Mixing
Generator matrix for Reed-Solomon code
$begingroup$
Given $,f(t)=t^3+t+1,,$ let $,mathbb{F}_{2^3}=frac{mathbb{F}_2left[tright]}{langle f,rangle},$ be a finite field and $,xi=t,left(textrm{mod},fright).$ Let also $,mathcal{C},$ be a cyclic code where $,g(x)=left(x-xiright)left(x-xi^2right)left(x-xi^3right)left(x-xi^4right),$ is his generator polynomial.
Question: Find a generator matrix of $,mathcal{C},$ in the basis $,B=left{1,xi,xi^2right},$ of $,mathbb{F}_{2^3},$ as an $,mathbb{F}_{2}$- vector space.
The fisrt thing they ask is to prove that $,xi,$ is a $,7^{,th},$ primitive root of the unity (quite easy). Then, prove that the length of $,mathcal{C},$ is $,n=7$.
I found out that $,mathcal{C},$ is a Reed-Solomon code, a type of BCH code whose length is always $,n=q-1,$ over a finite field $,mathbb{F}_q,,$ so this question is also easy to answer.
However, I do not know how to proceed to find the generator matrix of $,mathcal{C}.$ I tried to expand the generator polynomial
$$g(x)=g_0+g_1x+dots+g_4x^4$$
and define the generator matrix as
$$G=begin{pmatrix}g_0&g_1&g_2&g_3&g_4 & 0 & 0\
0 & g_0&g_1&g_2&g_3&g_4 & 0\
0 & 0 & g_0&g_1&g_2&g_3&g_4end{pmatrix}$$
in the basis $,B=left{1,xi,xi^2right},$ but I guess I failed.
Somebody please, help me! Thanks in advance.
abstract-algebra finite-fields coding-theory
asked Jan 11 at 11:01
CarlIOCarlIO
957
$endgroup$
|
show 6 more comments
$begingroup$
Given $,f(t)=t^3+t+1,,$ let $,mathbb{F}_{2^3}=frac{mathbb{F}_2left[tright]}{langle f,rangle},$ be a finite field and $,xi=t,left(textrm{mod},fright).$ Let also $,mathcal{C},$ be a cyclic code where $,g(x)=left(x-xiright)left(x-xi^2right)left(x-xi^3right)left(x-xi^4right),$ is his generator polynomial.
Question: Find a generator matrix of $,mathcal{C},$ in the basis $,B=left{1,xi,xi^2right},$ of $,mathbb{F}_{2^3},$ as an $,mathbb{F}_{2}$- vector space.
The fisrt thing they ask is to prove that $,xi,$ is a $,7^{,th},$ primitive root of the unity (quite easy). Then, prove that the length of $,mathcal{C},$ is $,n=7$.
I found out that $,mathcal{C},$ is a Reed-Solomon code, a type of BCH code whose length is always $,n=q-1,$ over a finite field $,mathbb{F}_q,,$ so this question is also easy to answer.
However, I do not know how to proceed to find the generator matrix of $,mathcal{C}.$ I tried to expand the generator polynomial
$$g(x)=g_0+g_1x+dots+g_4x^4$$
and define the generator matrix as
$$G=begin{pmatrix}g_0&g_1&g_2&g_3&g_4 & 0 & 0\
0 & g_0&g_1&g_2&g_3&g_4 & 0\
0 & 0 & g_0&g_1&g_2&g_3&g_4end{pmatrix}$$
in the basis $,B=left{1,xi,xi^2right},$ but I guess I failed.
Somebody please, help me! Thanks in advance.
abstract-algebra finite-fields coding-theory
asked Jan 11 at 11:01
CarlIOCarlIO
957
$endgroup$
$begingroup$
No, its fine. You need to calculate the coefficients $g_i$.
