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}
@@ -1692,6 +1697,218 @@ $$
 = \prod_{i=2}^n (x_i - x_1) ~~\cdot \prod_{2 \leq j < i \leq n} (x_i - x_j) ~~= \prod_{1 \leq j < i \leq n} (x_i - x_j)
 $$
 \end{proof}
+
+
+\vspace{1.0cm}
+
+\section{Cyclic Codes}
+
+\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,
+  $$F = F_q = \mathbb{F} = GF(q)$$
+
+  $F_q[x]$ 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}
+
+\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{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}
+
+
+\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 $C$.
+
+  Note: $C$ 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}
+
+\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}
+
+
 
 \bibliographystyle{unsrt}
 \bibliography{algebraic-coding-theory-notes.bib}

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 $r(X) = f(X) - q(X)\cdot p(X) \in I$