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.
Can someone explain why they are doing that calculation with the multiplicative inverses please?
polynomials
edited Dec 10 '18 at 21:22
Arthur
114k7115197
asked Dec 10 '18 at 20:56
J. Doe HueJ. Doe Hue
135
$endgroup$
add a comment |
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.
Can someone explain why they are doing that calculation with the multiplicative inverses please?
polynomials
edited Dec 10 '18 at 21:22
Arthur
114k7115197
asked Dec 10 '18 at 20:56
J. Doe HueJ. Doe Hue
135
$endgroup$
$begingroup$
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
add a comment |
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.
Can someone explain why they are doing that calculation with the multiplicative inverses please?
polynomials
edited Dec 10 '18 at 21:22
Arthur
114k7115197
asked Dec 10 '18 at 20:56
J. Doe HueJ. Doe Hue
135
$endgroup$
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.
Can someone explain why they are doing that calculation with the multiplicative inverses please?
polynomials
polynomials
edited Dec 10 '18 at 21:22
Arthur
114k7115197
asked Dec 10 '18 at 20:56
J. Doe HueJ. Doe Hue
135
edited Dec 10 '18 at 21:22
Arthur
114k7115197
asked Dec 10 '18 at 20:56
J. Doe HueJ. Doe Hue
135
edited Dec 10 '18 at 21:22
Arthur
114k7115197
edited Dec 10 '18 at 21:22
Arthur
114k7115197
edited Dec 10 '18 at 21:22
Arthur
114k7115197
114k7115197
asked Dec 10 '18 at 20:56
J. Doe HueJ. Doe Hue
135
asked Dec 10 '18 at 20:56
J. Doe HueJ. Doe Hue
135
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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We are in $GF(2^n)$, with $n=4$. It is the splitting field of $x^{16}-x$. Thus $g^{15}=1$ (or $g^{-3}=g^{12}$).
answered Dec 10 '18 at 21:40
Chris CusterChris Custer
12.9k3826
$endgroup$
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
add a comment |
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1 Answer
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1 Answer
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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 $g^{-3}=g^{12}$).
answered Dec 10 '18 at 21:40
Chris CusterChris Custer
12.9k3826
$endgroup$
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
add a comment |
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 $g^{-3}=g^{12}$).
answered Dec 10 '18 at 21:40
Chris CusterChris Custer
12.9k3826
$endgroup$
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
add a comment |
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 $g^{-3}=g^{12}$).
answered Dec 10 '18 at 21:40
Chris CusterChris Custer
12.9k3826
$endgroup$
We are in $GF(2^n)$, with $n=4$. It is the splitting field of $x^{16}-x$. Thus $g^{15}=1$ (or $g^{-3}=g^{12}$).
answered Dec 10 '18 at 21:40
Chris CusterChris Custer
12.9k3826
answered Dec 10 '18 at 21:40
Chris CusterChris Custer
12.9k3826
answered Dec 10 '18 at 21:40
Chris CusterChris Custer
12.9k3826
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
add a comment |
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
add a comment |
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$begingroup$
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