Mixing
Verifying if a given polynomial is primitive polynomial
$begingroup$
Given a polynomial: $f(x) = x^2 + 2x + 2$ over $GF(3)$. I want to know if i can use it to construct $GF(3^2)$.
My approach:
- This equation satisfies first condition: A primitive polynomial is irreducible.
- The second condition i'm trying to confirm is that i can use it generate $GF(3^2)$.
$alpha^2 + 2alpha + 2 rightarrow alpha^2 = alpha + 1$.
- $alpha^1 = alpha$
- $alpha^2 = alpha+1$
- $alpha^3 = alpha*alpha^2 rightarrow1 + 2alpha $
- $alpha^4 = 2 $
- $alpha^5 = 2alpha$
- $alpha^6 = alpha+2$
$alpha^7 = 1$
Now Here $alpha^8$ should be equal to $1$ instead of $alpha^7 .$
$alpha^8 = alpha$
What i am doing wrong , any help would be great.
polynomials galois-theory finite-fields cryptography coding-theory
asked Jan 27 at 4:05
Khan SaabKhan Saab
1979
$endgroup$
add a comment |
$begingroup$
Given a polynomial: $f(x) = x^2 + 2x + 2$ over $GF(3)$. I want to know if i can use it to construct $GF(3^2)$.
My approach:
- This equation satisfies first condition: A primitive polynomial is irreducible.
- The second condition i'm trying to confirm is that i can use it generate $GF(3^2)$.
$alpha^2 + 2alpha + 2 rightarrow alpha^2 = alpha + 1$.
- $alpha^1 = alpha$
- $alpha^2 = alpha+1$
- $alpha^3 = alpha*alpha^2 rightarrow1 + 2alpha $
- $alpha^4 = 2 $
- $alpha^5 = 2alpha$
- $alpha^6 = alpha+2$
$alpha^7 = 1$
Now Here $alpha^8$ should be equal to $1$ instead of $alpha^7 .$
$alpha^8 = alpha$
What i am doing wrong , any help would be great.
polynomials galois-theory finite-fields cryptography coding-theory
asked Jan 27 at 4:05
Khan SaabKhan Saab
1979
$endgroup$
$begingroup$
You made an error at $alpha^6$. See my answer. Cheers!
$endgroup$
– Robert Lewis
Jan 27 at 4:59
1
$begingroup$
You simply skipped $alpha^6$. You correctly found out that $alpha^4=2=-1$ is in the prime field. From this it follows that for all $i$ we have $alpha^{4+i}=-alpha^i$ (or $=2alpha^i$ if you prefer that). So you should have $alpha^6=2alpha^2=2alpha+2$, and only $alpha^7=2alpha^3=alpha+2$. Undoubtedly you multiplied by $alpha$ twice at that point, and forgot to record the result. Anyway, once you get back to the prime field you can use that to check the rest as above.
$endgroup$
– Jyrki Lahtonen
Jan 27 at 19:34
add a comment |
$begingroup$
Given a polynomial: $f(x) = x^2 + 2x + 2$ over $GF(3)$. I want to know if i can use it to construct $GF(3^2)$.
My approach:
- This equation satisfies first condition: A primitive polynomial is irreducible.
- The second condition i'm trying to confirm is that i can use it generate $GF(3^2)$.
$alpha^2 + 2alpha + 2 rightarrow alpha^2 = alpha + 1$.
- $alpha^1 = alpha$
- $alpha^2 = alpha+1$
- $alpha^3 = alpha*alpha^2 rightarrow1 + 2alpha $
- $alpha^4 = 2 $
- $alpha^5 = 2alpha$
- $alpha^6 = alpha+2$
$alpha^7 = 1$
Now Here $alpha^8$ should be equal to $1$ instead of $alpha^7 .$
$alpha^8 = alpha$
What i am doing wrong , any help would be great.
polynomials galois-theory finite-fields cryptography coding-theory
asked Jan 27 at 4:05
Khan SaabKhan Saab
1979
$endgroup$
Given a polynomial: $f(x) = x^2 + 2x + 2$ over $GF(3)$. I want to know if i can use it to construct $GF(3^2)$.
My approach:
- This equation satisfies first condition: A primitive polynomial is irreducible.
- The second condition i'm trying to confirm is that i can use it generate $GF(3^2)$.
$alpha^2 + 2alpha + 2 rightarrow alpha^2 = alpha + 1$.