$endgroup$
– Wuestenfux
Jan 11 at 11:31
$begingroup$
Basic Galois theory tells us that the zeros of $f(t)$ are $xi,xi^2$ and $xi^4$. So $$f(x)=x^3+x+1=(x-xi)(x-xi^2)(x-xi^4).$$ All you need to do is to calculate $$g(x)=f(x)(x-xi^3).$$ Observe that $xi^3=xi+1$ because $xi$ is a zero of $f$.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 11:41
$begingroup$
@Jyrki Lahtonen: I guess you mean $f(t)$ is the minimal polynomial of $xi$ over $mathbb{F}_{2}$. I didn't realize, thank you very much. However, when I calculate the coefficients of $g(x)$ as you said I get $$g(x)=x^4 - left(xi +1right)^3+x^2-xi x-left(xi+1right)$$ so, I can't express the coefficients of $x^3$ and $1$ in the basis $B$, or maybe I just don't know how.
$endgroup$
– CarlIO
Jan 11 at 17:45
$begingroup$
Given that $f(x)$ has no quadratic term the coefficient of $x^3$ in $g(x)$ should be just $xi^3=xi+1$, no?
$endgroup$
– Jyrki Lahtonen
Jan 11 at 17:51
1
$begingroup$
Effectively you would then replace each entry of the $3times7$ matrix with a $3times3$ block.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 18:30
|
show 6 more comments
$begingroup$
Given $,f(t)=t^3+t+1,,$ let $,mathbb{F}_{2^3}=frac{mathbb{F}_2left[tright]}{langle f,rangle},$ be a finite field and $,xi=t,left(textrm{mod},fright).$ Let also $,mathcal{C},$ be a cyclic code where $,g(x)=left(x-xiright)left(x-xi^2right)left(x-xi^3right)left(x-xi^4right),$ is his generator polynomial.
Question: Find a generator matrix of $,mathcal{C},$ in the basis $,B=left{1,xi,xi^2right},$ of $,mathbb{F}_{2^3},$ as an $,mathbb{F}_{2}$- vector space.
The fisrt thing they ask is to prove that $,xi,$ is a $,7^{,th},$ primitive root of the unity (quite easy). Then, prove that the length of $,mathcal{C},$ is $,n=7$.
I found out that $,mathcal{C},$ is a Reed-Solomon code, a type of BCH code whose length is always $,n=q-1,$ over a finite field $,mathbb{F}_q,,$ so this question is also easy to answer.
However, I do not know how to proceed to find the generator matrix of $,mathcal{C}.$ I tried to expand the generator polynomial
$$g(x)=g_0+g_1x+dots+g_4x^4$$
and define the generator matrix as
$$G=begin{pmatrix}g_0&g_1&g_2&g_3&g_4 & 0 & 0\
0 & g_0&g_1&g_2&g_3&g_4 & 0\
0 & 0 & g_0&g_1&g_2&g_3&g_4end{pmatrix}$$
in the basis $,B=left{1,xi,xi^2right},$ but I guess I failed.
Somebody please, help me! Thanks in advance.
abstract-algebra finite-fields coding-theory
asked Jan 11 at 11:01
CarlIOCarlIO
957
$endgroup$
Given $,f(t)=t^3+t+1,,$ let $,mathbb{F}_{2^3}=frac{mathbb{F}_2left[tright]}{langle f,rangle},$ be a finite field and $,xi=t,left(textrm{mod},fright).$ Let also $,mathcal{C},$ be a cyclic code where $,g(x)=left(x-xiright)left(x-xi^2right)left(x-xi^3right)left(x-xi^4right),$ is his generator polynomial.
Question: Find a generator matrix of $,mathcal{C},$ in the basis $,B=left{1,xi,xi^2right},$ of $,mathbb{F}_{2^3},$ as an $,mathbb{F}_{2}$- vector space.
The fisrt thing they ask is to prove that $,xi,$ is a $,7^{,th},$ primitive root of the unity (quite easy). Then, prove that the length of $,mathcal{C},$ is $,n=7$.
I found out that $,mathcal{C},$ is a Reed-Solomon code, a type of BCH code whose length is always $,n=q-1,$ over a finite field $,mathbb{F}_q,,$ so this question is also easy to answer.
However, I do not know how to proceed to find the generator matrix of $,mathcal{C}.$ I tried to expand the generator polynomial
$$g(x)=g_0+g_1x+dots+g_4x^4$$
and define the generator matrix as
$$G=begin{pmatrix}g_0&g_1&g_2&g_3&g_4 & 0 & 0\
0 & g_0&g_1&g_2&g_3&g_4 & 0\
0 & 0 & g_0&g_1&g_2&g_3&g_4end{pmatrix}$$
in the basis $,B=left{1,xi,xi^2right},$ but I guess I failed.
