Multiplicative inverse of a polynomial in $GF(8)$
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I am trying to find the inverse of $x ^3+x +1$ in $GF(8)$.
I have done the Euclidean algorithm but I am stuck in the forward process to get the inverse. Please explain how to do it from reverse.
number-theory galois-theory finite-fields cryptography
asked yesterday
Mariam Mashuryan
13
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Mariam Mashuryan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
0
down vote
favorite
I am trying to find the inverse of $x ^3+x +1$ in $GF(8)$.
I have done the Euclidean algorithm but I am stuck in the forward process to get the inverse. Please explain how to do it from reverse.
number-theory galois-theory finite-fields cryptography
asked yesterday
Mariam Mashuryan
13
New contributor
Mariam Mashuryan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Do it this way - much less painful and less error-prone.
– Bill Dubuque
yesterday
Normally $x$ stands for an indeterminate, when $x^3+x+1$ is a polynomial. Polynomials don't have inverses in a finite field. If, as I suspect, $x$ stands for some element of $GF(8)$ you need to tell us which element. Or at least, tell us its minimal polynomial. You see, very often the field $GF(8)$ is defined as the quotient ring $GF(2)[x]/langle x^3+x+1rangle$. But, in that field the coset of $x^3+x+1$ is equal to zero. Again implying that it has no inverse (can't divide by zero).
– Jyrki Lahtonen
21 hours ago
The (only) other alternative is to have defined $GF(8)$ as the quotien $GF(2)[x]/langle x^3+x^2+1rangle$. In that case the coset of $x^3+x+1$ equals the coset of $x^2+x$ (because $(x^3+x+1)-(x^3+x^2+1)=x^2+x$). You can find the inverse of that by the usual process involving extended Euclid. You see, to positively id an element of the quotient ring we customarily reduce them by the polynomial defining the field.
– Jyrki Lahtonen
21 hours ago
By saying finite field, i mean the polynomial of the field, which is x^8+ x^4+ x^3+x+1. So i have to find the inverse of the first polynomial modulus this one
– Mariam Mashuryan
18 hours ago
What do you mean by $x^8 + x^4 + x^3 + x + 1$ being the "polynomial of the field"?
– Tobias Kildetoft
16 hours ago
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to find the inverse of $x ^3+x +1$ in $GF(8)$.
I have done the Euclidean algorithm but I am stuck in the forward process to get the inverse. Please explain how to do it from reverse.
number-theory galois-theory finite-fields cryptography
asked yesterday
Mariam Mashuryan
13
New contributor
Mariam Mashuryan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I am trying to find the inverse of $x ^3+x +1$ in $GF(8)$.
I have done the Euclidean algorithm but I am stuck in the forward process to get the inverse. Please explain how to do it from reverse.
number-theory galois-theory finite-fields cryptography
number-theory galois-theory finite-fields cryptography
asked yesterday
Mariam Mashuryan
13
New contributor
Mariam Mashuryan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Mariam Mashuryan
13
New contributor
Mariam Mashuryan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 16 hours ago
asked yesterday
Mariam Mashuryan
13
New contributor
Mariam Mashuryan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Mariam Mashuryan
13
asked yesterday
Mariam Mashuryan
13
13
New contributor
Mariam Mashuryan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mariam Mashuryan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Mariam Mashuryan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Do it this way - much less painful and less error-prone.
– Bill Dubuque
yesterday
Normally $x$ stands for an indeterminate, when $x^3+x+1$ is a polynomial. Polynomials don't have inverses in a finite field. If, as I suspect, $x$ stands for some element of $GF(8)$ you need to tell us which element. Or at least, tell us its minimal polynomial. You see, very often the field $GF(8)$ is defined as the quotient ring $GF(2)[x]/langle x^3+x+1rangle$. But, in that field the coset of $x^3+x+1$ is equal to zero. Again implying that it has no inverse (can't divide by zero).
– Jyrki Lahtonen
21 hours ago
The (only) other alternative is to have defined $GF(8)$ as the quotien $GF(2)[x]/langle x^3+x^2+1rangle$. In that case the coset of $x^3+x+1$ equals the coset of $x^2+x$ (because $(x^3+x+1)-(x^3+x^2+1)=x^2+x$). You can find the inverse of that by the usual process involving extended Euclid. You see, to positively id an element of the quotient ring we customarily reduce them by the polynomial defining the field.
– Jyrki Lahtonen
21 hours ago
By saying finite field, i mean the polynomial of the field, which is x^8+ x^4+ x^3+x+1. So i have to find the inverse of the first polynomial modulus this one
– Mariam Mashuryan
18 hours ago
What do you mean by $x^8 + x^4 + x^3 + x + 1$ being the "polynomial of the field"?
– Tobias Kildetoft
16 hours ago
|
show 3 more comments
Do it this way - much less painful and less error-prone.
– Bill Dubuque
yesterday
Normally $x$ stands for an indeterminate, when $x^3+x+1$ is a polynomial. Polynomials don't have inverses in a finite field. If, as I suspect, $x$ stands for some element of $GF(8)$ you need to tell us which element. Or at least, tell us its minimal polynomial. You see, very often the field $GF(8)$ is defined as the quotient ring $GF(2)[x]/langle x^3+x+1rangle$. But, in that field the coset of $x^3+x+1$ is equal to zero. Again implying that it has no inverse (can't divide by zero).