- $alpha^1 = alpha$
- $alpha^2 = alpha+1$
- $alpha^3 = alpha*alpha^2 rightarrow1 + 2alpha $
- $alpha^4 = 2 $
- $alpha^5 = 2alpha$
- $alpha^6 = alpha+2$
$alpha^7 = 1$
Now Here $alpha^8$ should be equal to $1$ instead of $alpha^7 .$
$alpha^8 = alpha$
What i am doing wrong , any help would be great.
polynomials galois-theory finite-fields cryptography coding-theory
polynomials galois-theory finite-fields cryptography coding-theory
asked Jan 27 at 4:05
Khan SaabKhan Saab
1979
asked Jan 27 at 4:05
Khan SaabKhan Saab
1979
asked Jan 27 at 4:05
Khan SaabKhan Saab
1979
asked Jan 27 at 4:05
Khan SaabKhan Saab
1979
asked Jan 27 at 4:05
Khan SaabKhan Saab
1979
1979
$begingroup$
You made an error at $alpha^6$. See my answer. Cheers!
$endgroup$
– Robert Lewis
Jan 27 at 4:59
1
$begingroup$
You simply skipped $alpha^6$. You correctly found out that $alpha^4=2=-1$ is in the prime field. From this it follows that for all $i$ we have $alpha^{4+i}=-alpha^i$ (or $=2alpha^i$ if you prefer that). So you should have $alpha^6=2alpha^2=2alpha+2$, and only $alpha^7=2alpha^3=alpha+2$. Undoubtedly you multiplied by $alpha$ twice at that point, and forgot to record the result. Anyway, once you get back to the prime field you can use that to check the rest as above.
$endgroup$
– Jyrki Lahtonen
Jan 27 at 19:34
add a comment |
$begingroup$
You made an error at $alpha^6$. See my answer. Cheers!
$endgroup$
– Robert Lewis
Jan 27 at 4:59
1
$begingroup$
You simply skipped $alpha^6$. You correctly found out that $alpha^4=2=-1$ is in the prime field. From this it follows that for all $i$ we have $alpha^{4+i}=-alpha^i$ (or $=2alpha^i$ if you prefer that). So you should have $alpha^6=2alpha^2=2alpha+2$, and only $alpha^7=2alpha^3=alpha+2$. Undoubtedly you multiplied by $alpha$ twice at that point, and forgot to record the result. Anyway, once you get back to the prime field you can use that to check the rest as above.
$endgroup$
– Jyrki Lahtonen
Jan 27 at 19:34
$begingroup$
You made an error at $alpha^6$. See my answer. Cheers!
$endgroup$
– Robert Lewis
Jan 27 at 4:59
$begingroup$
You made an error at $alpha^6$. See my answer. Cheers!
$endgroup$
– Robert Lewis
Jan 27 at 4:59
1
1
$begingroup$
You simply skipped $alpha^6$. You correctly found out that $alpha^4=2=-1$ is in the prime field. From this it follows that for all $i$ we have $alpha^{4+i}=-alpha^i$ (or $=2alpha^i$ if you prefer that). So you should have $alpha^6=2alpha^2=2alpha+2$, and only $alpha^7=2alpha^3=alpha+2$. Undoubtedly you multiplied by $alpha$ twice at that point, and forgot to record the result. Anyway, once you get back to the prime field you can use that to check the rest as above.
$endgroup$
– Jyrki Lahtonen
Jan 27 at 19:34
$begingroup$
You simply skipped $alpha^6$. You correctly found out that $alpha^4=2=-1$ is in the prime field. From this it follows that for all $i$ we have $alpha^{4+i}=-alpha^i$ (or $=2alpha^i$ if you prefer that). So you should have $alpha^6=2alpha^2=2alpha+2$, and only $alpha^7=2alpha^3=alpha+2$. Undoubtedly you multiplied by $alpha$ twice at that point, and forgot to record the result. Anyway, once you get back to the prime field you can use that to check the rest as above.
$endgroup$
– Jyrki Lahtonen
Jan 27 at 19:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It strikes me that our OP Khan Saab's calculations of the powers of $alpha$ are OK through $alpha^5$, but there is an error in $alpha^6$. With
$alpha^2 = alpha + 1, tag 1$
we have, correctly,
$alpha^5 = 2 alpha; tag 2$
then
$alpha^6 = 2 alpha^2 = 2 alpha + 2, tag 3$
not $alpha^6 = alpha + 2$!
