Week 7

Section Week 7

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

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Subsection Tuesday 2/24

Coding Theorist of the Day.

Ron Roth, DSc 1988

At Technion in Haifa
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Broad work: MDS codes, rank metric, field structure, constrained coding
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Author of Introduction to Coding Theory
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Reminders/Announcements.

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

Proposition 36. Degree Mantra.

A polynomial of degree \(t\) over a field \(F\) has at most \(t\) roots in \(F\text{.}\)

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Proof.

Apply the division algorithm and induct on the degree.

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For \(F[x]/p(x)=E\text{;}\) \(\deg(p(x))>\deg(\bar{f}(x))\) for \(\bar{f}(x)\in E\text{.}\)
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In \(\F_3[x]/(x^2+1)\text{:}\) \(\alpha=x\) is not primitive, but \(\alpha+1\) is primitive
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Lemma 37.

Let \(F\) be a finite field.

For \(a \in F, a^{|F|}=a\)
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If \(a \in \GF(q^m)\text{,}\) then \(a\in \GF(q)\) if and only if \(a^q=a\text{.}\)
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Proof.

Sketch:

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*}

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Follows from (1) and Degree Mantra.
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Lemma 38. Order of an Element Divides \(q-1\).

Let \(a\in\GF(q)\text{.}\) Then \(|a|\) divides \(q-1\text{.}\)

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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 \(r=0\text{.}\)

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Theorem 39. Finite Fields have a Primitive Element.

Let \(F=\GF(q)\text{.}\) Then there exists a primitive element \(\gamma\in F\text{.}\)

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Proof.

Sketch: Show that the elements of maximum order are primitive.

Claim: If \(t\) is the max order of any \(\alpha \in F^*\) then for any \(\beta \in F^*\) we have \(|\beta|\mid t\text{.}\)
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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 \(bt\)
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- Show \(t\) and \(b\) divide \(|\alpha \beta^d|\) by looking at \((\alpha \beta^d)^{|\alpha\beta^d|b}\) and \((\alpha \beta^d)^{|\alpha\beta^d|t}\)
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- Conclude because of the GCDs that \(d=s\)
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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 \(q-1\).
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Subsection Thursday 2/26

Coding Theorist of the Day.

Heide Gluesing-Luerssen, PhD Bremen 1991

At University of Kentucky
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Transitioned research to coding theory from system control theory post PhD
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Weight enumerators, cyclic codes, rank metric & subspace coding
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My thesis advisor
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Reminders/Announcements.

Office hours today only until 4
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Group Contract information posted
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Problem Set M4 posted; due 3/10. Submit by 3/5 for feedback before exam
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Exam 2 on T 3/10 over modules 3 & 4 (through today)
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Goals.

Establish the existence of all finite fields
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Give a construction for a \(2\)-error correcting code (an alternant code)
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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*}

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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 \(f_m^{-1}\text{.}\)

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Theorem 40. Fields of Order \(p^n\) Exist.

The roots of \(Q(x)=x^{p^n}-x\in\F_p[x]\) in one of its splitting fields form a field of order \(p^n\text{.}\)

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Proof.

Sketch: Let \(\alpha,\beta\) be roots of \(Q(x)\) in a splitting field \(E\text{.}\)

\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*}

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Any finite subset of a field closed under field operations is a field (only requires checking inverses)
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\(Q'(x)=p^n x^{p^n}=1=-1\) so \(\gcd(Q(x),Q'(x))=1\) and hence \(Q(x)\) has no repeated roots.
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So the roots are a field of size \(p^n\text{.}\)

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A 2-error correcting code.

We build off a binary Hamming code. Remember \(C_{H,4}\) is the \([15,11,3]_2\) code with parity check matrix whose \(i\)th column is the binary representation of \(i\text{.}\)

\begin{equation*} H=\begin{bmatrix} 0 \amp 0 \amp \ldots \amp 1\\ 0 \amp 0 \amp \ldots \amp 1 \\ 0 \amp 1 \amp \ldots \amp 1\\ 1 \amp 0 \amp \ldots \amp 1 \end{bmatrix} =\begin{bmatrix} 1 \amp 2 \amp \ldots \amp 15\end{bmatrix}\text{.} \end{equation*}

Our idea is that \(C_H\) uses \(r\) bits to fix 1 error in \(2^r-1\) bits. So we should be able to use \(2r\) bits to fix 2 errors in \(2^r-1\) bits.

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We will make a new parity-check matrix \(H\) by coming up with a function \(f\) on the entries of the first row of our original matrix so that syndromes corresponding to 2-error patterns are decodable.

\begin{equation*} H=\begin{bmatrix} 1 \amp 2 \amp \ldots \amp 15 \\ f(1)\amp f(2)\amp \ldots \amp f(15)\end{bmatrix} \end{equation*}

which is \(8\times 15\) as a matrix over \(\F_2\) and \(2\times 15\) as a matrix over \(\F_{2^4}\text{.}\)

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A syndrome for an error of weight 2 is

\begin{equation*} \begin{bmatrix} s_1\\s_2\end{bmatrix}=H^i+H^j = \begin{bmatrix} i +j \\ f(i)+f(j)\end{bmatrix}\text{.} \end{equation*}

We will use the correspondence between \(\F_2^4\) and \(\F_{2^4}\) to write

\begin{equation*} H=\begin{bmatrix} \alpha_1 \amp \alpha_2 \amp \ldots \amp \alpha_15 \\ f(\alpha_1)\amp f(\alpha_2)\amp \ldots \amp f(\alpha_15)\end{bmatrix}\text{.} \end{equation*}

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