Somebody please, help me! Thanks in advance.
abstract-algebra finite-fields coding-theory
abstract-algebra finite-fields coding-theory
asked Jan 11 at 11:01
CarlIOCarlIO
957
asked Jan 11 at 11:01
CarlIOCarlIO
957
edited Jan 19 at 12:16
CarlIO
asked Jan 11 at 11:01
CarlIOCarlIO
957
asked Jan 11 at 11:01
CarlIOCarlIO
957
asked Jan 11 at 11:01
CarlIOCarlIO
957
957
$begingroup$
No, its fine. You need to calculate the coefficients $g_i$.
$endgroup$
– Wuestenfux
Jan 11 at 11:31
$begingroup$
Basic Galois theory tells us that the zeros of $f(t)$ are $xi,xi^2$ and $xi^4$. So $$f(x)=x^3+x+1=(x-xi)(x-xi^2)(x-xi^4).$$ All you need to do is to calculate $$g(x)=f(x)(x-xi^3).$$ Observe that $xi^3=xi+1$ because $xi$ is a zero of $f$.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 11:41
$begingroup$
@Jyrki Lahtonen: I guess you mean $f(t)$ is the minimal polynomial of $xi$ over $mathbb{F}_{2}$. I didn't realize, thank you very much. However, when I calculate the coefficients of $g(x)$ as you said I get $$g(x)=x^4 - left(xi +1right)^3+x^2-xi x-left(xi+1right)$$ so, I can't express the coefficients of $x^3$ and $1$ in the basis $B$, or maybe I just don't know how.
$endgroup$
– CarlIO
Jan 11 at 17:45
$begingroup$
Given that $f(x)$ has no quadratic term the coefficient of $x^3$ in $g(x)$ should be just $xi^3=xi+1$, no?
$endgroup$
– Jyrki Lahtonen
Jan 11 at 17:51
1
$begingroup$
Effectively you would then replace each entry of the $3times7$ matrix with a $3times3$ block.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 18:30
|
show 6 more comments
$begingroup$
No, its fine. You need to calculate the coefficients $g_i$.
$endgroup$
– Wuestenfux
Jan 11 at 11:31
$begingroup$
Basic Galois theory tells us that the zeros of $f(t)$ are $xi,xi^2$ and $xi^4$. So $$f(x)=x^3+x+1=(x-xi)(x-xi^2)(x-xi^4).$$ All you need to do is to calculate $$g(x)=f(x)(x-xi^3).$$ Observe that $xi^3=xi+1$ because $xi$ is a zero of $f$.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 11:41
$begingroup$
@Jyrki Lahtonen: I guess you mean $f(t)$ is the minimal polynomial of $xi$ over $mathbb{F}_{2}$. I didn't realize, thank you very much. However, when I calculate the coefficients of $g(x)$ as you said I get $$g(x)=x^4 - left(xi +1right)^3+x^2-xi x-left(xi+1right)$$ so, I can't express the coefficients of $x^3$ and $1$ in the basis $B$, or maybe I just don't know how.
$endgroup$
– CarlIO
Jan 11 at 17:45
$begingroup$
Given that $f(x)$ has no quadratic term the coefficient of $x^3$ in $g(x)$ should be just $xi^3=xi+1$, no?
$endgroup$
– Jyrki Lahtonen
Jan 11 at 17:51
1
$begingroup$
Effectively you would then replace each entry of the $3times7$ matrix with a $3times3$ block.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 18:30
$begingroup$
No, its fine. You need to calculate the coefficients $g_i$.
$endgroup$
– Wuestenfux
Jan 11 at 11:31
$begingroup$
No, its fine. You need to calculate the coefficients $g_i$.
