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Can someone explain to me how the minimal polynomials in page 4 of this document are obtained? Please help me.

http://web.ntpu.edu.tw/~yshan/BCH_code.pdf

It should be something standard about minimal polynomials you should only check the Field in the example.. I get $a^2$, but I am not sure about the other two.

Alternatively can someone show me how to do exercise 5 in http://www.ms.uky.edu/~corso/teaching/math362/Coding-Theory-3.pdf page 465 ??

vounoo
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  • What does $GF(2^m)$ mean? – Gregory Grant May 11 '15 at 16:57
  • Do you know anything about cyclotomic polynomials? – Git Gud May 11 '15 at 16:57
  • Never heard of cyclotomic polynomial... – vounoo May 11 '15 at 17:01
  • These polynomials are over $\mathbb{Z}_2$. They should be elements of the Ideal $\frac{\mathbb{Z}_2[x]}{x^4+x+1}$ where $a^4+a+1=0$ etc... – vounoo May 11 '15 at 17:03
  • @GregoryGrant $GF(q)$ is one of the standard notations for the unique (up to isomorphism) field of cardinality $q$. Even our tag wiki explains this ;-) – Jyrki Lahtonen May 11 '15 at 17:08
  • @JyrkiLahtonen Don't most people call it $\mathbb F_q$? I never saw $GF(q)$ before. – Gregory Grant May 11 '15 at 17:09
  • @JyrkiLahtonen In his defense it wasn't properly tagged at the time of the comment. – Git Gud May 11 '15 at 17:09
  • Is this really complicated ? I thought I was just retarded ... – vounoo May 11 '15 at 17:12
  • I have found this example in 5-6 different sources.. everyone seems to be just throwing it at you as if it is obvious – vounoo May 11 '15 at 17:13
  • http://www.ms.uky.edu/~corso/teaching/math362/Coding-Theory-3.pdf

    pg 460 (second page of the file)

     – vounoo May 11 '15 at 17:14
  • here is another such source – vounoo May 11 '15 at 17:14
  • @vounoo They are skipping the menial work, this is field theory so the authors choose not to waste time with it, since these are coding theory notes. It depends on what you know, what you can use... I wouldn't be surprised if you're supposed to know by heart a few of the irreducible polynomials of low degree over $\mathbb F_2$, in this case you'd just have to try until you find them. – Git Gud May 11 '15 at 17:14
  • Alternatively you can find the powers of $\alpha$ up to $\alpha^{15}$ and then look for patterns to try and find a annihilating polynomial. The simpler way would be to use cycloctomic classes which you don't know about. – Git Gud May 11 '15 at 17:16
  • if you see the second source I commented he actually takes $a^1$ for granted shows $a^2$ and asks you to do $a^3$ as an exercise, I tried to do it like he does but I either get something ugly or $a^3=a^3$ – vounoo May 11 '15 at 17:16
  • @GitGud looks like what he does in http://www.ms.uky.edu/~corso/teaching/math362/Coding-Theory-3.pdf – vounoo May 11 '15 at 17:17
  • Part of this is actually an exercise in http://www.ms.uky.edu/~corso/teaching/math362/Coding-Theory-3.pdf This is actually an exercise – vounoo May 11 '15 at 17:18
  • @Gregory, Git Gud: Sorry about making the facetious reference to our tag wiki. I edited that wiki myself specifically because after umpteen rounds I got tired of explaining this. The $GF(q)$ notation is standard in coding theory community, and it was also used in the undergraduate algebra courses I took back in the day (the lecture notes were written by a number theorist, so it was not only coding theory). I defaulted to $GF(q)$-notation myself before I started out here in MSE four years ago. Not for lack of exposure I might add. – Jyrki Lahtonen May 11 '15 at 17:25
  • @JyrkiLahtonen No harm done, I wasn't even aware of the so-called tag wiki until now... so I know where to look next time. I somehow got all the way through a PhD in algebraic number theory without encountering the $GF(q)$ notation. At first I thought it must be some ring of matrices. – Gregory Grant May 11 '15 at 18:40

1 Answers1

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Those polynomials can be easily explained. The minimal polynomial of $\alpha$ is $$\phi_1(x)=x^4+x+1,$$ because we defined $\alpha$ to be one of the zeros of this polynomial.

Because $\alpha$ is a primitive element, it is of order $15$. Therefore $\beta=\alpha^3$ is of order five - in other words $\beta^5=1$. This implies that $$ 0=\beta^5-1=(\beta-1)(\beta^4+\beta^3+\beta^2+\beta+1). $$ Because $\beta\neq1$ we see that $\beta$ is a zero of $$ \phi_2(x)=x^4+x^3+x^2+x+1. $$ On the other hand $\beta$ generates the field $GF(16)$, so its minimal polynomial has to be of degree four. Therefore $\phi_2(x)$ is the minimal polynomial. Alternatively we can see that $\beta$ has as its conjugates (act on it by the Frobenius automorphism) $\beta^2=\alpha^6$, $\beta^4=\alpha^{12}$ and $\beta^8=\beta^3=\alpha^9$ four distinct elements of $GF(16)$.

On the other hand $\gamma=\alpha^5$ is of order three ($=15/5$) only. Therefore it belongs to the subfield $GF(4)$ (that contains precisely the roots of unity of order $4-1$). Hence its minimal polynomial is the only irreducible quadratic polynomial over $GF(2)$, namely $$ \phi_3(x)=x^2+x+1. $$ We could also argue as above starting from $$ 0=\gamma^3-1=(\gamma-1)(\gamma^2+\gamma+1). $$

If we want to continue and find the minimal polynomial of $\alpha^7$ it is simplest to observe that $\alpha^7$ is a conjugate of $\alpha^{2\cdot7}=\alpha^{14}=\alpha^{-1}$. But obviously $\alpha^{-1}$ is a zero of the polynomial $$ x^4\phi_1(\frac1x)=x^4(\frac1{x^4}+\frac1x+1)=1+x^3+x^4. $$ Admittedly that is not much use for the purposes of constructing larger minimum distance BCH-codes because once we get to $\alpha^7$ we have exhausted all the elements of $GF(16)\setminus GF(2)$ as zeros of the generator polynomial. Thus the resulting code is kinda tiny. It is the repetion code of length 15.

Jyrki Lahtonen
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