)\cdot g(x)$ for $q(x) \in \mathbb{F}_q[x]$.
+
+  Since $deg(a(x)) <n,~~ deg(q(x)) < n-r$.
+
+  Hence
+  \begin{align*}
+    q(x)g(x) &= (q_0+q_1 x + \ldots + q_{n-r-1} x^{n-r-q}) \cdot g(x)\\
+             &= q_0 g(x)+q_1 x g(x) + \ldots + q_{n-r-1} x^{n-r-q} g(x)
+  \end{align*}
+  which is the desired linear combination.
+\end{proof}
+
+
+\begin{defn}{}[check polynomial]
+  For generator polynomial $g(x)$, which we know is a factor of $x^n -1$.
+
+  Then, the \emph{check polynomial} $h(x)$ is such that
+  $$x^n-1 = g(x) h(x)$$
+
+  Since $g(x)$ is monic, so also is $h(x)$. By Theorem \ref{12.12},
+  $$deg(g(x)) = n-k ~~\Longrightarrow~~ deg(h(x))=k$$
+\end{defn}
+
+\begin{thm}{12.14} \label{12.14}
+  $c(x) \in R_n$ is $c(x) \in C$ iff $c(x) \cdot h(x) =0$.
+\end{thm}
+\begin{proof}
+  in $R_n: ~ g(x) \cdot h(x) = x^n -1 = 0 \bmod{x_n -1}$,\\
+  hence
+  $$c(x) \in C ~\stackrel{implies}{\Longrightarrow}~ c(x) = a(x) \cdot g(x)$$
+  for some $a(x) \in R_n$.
+
+  \begin{align*}
+    \Longrightarrow~ c(x) \cdot h(x) &= \overbrace{a(x) \cdot g(x)}^{c(x)} \cdot h(x)\\
+    (\text{note that}~ g(x) &\cdot h(x)=0)\\
+                                     &= a(x) \cdot 0 = 0
+  \end{align*}
+  $\Longrightarrow~ c(x) \cdot h(x) = 0$ as stated.
+
+
+  \vspace{0.6cm}
+  Conversely, suppose $c(x) \cdot h(x) = 0$.
+
+  By division algorithm, $c(x) = q(x) \cdot g(x) + r(x)$ with $deg(r) < n-k$.
+
+  Then, $c(x) \cdot h(x) = 0$ implies
+  \begin{align*}
+    (q(x) \cdot g(x) + r(x)) \cdot h(x) &= 0\\
+    q(x) \cdot g(x) \cdot h(x) + r(x) \cdot h(x) &= 0\\
+    \text{recall}~ g(x)\cdot h(x) = x^n -1,~ \text{thus,}&\\
+    0 + r(x) \cdot h(x) &=0\\
+    r(x) \cdot h(x) &= 0\\
+    \text{ie.} r(x) \cdot h(x) = 0 \pmod{x^n-1}
+  \end{align*}
+
+  But $deg(r(x) \cdot h(x)) < n-k +k = n$, so $r(x) \cdot h(x)= 0$ in $\mathbb{F}_q$.
+
+  Hence $r(x)=0$, thus
+  \begin{align*}
+    c(x) &= q(x) \cdot g(x) \in C\\
+         &\Longrightarrow~ c(x) \in C
+  \end{align*}
+
+  since $C$ is an ideal of $R_n$ and $g(x) \in C$ and $q(x) \in R_n$.
+\end{proof}
+
+\subsection{Maximal and minimal cyclic codes}
+
+Let $x^n-1 = f_1(x) \cdot f_2(x) \cdot \ldots \cdot f_t(x)$, with $f_i(x)$ irreducible polynomials.
+
+\begin{defn}{6.1.6}[maximal and minimal cyclic codes]
+  the cyclic code generated by $f_i(x)$ is called a \emph{maximal cyclic code} (since it's a maximal ideal), denoted $M_i^+$.
+
+  The code generated by $(x^n-1)/f_i(x)$ is called a \emph{minimal cyclic code}, denoted $M_i^-$. Also named \emph{irreducible code}.
+\end{defn}
+
+\vspace{0.5cm}
+
+\begin{defn}{}[Zeros of a cyclic code]
+  Let $x^n-1= f_1(x) \cdot \ldots \cdot f_t(x)$. Let $\beta_i$ be a zero of $f_i(x)$ in some extension field of $\mathbb{F}_q$\\
+  ie. $f_i(\beta)=0$, and $f_i(x)$ is the minimal polynomial of $\beta_i$.
+\end{defn}
+
+$\Longrightarrow~ M_i^+$ is the set of polynomials $c(x)$ for which $c(\beta_i)=0$.
+
+In general, a cyclic code can be specified by requiring that all codewords have certain zeros: require that all codewords have $\beta_i$ points as zeros. For example:
+
+Start by any set $\alpha_1, \alpha_2, \ldots, \alpha_s$. Define the code $C$ by
+$$c(x) \in C ~\text{iff}~ c(\alpha_i)=0 ~\forall i=1,2, \ldots, s$$
+then $C$ is cyclic, and the generator polynomial is the least common multiple (lcm) of the minimal polynomials of $\alpha_1, \alpha_2, \ldots, \alpha_s$.
+
+\subsection{Hamming cyclic codes}
+\begin{thm}{6.3.1}
+  let $n := (q^m - 1)/(q-1)$, let $\beta$ a primitive $n$-th root of unity in $\mathbb{F}_{q^m}$. Let $(m, q-1)=1$.
+
+  The cyclic code $C:= \{ c(x) | c(\beta)=0 \}$ is equivalent to the $[, n-m]$ Hamming code over $\mathbb{F}_q$.
+\end{thm}
+\begin{proof}
+  since $n=(q-1) \cdot (q^{m-2} + 2 q^{m-3} + \ldots + m-1) + m$ have $(n, q-1) = (m, q-1) = 1$.
+
+  Therefore $\beta^{i(q-1)} \neq 1 ~\forall i \in [n]$, ie. $\beta^i \neq \mathbb{F}_q ~\forall i \in [n]$.
+
+  Columns of $H$, which are the representations of $1, \beta, \beta^2, \ldots, \beta^{n-1}$ as vectors in $\mathbb{F}_{q^m}$, are pairwise linearly independent over $\mathbb{F}_q$.
+
+  So $H$ is the PCM of an $[n, n-m]$ Hamming code.
+\end{proof}
+
+\begin{eg}{6.3.2}
+  smallest extension of $\mathbb{F}_2$ which contains a primitive 9-th root of unity is $\mathbb{F}_{2^6}$.
+
+  Let $\alpha$ a primitive element of the field, then $\alpha^63=1$ ,and $\beta := \alpha^7$ is a primitive 9-th root of unity.
+
+  Then, the minimal polynomial of $\beta$ has the zeros
+  $$\beta, \beta^2, \beta^4, \beta^8, \beta^{16}=\beta^7, \beta^{14}=\beta^5$$
+
+  $\Longrightarrow$ min. polynomial: $(x^9 - 1)/(x^3 -1) = x^6 + x^3 + 1$.
+
+  So $(x^9-1) = \underbrace{(x-1)}_{f_1(x)} \underbrace{(x^2+x+1)}_{f_2(x)} \underbrace{(x^6+x^3 + 1)}_{f_3(x)}$.
+
+  The code $M_3^+$ has pairwise independent columns in $H$, ie. min.distance $\geq 3$.
+
+  $M_3^+$ consists of codewords $(c_0 c_1 c_2 ~~ c_0 c_1 c_2 ~~ c_0 c_1 c_2)$, so $d=3$.
+
+  The code $M_3^-$ has check polynomial $x^6 + x^3 + 1$, so it's a $[9, 6]$-code.
+
+  Since $x^3-1 \in M_3^-$ (is a codeword), the distance is $2$.
+\end{eg}
+
+
+\subsection{Idempotent of a cyclic code}
+
+\begin{thm}{6.4.1}
+  $\exists!~ c(x) \in C$ which is an identity element for $C$.
+\end{thm}
+\begin{proof}
+  let $g(x)$ generator polynomial, $h(x)$ check polynomial.
+
+  ie. $g(x) \cdot h(x) = x^n -1$ in $\mathbb{F}_q[x]$.
+
+  Since $x^n -1$ has no multiple zeros (ie. has simple zeros),\\
+  \hspace*{2em} we have $(g(x), h(x))=1$, hence by the Bezout identity:
+  $$\exists~ a(x), b(x) \in \mathbb{F}_q[x] ~\text{s.th.}~ a(x) g(x) + b(x) h(x) = gcd(g, h)=1$$
+
+  Define $c(x) := a(x) g(x) = 1-b(x) h(x)$.
+
+  Clearly $c(x) \in C$. Furthermore, if $p(x) g(x) \in C$ (ie. $p(x)\cdot g(x)$ is any codeword in $C$), then
+  \begin{align*}
+    c(x) p(x) g(x) &= \underbrace{(1-b(x)h(x))}_{c(x)} p(x) g(x)\\
+                   &= p(x)g(x) - b(x)h(x)p(x)g(x)\\
+                   &(recall: g(x)h(x)=x^n-1)\\
+                   &= p(x) g(x) - b(x) p(x) \underbrace{g(x) h(x)}_{=x^n-1}\\
+                   &= p(x) g(x) \pmod{x^n -1}
+  \end{align*}
+