$endgroup$
– Wuestenfux
Jan 11 at 11:31
$begingroup$
Basic Galois theory tells us that the zeros of $f(t)$ are $xi,xi^2$ and $xi^4$. So $$f(x)=x^3+x+1=(x-xi)(x-xi^2)(x-xi^4).$$ All you need to do is to calculate $$g(x)=f(x)(x-xi^3).$$ Observe that $xi^3=xi+1$ because $xi$ is a zero of $f$.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 11:41
$begingroup$
Basic Galois theory tells us that the zeros of $f(t)$ are $xi,xi^2$ and $xi^4$. So $$f(x)=x^3+x+1=(x-xi)(x-xi^2)(x-xi^4).$$ All you need to do is to calculate $$g(x)=f(x)(x-xi^3).$$ Observe that $xi^3=xi+1$ because $xi$ is a zero of $f$.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 11:41
$begingroup$
@Jyrki Lahtonen: I guess you mean $f(t)$ is the minimal polynomial of $xi$ over $mathbb{F}_{2}$. I didn't realize, thank you very much. However, when I calculate the coefficients of $g(x)$ as you said I get $$g(x)=x^4 - left(xi +1right)^3+x^2-xi x-left(xi+1right)$$ so, I can't express the coefficients of $x^3$ and $1$ in the basis $B$, or maybe I just don't know how.
$endgroup$
– CarlIO
Jan 11 at 17:45
$begingroup$
@Jyrki Lahtonen: I guess you mean $f(t)$ is the minimal polynomial of $xi$ over $mathbb{F}_{2}$. I didn't realize, thank you very much. However, when I calculate the coefficients of $g(x)$ as you said I get $$g(x)=x^4 - left(xi +1right)^3+x^2-xi x-left(xi+1right)$$ so, I can't express the coefficients of $x^3$ and $1$ in the basis $B$, or maybe I just don't know how.
$endgroup$
– CarlIO
Jan 11 at 17:45
$begingroup$
Given that $f(x)$ has no quadratic term the coefficient of $x^3$ in $g(x)$ should be just $xi^3=xi+1$, no?
$endgroup$
– Jyrki Lahtonen
Jan 11 at 17:51
$begingroup$
Given that $f(x)$ has no quadratic term the coefficient of $x^3$ in $g(x)$ should be just $xi^3=xi+1$, no?
$endgroup$
– Jyrki Lahtonen
Jan 11 at 17:51
1
1
$begingroup$
Effectively you would then replace each entry of the $3times7$ matrix with a $3times3$ block.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 18:30
$begingroup$
Effectively you would then replace each entry of the $3times7$ matrix with a $3times3$ block.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 18:30
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Lots of comments. We have
$$g = (x-xi)(x-xi^2)(x-xi^3)(x-xi^4).$$
If $xi$ is a zero of the primitive polynomial $f=x^3+x+1$, then $f$ has the zeros $xi,xi^2,(xi^2)^2=xi^4$. Thus
$$g = (x^3+x+1)(x+xi^3) = x^4+x^2+x+xi^3x^3+xi^3 x +xi^3\
= x^4 + xi^3 x^3 +x^2+(xi^3+1)x+xi^3.$$
Since $xi$ is a zero of $f$, $xi^3+xi+1=0$ and so $xi=xi^3+1$
Hence,
$$g= = x^4 + xi^3 x^3 +x^2+xi x+xi^3.$$
answered Jan 13 at 8:43
WuestenfuxWuestenfux
4,5161413
$endgroup$
add a comment |
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$begingroup$
Lots of comments. We have
$$g = (x-xi)(x-xi^2)(x-xi^3)(x-xi^4).$$
If $xi$ is a zero of the primitive polynomial $f=x^3+x+1$, then $f$ has the zeros $xi,xi^2,(xi^2)^2=xi^4$. Thus
$$g = (x^3+x+1)(x+xi^3) = x^4+x^2+x+xi^3x^3+xi^3 x +xi^3\
= x^4 + xi^3 x^3 +x^2+(xi^3+1)x+xi^3.$$