– Jyrki Lahtonen
21 hours ago
The (only) other alternative is to have defined $GF(8)$ as the quotien $GF(2)[x]/langle x^3+x^2+1rangle$. In that case the coset of $x^3+x+1$ equals the coset of $x^2+x$ (because $(x^3+x+1)-(x^3+x^2+1)=x^2+x$). You can find the inverse of that by the usual process involving extended Euclid. You see, to positively id an element of the quotient ring we customarily reduce them by the polynomial defining the field.
– Jyrki Lahtonen
21 hours ago
By saying finite field, i mean the polynomial of the field, which is x^8+ x^4+ x^3+x+1. So i have to find the inverse of the first polynomial modulus this one
– Mariam Mashuryan
18 hours ago
What do you mean by $x^8 + x^4 + x^3 + x + 1$ being the "polynomial of the field"?
– Tobias Kildetoft
16 hours ago
Do it this way - much less painful and less error-prone.
– Bill Dubuque
yesterday
Do it this way - much less painful and less error-prone.
– Bill Dubuque
yesterday
Normally $x$ stands for an indeterminate, when $x^3+x+1$ is a polynomial. Polynomials don't have inverses in a finite field. If, as I suspect, $x$ stands for some element of $GF(8)$ you need to tell us which element. Or at least, tell us its minimal polynomial. You see, very often the field $GF(8)$ is defined as the quotient ring $GF(2)[x]/langle x^3+x+1rangle$. But, in that field the coset of $x^3+x+1$ is equal to zero. Again implying that it has no inverse (can't divide by zero).
– Jyrki Lahtonen
21 hours ago
Normally $x$ stands for an indeterminate, when $x^3+x+1$ is a polynomial. Polynomials don't have inverses in a finite field. If, as I suspect, $x$ stands for some element of $GF(8)$ you need to tell us which element. Or at least, tell us its minimal polynomial. You see, very often the field $GF(8)$ is defined as the quotient ring $GF(2)[x]/langle x^3+x+1rangle$. But, in that field the coset of $x^3+x+1$ is equal to zero. Again implying that it has no inverse (can't divide by zero).
– Jyrki Lahtonen
21 hours ago
The (only) other alternative is to have defined $GF(8)$ as the quotien $GF(2)[x]/langle x^3+x^2+1rangle$. In that case the coset of $x^3+x+1$ equals the coset of $x^2+x$ (because $(x^3+x+1)-(x^3+x^2+1)=x^2+x$). You can find the inverse of that by the usual process involving extended Euclid. You see, to positively id an element of the quotient ring we customarily reduce them by the polynomial defining the field.
– Jyrki Lahtonen
21 hours ago
The (only) other alternative is to have defined $GF(8)$ as the quotien $GF(2)[x]/langle x^3+x^2+1rangle$. In that case the coset of $x^3+x+1$ equals the coset of $x^2+x$ (because $(x^3+x+1)-(x^3+x^2+1)=x^2+x$). You can find the inverse of that by the usual process involving extended Euclid. You see, to positively id an element of the quotient ring we customarily reduce them by the polynomial defining the field.
– Jyrki Lahtonen
21 hours ago
By saying finite field, i mean the polynomial of the field, which is x^8+ x^4+ x^3+x+1. So i have to find the inverse of the first polynomial modulus this one
– Mariam Mashuryan
18 hours ago
By saying finite field, i mean the polynomial of the field, which is x^8+ x^4+ x^3+x+1. So i have to find the inverse of the first polynomial modulus this one
– Mariam Mashuryan
18 hours ago
What do you mean by $x^8 + x^4 + x^3 + x + 1$ being the "polynomial of the field"?
– Tobias Kildetoft
16 hours ago
What do you mean by $x^8 + x^4 + x^3 + x + 1$ being the "polynomial of the field"?
– Tobias Kildetoft
16 hours ago
|
show 3 more comments
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Do it this way - much less painful and less error-prone.
– Bill Dubuque
yesterday
Normally $x$ stands for an indeterminate, when $x^3+x+1$ is a polynomial. Polynomials don't have inverses in a finite field. If, as I suspect, $x$ stands for some element of $GF(8)$ you need to tell us which element. Or at least, tell us its minimal polynomial. You see, very often the field $GF(8)$ is defined as the quotient ring $GF(2)[x]/langle x^3+x+1rangle$. But, in that field the coset of $x^3+x+1$ is equal to zero. Again implying that it has no inverse (can't divide by zero).
– Jyrki Lahtonen
21 hours ago
The (only) other alternative is to have defined $GF(8)$ as the quotien $GF(2)[x]/langle x^3+x^2+1rangle$. In that case the coset of $x^3+x+1$ equals the coset of $x^2+x$ (because $(x^3+x+1)-(x^3+x^2+1)=x^2+x$). You can find the inverse of that by the usual process involving extended Euclid. You see, to positively id an element of the quotient ring we customarily reduce them by the polynomial defining the field.
– Jyrki Lahtonen
21 hours ago
By saying finite field, i mean the polynomial of the field, which is x^8+ x^4+ x^3+x+1. So i have to find the inverse of the first polynomial modulus this one
– Mariam Mashuryan
18 hours ago
What do you mean by $x^8 + x^4 + x^3 + x + 1$ being the "polynomial of the field"?
– Tobias Kildetoft
16 hours ago