Continuing:
$alpha^7 = 2alpha^2 + 2alpha = 2(alpha + 1) + 2alpha = alpha + 2; tag 4$
$alpha^8 = alpha^2 + 2alpha = alpha + 1 + 2alpha = 1! tag 5$
easy, you know the way it's supposed to be!
answered Jan 27 at 4:43
Robert LewisRobert Lewis
48.3k23167
$endgroup$
1
$begingroup$
Thank you very much :)
$endgroup$
– Khan Saab
Jan 27 at 5:05
$begingroup$
@KhanSaab: my pleasure sir! And thank you for the "acceptance"!
$endgroup$
– Robert Lewis
Jan 27 at 5:06
1
$begingroup$
Good job. ${}{}$
$endgroup$
– Jyrki Lahtonen
Jan 27 at 19:35
$begingroup$
@JyrkiLahtonen: Quite a compliment considering the source! Thanks, Jyrki.
$endgroup$
– Robert Lewis
Jan 27 at 19:36
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
It strikes me that our OP Khan Saab's calculations of the powers of $alpha$ are OK through $alpha^5$, but there is an error in $alpha^6$. With
$alpha^2 = alpha + 1, tag 1$
we have, correctly,
$alpha^5 = 2 alpha; tag 2$
then
$alpha^6 = 2 alpha^2 = 2 alpha + 2, tag 3$
not $alpha^6 = alpha + 2$!
Continuing:
$alpha^7 = 2alpha^2 + 2alpha = 2(alpha + 1) + 2alpha = alpha + 2; tag 4$
$alpha^8 = alpha^2 + 2alpha = alpha + 1 + 2alpha = 1! tag 5$
easy, you know the way it's supposed to be!
answered Jan 27 at 4:43
Robert LewisRobert Lewis
48.3k23167
$endgroup$
1
$begingroup$
Thank you very much :)
$endgroup$
– Khan Saab
Jan 27 at 5:05
$begingroup$
@KhanSaab: my pleasure sir! And thank you for the "acceptance"!
$endgroup$
– Robert Lewis
Jan 27 at 5:06
1
$begingroup$
Good job. ${}{}$
$endgroup$
– Jyrki Lahtonen
Jan 27 at 19:35
$begingroup$
@JyrkiLahtonen: Quite a compliment considering the source! Thanks, Jyrki.
$endgroup$
– Robert Lewis
Jan 27 at 19:36
add a comment |
$begingroup$
It strikes me that our OP Khan Saab's calculations of the powers of $alpha$ are OK through $alpha^5$, but there is an error in $alpha^6$. With
$alpha^2 = alpha + 1, tag 1$
we have, correctly,
$alpha^5 = 2 alpha; tag 2$
then
$alpha^6 = 2 alpha^2 = 2 alpha + 2, tag 3$
not $alpha^6 = alpha + 2$!
Continuing:
$alpha^7 = 2alpha^2 + 2alpha = 2(alpha + 1) + 2alpha = alpha + 2; tag 4$
$alpha^8 = alpha^2 + 2alpha = alpha + 1 + 2alpha = 1! tag 5$
easy, you know the way it's supposed to be!
answered Jan 27 at 4:43
Robert LewisRobert Lewis
48.3k23167
$endgroup$
1
$begingroup$
Thank you very much :)
$endgroup$
– Khan Saab
Jan 27 at 5:05
$begingroup$
@KhanSaab: my pleasure sir! And thank you for the "acceptance"!
$endgroup$
– Robert Lewis
Jan 27 at 5:06
1
$begingroup$
Good job. ${}{}$
$endgroup$
– Jyrki Lahtonen
Jan 27 at 19:35
$begingroup$
@JyrkiLahtonen: Quite a compliment considering the source! Thanks, Jyrki.
$endgroup$
– Robert Lewis
Jan 27 at 19:36
add a comment |
$begingroup$
It strikes me that our OP Khan Saab's calculations of the powers of $alpha$ are OK through $alpha^5$, but there is an error in $alpha^6$. With
$alpha^2 = alpha + 1, tag 1$
we have, correctly,
$alpha^5 = 2 alpha; tag 2$
then
$alpha^6 = 2 alpha^2 = 2 alpha + 2, tag 3$
not $alpha^6 = alpha + 2$!
