Generators for polynomials annd multiplicative inverses, not sure what is going on here $X^{-3}$ becomes...












1












$begingroup$


I am confused on this problem. I am not sure what is happening with the (-3) for it to become (12), to then become a larger polynomial in a GF type of problem.



Here is more information about the problem. I hope it helps, but I don't have an actual problem per se. It's more of a concept I want to understand.



enter image description here



Can someone explain why they are doing that calculation with the multiplicative inverses please?



enter image description here



enter image description here






polynomials







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think we may need to read the actual problem text as well. Without that, we can guess, but even we will be somewhat uncertain as to what's actually going on.
    $endgroup$
    – Arthur
    Dec 10 '18 at 21:01












  • $begingroup$
    Let me see if I can find it in the book. Thank you.
    $endgroup$
    – J. Doe Hue
    Dec 10 '18 at 21:06






  • 1




    $begingroup$
    I added more context for the problem. Hopefully it helps.
    $endgroup$
    – J. Doe Hue
    Dec 10 '18 at 21:10
















1












$begingroup$


I am confused on this problem. I am not sure what is happening with the (-3) for it to become (12), to then become a larger polynomial in a GF type of problem.



Here is more information about the problem. I hope it helps, but I don't have an actual problem per se. It's more of a concept I want to understand.



enter image description here



Can someone explain why they are doing that calculation with the multiplicative inverses please?



enter image description here



enter image description here






polynomials







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think we may need to read the actual problem text as well. Without that, we can guess, but even we will be somewhat uncertain as to what's actually going on.
    $endgroup$
    – Arthur
    Dec 10 '18 at 21:01












  • $begingroup$
    Let me see if I can find it in the book. Thank you.
    $endgroup$
    – J. Doe Hue
    Dec 10 '18 at 21:06






  • 1




    $begingroup$
    I added more context for the problem. Hopefully it helps.
    $endgroup$
    – J. Doe Hue
    Dec 10 '18 at 21:10














1












1








1





$begingroup$


I am confused on this problem. I am not sure what is happening with the (-3) for it to become (12), to then become a larger polynomial in a GF type of problem.



Here is more information about the problem. I hope it helps, but I don't have an actual problem per se. It's more of a concept I want to understand.



enter image description here



Can someone explain why they are doing that calculation with the multiplicative inverses please?



enter image description here



enter image description here






polynomials







share|cite|improve this question











$endgroup$




I am confused on this problem. I am not sure what is happening with the (-3) for it to become (12), to then become a larger polynomial in a GF type of problem.



Here is more information about the problem. I hope it helps, but I don't have an actual problem per se. It's more of a concept I want to understand.



enter image description here



Can someone explain why they are doing that calculation with the multiplicative inverses please?



enter image description here



enter image description here






polynomials




polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 21:22









Arthur

114k7115197




114k7115197










asked Dec 10 '18 at 20:56









J. Doe HueJ. Doe Hue

135




135












  • $begingroup$
    I think we may need to read the actual problem text as well. Without that, we can guess, but even we will be somewhat uncertain as to what's actually going on.
    $endgroup$
    – Arthur
    Dec 10 '18 at 21:01












  • $begingroup$
    Let me see if I can find it in the book. Thank you.
    $endgroup$
    – J. Doe Hue
    Dec 10 '18 at 21:06






  • 1




    $begingroup$
    I added more context for the problem. Hopefully it helps.
    $endgroup$
    – J. Doe Hue
    Dec 10 '18 at 21:10


















  • $begingroup$
    I think we may need to read the actual problem text as well. Without that, we can guess, but even we will be somewhat uncertain as to what's actually going on.
    $endgroup$
    – Arthur
    Dec 10 '18 at 21:01












  • $begingroup$
    Let me see if I can find it in the book. Thank you.
    $endgroup$
    – J. Doe Hue
    Dec 10 '18 at 21:06






  • 1




    $begingroup$
    I added more context for the problem. Hopefully it helps.
    $endgroup$
    – J. Doe Hue
    Dec 10 '18 at 21:10
















$begingroup$
I think we may need to read the actual problem text as well. Without that, we can guess, but even we will be somewhat uncertain as to what's actually going on.
$endgroup$
– Arthur
Dec 10 '18 at 21:01






$begingroup$
I think we may need to read the actual problem text as well. Without that, we can guess, but even we will be somewhat uncertain as to what's actually going on.
$endgroup$
– Arthur
Dec 10 '18 at 21:01














$begingroup$
Let me see if I can find it in the book. Thank you.
$endgroup$
– J. Doe Hue
Dec 10 '18 at 21:06




$begingroup$
Let me see if I can find it in the book. Thank you.
$endgroup$
– J. Doe Hue
Dec 10 '18 at 21:06




1




1




$begingroup$
I added more context for the problem. Hopefully it helps.
$endgroup$
– J. Doe Hue
Dec 10 '18 at 21:10




$begingroup$
I added more context for the problem. Hopefully it helps.
$endgroup$
– J. Doe Hue
Dec 10 '18 at 21:10










1 Answer
1






active

oldest

votes


















2












$begingroup$

We are in $GF(2^n)$, with $n=4$. It is the splitting field of $x^{16}-x$. Thus $g^{15}=1$ (or $g^{-3}=g^{12}$).