+  So $c(x)$ is an identity element for $C$, and hence is unique.
+\end{proof}
+
+Recall:
+\begin{itemize}
+  \item[nilpotent] $a \in A$ s.th. $a^n=0$ for $n>0$
+  \item[idempotent] $e \in A$ s.th. $e^2 = e$
+\end{itemize}
+
+\begin{defn}{6.4.2}[primitive idempotent]
+  the idempotent of an irreducible cyclic code $M_i^-$ is called a \emph{primitive idempotent}, denoted by $\theta_i(x)$.
+\end{defn}
+
+\begin{thm}{6.4.3}
+  if $C_1, C_2$ are cyclic codes with idempotents $c_1(x), c_2(x)$, then
+  \begin{enumerate}[i.]
+    \item $C_1 \cap C_2$ has idempotent $c_1(x) \cdot c_2(x)$
+    \item $C_1 + C_2$, ie. the set of all words $a+b$ with $a \in C_1,~ b \in C_2$, has idempotent $c_1(x) + c_2(x) + c_1(x) \cdot c_2(x) \in C_1+C_2$.
+  \end{enumerate}
+\end{thm}
+
+\begin{thm}{6.4.4}
+  for the primitive idempotent we have:
+  \begin{enumerate}[i.]
+    \item $\theta_i(x) \theta_j(x)=0$ if $i \neq j$
+    \item $\sum_{i=1}^t \theta_i(x) = 1$
+    \item $1+ \theta_{i_1}(x) + \theta_{i_2}(x) + \ldots + \theta_{i_r}(x)$ is the idempotent of the code with generator $f_{i_1}(x) \cdot f_{i_2}(x) \cdot \ldots \cdot f_{i_r}(x)$.
+  \end{enumerate}
+\end{thm}
+
+\subsection{Trace representation}
+
+\begin{defn}{1.1.29}
+  $q=p^r$, trace: $Tr: \mathbb{F}_q \longrightarrow \mathbb{F}_p$.
+
+  If $\zeta \in \mathbb{F}_q$, then
+  $$Tr(\zeta) := \zeta + \zeta^p + \zeta^{p^2} + \ldots + \zeta^{p^{r-1}} \in \mathbb{F}_p$$
+\end{defn}
+
+\begin{thm}{1.1.30} \label{1.1.30}
+  properties of the trace function:
+  \begin{enumerate}[i.]
+    \item $\forall \zeta \in \mathbb{F}_q, ~~ Tr(\zeta) \in \mathbb{F}_p$
+    \item there are elements $\theta \in \mathbb{F}_p$ such that $Tr(\zeta) \neq 0$
+    \item $Tr$ is a linear mapping.
+  \end{enumerate}
+\end{thm}
+\begin{proof}
+  \begin{enumerate}[i.]
+    \item by definition $(Tr(\zeta))^p = Tr(\zeta)$
+    \item the equation $x + x^p + \ldots + x^{p^{r-1}}=0$ can not have $q$ roots in $\mathbb{F}_q$
+    \item since $(\zeta + \eta)^p = \zeta^p + \eta^p$ and $\forall a \in \mathbb{F}_p$ we have $a^p = a$
+  \end{enumerate}
+\end{proof}
+
+\begin{cor}{}
+  $Tr$ takes every value $p^{-1}q$ times, and the polynomial $x + x^p + \ldots + x^{p^{r-1}}$ is a product of minimal polynomials.
+\end{cor}
+
+\begin{thm}{6.5.1}
+  let $k$ be the multiplicative order of $p mod n$, $q = p^k$, $\beta$ a primitive n-th root of unity in $\mathbb{F}_q$.
+
+  Then the set
+  $$V := \{ c(\zeta) := ( Tr(\zeta),~ Tr(\zeta \beta),~ \ldots,~ Tr(\zeta \beta^{n-1})) ~|~ \zeta \in \mathbb{F}_q \}$$
+  is an $[n, k]$ irreducible code over $\mathbb{F}_p$.
+\end{thm}
+\begin{proof}
+  by Theorem \ref{1.1.30}, $V$ is a linear code, and $c(\zeta \beta^-1)$ is a cyclic shift of $c(\zeta)$; hence $V$ is a cyclic code.
+
+  Since $\beta$ is in no subfield of $\mathbb{F}_q$, we know that $\beta$ is a zero of an irreducible polynomial $h(x) = h_0 + h_1 x + \ldots + h_k x^k$.
+
+  If $c(\zeta) = (c_0, c_1, \ldots, c_{n-1})$, then
+  $$\sum_{i=0}^k c_i h_i = Tr(\zeta h(\beta)) = Tr(0) = 0$$
+  since $\beta$ is a zero of $h(x)$, so that $h(\beta)=0$.
+
