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

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

@@ -1703,9 +1703,60 @@ \vspace{1.0cm}
 
 \section{Cyclic Codes}
 
+\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}
+\begin{defn}{}[cyclic code]
   a code $C$ is cyclic if
   \begin{enumerate}[i.]
     \item $C$ is a linear code
@@ -1717,7 +1768,7 @@ \end{defn}
 
 \begin{remark}{}[Notation]
   From now on,
-  $$F = F_q = \mathbb{F} = GF(q)$$
+  $$F_q = \mathbb{F}_q = GF(q)$$
 
   $F_q[x]$ is a ring; and for $f(x)$ irreducible in $F_q[x]$, then $\frac{F_q[x]}{f(x)}$ is a field.
 
@@ -1778,6 +1829,7 @@ 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 $R_n$.
@@ -1816,6 +1868,8 @@ 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.]
@@ -1857,58 +1911,124 @@
 \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 $C$.
 
-  Note: $C$ may contain other polynomials than the generator polynomial which also generate $C$.
+  Remark $C$ may contain other polynomials than the generator polynomial which also generate $C$.
 \end{defn}
 
-\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}
+\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 $x^3-1$.
 
-\vspace{0.6cm}
+  $$x^3-1 = (x+1) \cdot (x^2+x+1)$$
+  so by Theorem \ref{12.9}(ii):
 
-\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{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}
 
-\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}
+  Notice that the 2nd column contains the multiples $\pmod{x^3-1}$ of the polynomials at the 1st column.
+\end{eg}
 
-\begin{defn}{}[principal ideal]
-  generated by a single element, $(a)$.
+\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 $r-1$.
 
-  $(a)$: principal ideal, the set of all multiples $xa$ with $x \in R$.
-\end{defn}
+  This contradicts minimality of $deg(g(x))$; thus $g_0 \neq 0$.
+\end{proof}
 
-\begin{defn}{}[prime spectrum - $Spec(A)$]
-  set of prime ideals of $A$. ie.
+\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.
 
-  $$Spec(A) = \{ P ~|~ P \subset A~ \text{is a prime ideal} \}$$
-\end{defn}
+  These $n-r$ represent the codewords $g(x), x \cdot g(x), x^2 g(x), \ldots, x^{n-r-1}g(x)$.
 
-\begin{defn}{}[integral domain]
-  Ring in which the product of any two nonzero elements is nonzero.
+  It remains to show that every codeword in $C$ can be expressed as a linear combination of them.
 
-  ie. no zerodivisors.
+  From Theorem \ref{12.9}.(ii), if $a(x) \in C$ then $a(x)=q(x)\cdot g(x)$ for $q(x) \in \mathbb{F}_q[x]$.
 
-  ie. $\forall~ 0 \neq a,~ 0 \neq b \in A,~ ab \neq 0 \in A$.
+  Since $deg(a(x)) <n,~~ deg(q(x)) < n-r$.
 
-  Every field is an integral domain, not the converse.
-\end{defn}
+  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}{}[principal ideal domain - PID]
-  integral domain in which every ideal is principal. ie.
-  ie. $\forall I \subset R,~ \exists~ a \in I$ such that $I = (a) = \{ ra ~|~ r \in R \}$.
+
+\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}
 
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