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    One doesn't actually need the notion of splitting field for this. In a field of $q$ elements, the multiplicative group of non-zero elements has $q-1$ elements. So by Lagrange's theorem from group theory, all its elements satisfy $x^{q-1}=1$.
    $endgroup$
    – Andreas Blass
    Dec 10 '18 at 21:49












  • $begingroup$
    @AndreasBlass Nice. Thanks for that insight.
    $endgroup$
    – Chris Custer
    Dec 10 '18 at 21:59










  • $begingroup$
    Thanks for the explanation.
    $endgroup$
    – J. Doe Hue
    Dec 10 '18 at 22:05











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

We are in $GF(2^n)$, with $n=4$. It is the splitting field of $x^{16}-x$. Thus $g^{15}=1$ (or $g^{-3}=g^{12}$).






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    One doesn't actually need the notion of splitting field for this. In a field of $q$ elements, the multiplicative group of non-zero elements has $q-1$ elements. So by Lagrange's theorem from group theory, all its elements satisfy $x^{q-1}=1$.
    $endgroup$
    – Andreas Blass
    Dec 10 '18 at 21:49












  • $begingroup$
    @AndreasBlass Nice. Thanks for that insight.
    $endgroup$
    – Chris Custer
    Dec 10 '18 at 21:59










  • $begingroup$
    Thanks for the explanation.
    $endgroup$
    – J. Doe Hue
    Dec 10 '18 at 22:05
















2












$begingroup$

We are in $GF(2^n)$, with $n=4$. It is the splitting field of $x^{16}-x$. Thus $g^{15}=1$ (or $g^{-3}=g^{12}$).






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    One doesn't actually need the notion of splitting field for this. In a field of $q$ elements, the multiplicative group of non-zero elements has $q-1$ elements. So by Lagrange's theorem from group theory, all its elements satisfy $x^{q-1}=1$.
    $endgroup$
    – Andreas Blass
    Dec 10 '18 at 21:49












  • $begingroup$
    @AndreasBlass Nice. Thanks for that insight.
    $endgroup$
    – Chris Custer
    Dec 10 '18 at 21:59










  • $begingroup$
    Thanks for the explanation.
    $endgroup$
    – J. Doe Hue
    Dec 10 '18 at 22:05














2












2








2





$begingroup$

We are in $GF(2^n)$, with $n=4$. It is the splitting field of $x^{16}-x$. Thus $g^{15}=1$ (or $g^{-3}=g^{12}$).






share|cite|improve this answer









$endgroup$



We are in $GF(2^n)$, with $n=4$. It is the splitting field of $x^{16}-x$. Thus $g^{15}=1$ (or $g^{-3}=g^{12}$).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 21:40









Chris CusterChris Custer

12.9k3826




12.9k3826








  • 1




    $begingroup$
    One doesn't actually need the notion of splitting field for this. In a field of $q$ elements, the multiplicative group of non-zero elements has $q-1$ elements. So by Lagrange's theorem from group theory, all its elements satisfy $x^{q-1}=1$.
    $endgroup$
    – Andreas Blass
    Dec 10 '18 at 21:49












  • $begingroup$
    @AndreasBlass Nice. Thanks for that insight.
    $endgroup$
    – Chris Custer
    Dec 10 '18 at 21:59










  • $begingroup$
    Thanks for the explanation.
    $endgroup$
    – J. Doe Hue
    Dec 10 '18 at 22:05














  • 1




    $begingroup$
    One doesn't actually need the notion of splitting field for this. In a field of $q$ elements, the multiplicative group of non-zero elements has $q-1$ elements. So by Lagrange's theorem from group theory, all its elements satisfy $x^{q-1}=1$.
    $endgroup$
    – Andreas Blass
    Dec 10 '18 at 21:49












  • $begingroup$
    @AndreasBlass Nice. Thanks for that insight.
    $endgroup$
    – Chris Custer
    Dec 10 '18 at 21:59










  • $begingroup$
    Thanks for the explanation.
    $endgroup$
    – J. Doe Hue
    Dec 10 '18 at 22:05








1




1




$begingroup$
One doesn't actually need the notion of splitting field for this. In a field of $q$ elements, the multiplicative group of non-zero elements has $q-1$ elements. So by Lagrange's theorem from group theory, all its elements satisfy $x^{q-1}=1$.
$endgroup$
– Andreas Blass
Dec 10 '18 at 21:49






$begingroup$
One doesn't actually need the notion of splitting field for this. In a field of $q$ elements, the multiplicative group of non-zero elements has $q-1$ elements. So by Lagrange's theorem from group theory, all its elements satisfy $x^{q-1}=1$.
$endgroup$
– Andreas Blass
Dec 10 '18 at 21:49














$begingroup$
@AndreasBlass Nice. Thanks for that insight.
$endgroup$
– Chris Custer
Dec 10 '18 at 21:59




$begingroup$
@AndreasBlass Nice. Thanks for that insight.
$endgroup$
– Chris Custer
Dec 10 '18 at 21:59












$begingroup$
Thanks for the explanation.
$endgroup$
– J. Doe Hue
Dec 10 '18 at 22:05




$begingroup$
Thanks for the explanation.
$endgroup$
– J. Doe Hue
Dec 10 '18 at 22:05


















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