+
+  So we observer $c(x) h(x) = 0$, from Theorem \ref{12.14} this means that $h(x)$ is the check polynomial,\\
+  ie. we have a parity check equation for the code $V$.
+
+  Since $h(x)$ irreducible, $x^k \cdot h(x^{-1})$ is the check polynomial for $V$, therefore, $V$ is an irreducible cyclic $[n, k]$-code.
+\end{proof}
+
+\subsection{Mattson-Solomon Polynomial}
+
+\begin{defn}{}[Mattson-Solomon polynomial]
+Let $\beta$ primitive n-th root of unity in the extension field $E$ of $\mathbb{F}_q$.
+
+Let $T$ the set of polynomials over $E$ with degree at most $n-1$, ie.
+$$T = \{ p(x) \in E ~|~ deg(p(x)) \leq n-1 \}$$
+
+Define $\Phi: T \longrightarrow T$ as:
+
+let $a(x) \in T$, then $A(X) = (\Phi_a)(X)$ is defined by
+$$A(X) := \sum_{j=1}^n a(\beta^j) X^{n-j}$$
+
+If $a=(a_0, a_1, \ldots, a_{n-1})$, the polynomial $A(X)$ obtained from $a_0 + a_1 x + \ldots a_{n-1} x^{n-1}$ is called the \emph{Mattson-Solomon polynomial} of the vector $a$.
+\end{defn}
+
+\begin{lemma}{6.5.3} \label{6.5.3}
+  the inverse of $\Phi$ is given by
+  $$a(x) = n^{-1}(\Phi A)(x^{-1}) \pmod{x^n -1}$$
+\end{lemma}
+\begin{proof}
+  $$A(\beta^k) = \sum_{j=1}^n \sum_{i=0}^{n-1} a_i \beta^{ij} \beta^{-kj} = \sum_{i=0}^{n-1} a_i \sum_{j=1}^n \beta^{(i-k)j}=n a_k$$
+\end{proof}
+
+\begin{remark}{}
+  Notation:
+  \begin{itemize}
+    \item[] $\circ$: multiplication of polynomials $mod (x^n -1)$
+    \item[] $*$: defined by
+      $$(\sum a_i x^i) * (\sum b_i x^i) := \sum a_i b_i x^i$$
+  \end{itemize}
+
+  $\Phi$ is an isomorphism of the ring $(T, +, \circ)$ onto the ring $(T, +, *)$.
+\end{remark}
+
+\begin{lemma}{6.5.4}
+  let $V$ a cyclic code over $\mathbb{F}_q$ generated by
+  $$g(x) = \prod_{k \in K} (x - \beta^k)$$
+
+  Suppose $\{1, 2, \ldots, d-1 \} \subset K$ and $a \in V$.
+
+  Then the degree of the Mattson-Solomon polynomial $A$ of $a$ is at most $n-d$.
+\end{lemma}
+\begin{proof}
+  $a(\beta^j) = 0$ for $j \in [d]$ since $g(x)|a(x)$.
+
+  Thus at the equation $A(X) := \sum_{j=1}^n a(\beta^j) X^{n-j}$, it will be $0$  between $1 \leq j \leq d-1$, so that degree is $n-d$.
+\end{proof}
+
+\begin{thm}{6.5.5}
+  if there are $r$ n-th roots of unity which are zeros of the Mattson-Solomon polynomial $A$ of a word $a$, then $w(a) = n-r$.
+\end{thm}
+\begin{proof}
+  consequence of Lemma \ref{6.5.3}.
 \end{proof}
 
 \bibliographystyle{unsrt}

M commutative-algebra-notes.pdf → commutative-algebra-notes.pdf

M commutative-algebra-notes.tex → commutative-algebra-notes.tex

@@ -225,7 +225,7 @@ Let $p(X) \in I$ be of minimal degree among nonzero elements of $I$.
 
   Observe that $(p(X)) \subseteq I$, because $p(X) \in I$ and $I$ is an ideal.
 
-  Let $f(X) \in I$. By Euclidean division algorithm in $K[X]$, $\exists q, r \in K[X]$ such that $f(X) = q(X) \cdot p(X) + r(X)$ with eithr $r(X)=0$ or $deg(r) < deg(p)$.
+  Let $f(X) \in I$. By Euclidean division algorithm in $K[X]$, $\exists q, r \in K[X]$ such that $f(X) = q(X) \cdot p(X) + r(X)$ with either $r(X)=0$ or $deg(r) < deg(p)$.
 
   Since $f,p \in I$, then $r(X) = f(X) - q(X)\cdot p(X) \in I$