Continuing:
$alpha^7 = 2alpha^2 + 2alpha = 2(alpha + 1) + 2alpha = alpha + 2; tag 4$
$alpha^8 = alpha^2 + 2alpha = alpha + 1 + 2alpha = 1! tag 5$
easy, you know the way it's supposed to be!
answered Jan 27 at 4:43
Robert LewisRobert Lewis
48.3k23167
$endgroup$
It strikes me that our OP Khan Saab's calculations of the powers of $alpha$ are OK through $alpha^5$, but there is an error in $alpha^6$. With
$alpha^2 = alpha + 1, tag 1$
we have, correctly,
$alpha^5 = 2 alpha; tag 2$
then
$alpha^6 = 2 alpha^2 = 2 alpha + 2, tag 3$
not $alpha^6 = alpha + 2$!
Continuing:
$alpha^7 = 2alpha^2 + 2alpha = 2(alpha + 1) + 2alpha = alpha + 2; tag 4$
$alpha^8 = alpha^2 + 2alpha = alpha + 1 + 2alpha = 1! tag 5$
easy, you know the way it's supposed to be!
answered Jan 27 at 4:43
Robert LewisRobert Lewis
48.3k23167
edited Jan 27 at 4:50
answered Jan 27 at 4:43
Robert LewisRobert Lewis
48.3k23167
answered Jan 27 at 4:43
Robert LewisRobert Lewis
48.3k23167
answered Jan 27 at 4:43
Robert LewisRobert Lewis
48.3k23167
48.3k23167
1
$begingroup$
Thank you very much :)
$endgroup$
– Khan Saab
Jan 27 at 5:05
$begingroup$
@KhanSaab: my pleasure sir! And thank you for the "acceptance"!
$endgroup$
– Robert Lewis
Jan 27 at 5:06
1
$begingroup$
Good job. ${}{}$
$endgroup$
– Jyrki Lahtonen
Jan 27 at 19:35
$begingroup$
@JyrkiLahtonen: Quite a compliment considering the source! Thanks, Jyrki.
$endgroup$
– Robert Lewis
Jan 27 at 19:36
add a comment |
1
$begingroup$
Thank you very much :)
$endgroup$
– Khan Saab
Jan 27 at 5:05
$begingroup$
@KhanSaab: my pleasure sir! And thank you for the "acceptance"!
$endgroup$
– Robert Lewis
Jan 27 at 5:06
1
$begingroup$
Good job. ${}{}$
$endgroup$
– Jyrki Lahtonen
Jan 27 at 19:35
$begingroup$
@JyrkiLahtonen: Quite a compliment considering the source! Thanks, Jyrki.
$endgroup$
– Robert Lewis
Jan 27 at 19:36
1
1
$begingroup$
Thank you very much :)
$endgroup$
– Khan Saab
Jan 27 at 5:05
$begingroup$
Thank you very much :)
$endgroup$
– Khan Saab
Jan 27 at 5:05
$begingroup$
@KhanSaab: my pleasure sir! And thank you for the "acceptance"!
$endgroup$
– Robert Lewis
Jan 27 at 5:06
$begingroup$
@KhanSaab: my pleasure sir! And thank you for the "acceptance"!
$endgroup$
– Robert Lewis
Jan 27 at 5:06
1
1
$begingroup$
Good job. ${}{}$
$endgroup$
– Jyrki Lahtonen
Jan 27 at 19:35
$begingroup$
Good job. ${}{}$
$endgroup$
– Jyrki Lahtonen
Jan 27 at 19:35
$begingroup$
@JyrkiLahtonen: Quite a compliment considering the source! Thanks, Jyrki.
$endgroup$
– Robert Lewis
Jan 27 at 19:36
$begingroup$
@JyrkiLahtonen: Quite a compliment considering the source! Thanks, Jyrki.
$endgroup$
– Robert Lewis
Jan 27 at 19:36
add a comment |
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$begingroup$
You made an error at $alpha^6$. See my answer. Cheers!
$endgroup$
– Robert Lewis
Jan 27 at 4:59
1
$begingroup$
You simply skipped $alpha^6$. You correctly found out that $alpha^4=2=-1$ is in the prime field. From this it follows that for all $i$ we have $alpha^{4+i}=-alpha^i$ (or $=2alpha^i$ if you prefer that). So you should have $alpha^6=2alpha^2=2alpha+2$, and only $alpha^7=2alpha^3=alpha+2$. Undoubtedly you multiplied by $alpha$ twice at that point, and forgot to record the result. Anyway, once you get back to the prime field you can use that to check the rest as above.
$endgroup$
– Jyrki Lahtonen
Jan 27 at 19:34