Since $xi$ is a zero of $f$, $xi^3+xi+1=0$ and so $xi=xi^3+1$
Hence,
$$g= = x^4 + xi^3 x^3 +x^2+xi x+xi^3.$$
answered Jan 13 at 8:43
WuestenfuxWuestenfux
4,5161413
$endgroup$
add a comment |
$begingroup$
Lots of comments. We have
$$g = (x-xi)(x-xi^2)(x-xi^3)(x-xi^4).$$
If $xi$ is a zero of the primitive polynomial $f=x^3+x+1$, then $f$ has the zeros $xi,xi^2,(xi^2)^2=xi^4$. Thus
$$g = (x^3+x+1)(x+xi^3) = x^4+x^2+x+xi^3x^3+xi^3 x +xi^3\
= x^4 + xi^3 x^3 +x^2+(xi^3+1)x+xi^3.$$
Since $xi$ is a zero of $f$, $xi^3+xi+1=0$ and so $xi=xi^3+1$
Hence,
$$g= = x^4 + xi^3 x^3 +x^2+xi x+xi^3.$$
answered Jan 13 at 8:43
WuestenfuxWuestenfux
4,5161413
$endgroup$
add a comment |
$begingroup$
Lots of comments. We have
$$g = (x-xi)(x-xi^2)(x-xi^3)(x-xi^4).$$
If $xi$ is a zero of the primitive polynomial $f=x^3+x+1$, then $f$ has the zeros $xi,xi^2,(xi^2)^2=xi^4$. Thus
$$g = (x^3+x+1)(x+xi^3) = x^4+x^2+x+xi^3x^3+xi^3 x +xi^3\
= x^4 + xi^3 x^3 +x^2+(xi^3+1)x+xi^3.$$
Since $xi$ is a zero of $f$, $xi^3+xi+1=0$ and so $xi=xi^3+1$
Hence,
$$g= = x^4 + xi^3 x^3 +x^2+xi x+xi^3.$$
answered Jan 13 at 8:43
WuestenfuxWuestenfux
4,5161413
$endgroup$
Lots of comments. We have
$$g = (x-xi)(x-xi^2)(x-xi^3)(x-xi^4).$$
If $xi$ is a zero of the primitive polynomial $f=x^3+x+1$, then $f$ has the zeros $xi,xi^2,(xi^2)^2=xi^4$. Thus
$$g = (x^3+x+1)(x+xi^3) = x^4+x^2+x+xi^3x^3+xi^3 x +xi^3\
= x^4 + xi^3 x^3 +x^2+(xi^3+1)x+xi^3.$$
Since $xi$ is a zero of $f$, $xi^3+xi+1=0$ and so $xi=xi^3+1$
Hence,
$$g= = x^4 + xi^3 x^3 +x^2+xi x+xi^3.$$
answered Jan 13 at 8:43
WuestenfuxWuestenfux
4,5161413
answered Jan 13 at 8:43
WuestenfuxWuestenfux
4,5161413
answered Jan 13 at 8:43
WuestenfuxWuestenfux
4,5161413
answered Jan 13 at 8:43
WuestenfuxWuestenfux
4,5161413
4,5161413
add a comment |
add a comment |
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$begingroup$
No, its fine. You need to calculate the coefficients $g_i$.
$endgroup$
– Wuestenfux
Jan 11 at 11:31
$begingroup$
Basic Galois theory tells us that the zeros of $f(t)$ are $xi,xi^2$ and $xi^4$. So $$f(x)=x^3+x+1=(x-xi)(x-xi^2)(x-xi^4).$$ All you need to do is to calculate $$g(x)=f(x)(x-xi^3).$$ Observe that $xi^3=xi+1$ because $xi$ is a zero of $f$.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 11:41
$begingroup$
@Jyrki Lahtonen: I guess you mean $f(t)$ is the minimal polynomial of $xi$ over $mathbb{F}_{2}$. I didn't realize, thank you very much. However, when I calculate the coefficients of $g(x)$ as you said I get $$g(x)=x^4 - left(xi +1right)^3+x^2-xi x-left(xi+1right)$$ so, I can't express the coefficients of $x^3$ and $1$ in the basis $B$, or maybe I just don't know how.
$endgroup$
– CarlIO
Jan 11 at 17:45
$begingroup$
Given that $f(x)$ has no quadratic term the coefficient of $x^3$ in $g(x)$ should be just $xi^3=xi+1$, no?
$endgroup$
– Jyrki Lahtonen
Jan 11 at 17:51
1
$begingroup$
Effectively you would then replace each entry of the $3times7$ matrix with a $3times3$ block.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 18:30