Proposition 34. Properties of Irreducibles.
If \(p(x)\) is irreducible over a field \(F\) then
Section Week 6
This is an outline of the topics we covered in the sixth week of class.
Subsection Tuesday 2/17
Coding Theorist of the Day.
Christine Kelley, PhD 2006
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At Univ. Nebraska Lincoln
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Low density parity check codes
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Had a rough start in Calculus I but was encouraged to major in math and go to grad school by Rob Beezer (lead developer of PreTeXt - the typesetting system for these notes)
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Director of MAA Project NExT
Announcements.
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Exam 1 reattempt Thursday in class. Roughly comparable difficulty and theme
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Exam Reattempt Survey available
Notes.
Goal: Build and do computations in finite fields of non-prime order.
Idea: Mimic the construction from \(\Z\) of \(\Z_p\) but with polynomials \(F[x]\text{.}\) They are almost a field except for lacking inverses for most elements.
Activity Problems.
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\(f(x)=x^4+x^2+x+1, g(x)=x^3+1\text{.}\) We have\begin{align*} x^4+x^2+x+1 \amp = x(x^3+1)+x^2+1\\ x^3+1 \amp =x(x^2+1)+x+1\\ x^2+1 \amp =(x+1)(x+1)+0 \end{align*}so \(\gcd(f,g)=x+1\)
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Now we reverse this to find \(t(x),s(x)\text{.}\)\begin{align*} x+1 \amp = x^3+1+x(x^2+1)\\ \amp = x^3+1+x((x^4+x^2+x+1)+x(x^3+1))\\ \amp = (x^3+1)(x^2+1)+x(x^4+x^2+x+1) \end{align*}so \(s(x)=x\) and \(t(x)=x^3+1\text{.}\)
Making a New Field.
Why? If \(\gcd(f,g)=1\text{,}\) then
\begin{align*}
1 \amp = s(x)f(x)+t(x)g(x)\\
1 \amp - t(x)g(x) = s(x)f(x)\\
1 \amp \equiv t(x)g(x) \pmod{f(x)}
\end{align*}
So \(g\) is invertible modulo \(f\text{.}\) Question: How to make all \(g\) have \(\gcd(f,g)=1\text{?}\)
Idea: take \(F[x]/p(x)\text{:}\) the set of equivalence classes of polynomials modulo p. If \(\deg(p(x))=s\) then
\begin{equation*}
F[x]/p(x)=\left\{ \sum_{i=0}^{s-1} a_ix^i \mid a_i\in F\right\}
\end{equation*}
since any two such polynomials have a difference which has lower degree than \(p(x)\) and is therefore not equivalent to \(p(x)\) and any polynomial of degree \(s\) or higher can be reduced mod \(p\) to one of these.
Claim 35. Polynomials Mod an Irreducible Form a Field.
For any field \(F\) and irreducible \(p(x)\text{,}\) \(F[x]/p(x)\) is a field. If \(|F|=q\) then \(|F[x]/p(x)|=q^s\text{.}\)
Question: Do there exist irreducible polynomials for every degree over every field? Answer: In general no, but yes when \(F\) is finite.
Subsection Thursday 2/19
It was an exam. No notes today.
