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Section Week 7

This is an outline of the topics we covered in the seventh week of class.

Subsection Tuesday 2/24

Coding Theorist of the Day.

Ron Roth, DSc 1988

Reminders/Announcements.

  • Group Contract Submission postponed to next R
  • Today: Do more with operations in finite fields; introduce characteristic, order, primitive elements
  • Thursday: Splitting fields and a \(2\)-error correcting code

Results on Order and Primitive Elements.

Proof.
  • For \(F[x]/p(x)=E\text{;}\) \(\deg(p(x))>\deg(\bar{f}(x))\) for \(\bar{f}(x)\in E\text{.}\)
  • In \(\F_3[x]/(x^2+1)\text{:}\) \(\alpha=x\) is not primitive, but \(\alpha+1\) is primitive
Proof.
Sketch:
  1. If \(a=0\text{,}\) clear. If \(a\neq 0\) then
    \begin{equation*} \{a\alpha \mid \alpha \in F^*\} = \{\alpha \mid \alpha \in F^*\} \end{equation*}
    since multiplication by \(\alpha\) is invertible. So
    \begin{align*} \prod_{\alpha \in F^*} (a\alpha)\amp=\prod_{\alpha \in F^*} (\alpha)\\ a^{|\F^*|} \amp = 1 \end{align*}
  2. Follows from (1) and Degree Mantra.
Proof.
Use the Division Algorithm: \(q-1=b|a|+r\) with \(r\lt|a|\text{.}\) Then
\begin{align*} 1 \amp = a^{q-1}\\ \amp = a^{b|a|+r}\\ \amp = (a^{|a|})^b a^r\\ \amp = a^r \end{align*}
and since \(r\lt|a|\) we must have \(r=0\text{.}\)
Proof.
Sketch: Show that the elements of maximum order are primitive.
  1. Claim: If \(t\) is the max order of any \(\alpha \in F^*\) then for any \(\beta \in F^*\) we have \(|\beta|\mid t\text{.}\)
    Idea: Take \(\alpha\) with \(|\alpha|=t\text{.}\) Let \(|\beta|=s, d=\gcd(s,t)\) and write \(s=bd\) with \(\gcd(b,d)=1\text{.}\) Then
    • Show \(|\alpha \beta^d|\) divides \(bt\)
    • Show \(t\) and \(b\) divide \(|\alpha \beta^d|\) by looking at \((\alpha \beta^d)^{|\alpha\beta^d|b}\) and \((\alpha \beta^d)^{|\alpha\beta^d|t}\)
    • Conclude because of the GCDs that \(d=s\)
  2. Now count: all \(\beta\in F^*\) are roots of \(x^t-x\text{,}\) so we have \(t\geq q-1\text{.}\) But also \(t\leq q-1\) by Order of an Element Divides \(q-1\).

Subsection Thursday 2/26

Coding Theorist of the Day.

Heide Gluesing-Luerssen, PhD Bremen 1991

Reminders/Announcements.

Goals.

Characteristic, Frobenius, Splitting Field.

Recall the binomial theorem:
\begin{equation*} (x+y)^p=\sum_{i=0}^p \binom{p}{i} x^i y^{p-i} \end{equation*}
For the Frobenius map \(f_m: F \to F, \alpha \mapsto \alpha^{p^m}\) where \(\char{F}=p\text{,}\) we have that if \(|F|=p^n\) then \(f_{n-m}\) is \(f_m^{-1}\text{.}\)
Proof.
Sketch: Let \(\alpha,\beta\) be roots of \(Q(x)\) in a splitting field \(E\text{.}\)
  1. \begin{align*} (\alpha + \beta)^{p^n} \amp = \alpha^{p^n}+\beta^{p^n}=\alpha+\beta \\ (\alpha \beta)^{p^n} \amp = \alpha^{p^n}\beta^{p^n}=\alpha\beta \end{align*}
  2. Any finite subset of a field closed under field operations is a field (only requires checking inverses)
  3. \(Q'(x)=p^n x^{p^n}=1=-1\) so \(\gcd(Q(x),Q'(x))=1\) and hence \(Q(x)\) has no repeated roots.
So the roots are a field of size \(p^n\text{.}\)

A 2-error correcting code.

We build off a binary Hamming code.