Mixing
Roots of Unity and Primitive roots
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For $mathbb{Z}/13$, I want to find its primitive roots and 4th roots of unity.
For $g$ to be a primitive root, we must have that $g^6 neq 1 pmod{13}$ and $g^4 neq 1 pmod{13}$. $2$ satisfies this. Do I just go through all the numbers $0,1,2, cdots, 12$ and check this? I'm not too clear what a primitive root is.
Then for the 4th roots of unity, do we just go through each number raised to the fourth power and see what gets us $1$?
So, $1,5, 8, 12$ are the 4th roots of unity.
abstract-algebra
asked Mar 2 '15 at 6:05
aldnoah.Algebraaldnoah.Algebra
62
$endgroup$
add a comment |
$begingroup$
For $mathbb{Z}/13$, I want to find its primitive roots and 4th roots of unity.
For $g$ to be a primitive root, we must have that $g^6 neq 1 pmod{13}$ and $g^4 neq 1 pmod{13}$. $2$ satisfies this. Do I just go through all the numbers $0,1,2, cdots, 12$ and check this? I'm not too clear what a primitive root is.
Then for the 4th roots of unity, do we just go through each number raised to the fourth power and see what gets us $1$?
So, $1,5, 8, 12$ are the 4th roots of unity.
abstract-algebra
asked Mar 2 '15 at 6:05
aldnoah.Algebraaldnoah.Algebra
62
$endgroup$
$begingroup$
The command bmod (or pmod, depending on your preferences) renders the mod notation nicely.
$endgroup$
– Travis
Mar 2 '15 at 6:14
$begingroup$
Thank you. I've edited the post appropriately.
$endgroup$
– aldnoah.Algebra
Mar 2 '15 at 6:21
$begingroup$
Note that if $g$ is a primitive root then $g^3$ and its powers are 4th roots of unity.
$endgroup$
– Gerry Myerson
Mar 2 '15 at 6:27
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Yes, this is a very small field, you can check the 12 non-zero elements by hand, just multiply and reduce mod 13 as you go. A primitive root of unity for $mathbb Z/13mathbb Z$ is one where its powers give you all 12 non-zero elements. It's enough to raise it to the 6th power, if none of those are one then it must be primitive.
$endgroup$
– Gregory Grant
Mar 2 '15 at 6:33
$begingroup$
if $(i,phi(n))=1$and g is a primitive roots $g^i$is primitive roots and they are all primitve roots
$endgroup$
– ali
Mar 2 '15 at 18:14
add a comment |
$begingroup$
For $mathbb{Z}/13$, I want to find its primitive roots and 4th roots of unity.
For $g$ to be a primitive root, we must have that $g^6 neq 1 pmod{13}$ and $g^4 neq 1 pmod{13}$. $2$ satisfies this. Do I just go through all the numbers $0,1,2, cdots, 12$ and check this? I'm not too clear what a primitive root is.
Then for the 4th roots of unity, do we just go through each number raised to the fourth power and see what gets us $1$?
So, $1,5, 8, 12$ are the 4th roots of unity.
abstract-algebra
asked Mar 2 '15 at 6:05
aldnoah.Algebraaldnoah.Algebra
62
$endgroup$
For $mathbb{Z}/13$, I want to find its primitive roots and 4th roots of unity.
For $g$ to be a primitive root, we must have that $g^6 neq 1 pmod{13}$ and $g^4 neq 1 pmod{13}$. $2$ satisfies this. Do I just go through all the numbers $0,1,2, cdots, 12$ and check this? I'm not too clear what a primitive root is.
Then for the 4th roots of unity, do we just go through each number raised to the fourth power and see what gets us $1$?
So, $1,5, 8, 12$ are the 4th roots of unity.
abstract-algebra
abstract-algebra
asked Mar 2 '15 at 6:05
aldnoah.Algebraaldnoah.Algebra
62
asked Mar 2 '15 at 6:05
aldnoah.Algebraaldnoah.Algebra
62
edited Mar 2 '15 at 6:21
aldnoah.Algebra
asked Mar 2 '15 at 6:05
aldnoah.Algebraaldnoah.Algebra
62
asked Mar 2 '15 at 6:05
aldnoah.Algebraaldnoah.Algebra
62
asked Mar 2 '15 at 6:05
aldnoah.Algebraaldnoah.Algebra
62
62
$begingroup$
The command bmod (or pmod, depending on your preferences) renders the mod notation nicely.
$endgroup$
– Travis
Mar 2 '15 at 6:14
$begingroup$
Thank you. I've edited the post appropriately.
$endgroup$
– aldnoah.Algebra
Mar 2 '15 at 6:21
$begingroup$
Note that if $g$ is a primitive root then $g^3$ and its powers are 4th roots of unity.
$endgroup$
– Gerry Myerson
Mar 2 '15 at 6:27
$begingroup$
Yes, this is a very small field, you can check the 12 non-zero elements by hand, just multiply and reduce mod 13 as you go. A primitive root of unity for $mathbb Z/13mathbb Z$ is one where its powers give you all 12 non-zero elements. It's enough to raise it to the 6th power, if none of those are one then it must be primitive.
$endgroup$
– Gregory Grant
Mar 2 '15 at 6:33
$begingroup$
if $(i,phi(n))=1$and g is a primitive roots $g^i$is primitive roots and they are all primitve roots
$endgroup$
– ali
Mar 2 '15 at 18:14
add a comment |
$begingroup$
The command bmod (or pmod, depending on your preferences) renders the mod notation nicely.
$endgroup$
– Travis
Mar 2 '15 at 6:14
$begingroup$
Thank you. I've edited the post appropriately.
$endgroup$
– aldnoah.Algebra
Mar 2 '15 at 6:21
$begingroup$
Note that if $g$ is a primitive root then $g^3$ and its powers are 4th roots of unity.
$endgroup$
– Gerry Myerson
Mar 2 '15 at 6:27
$begingroup$
Yes, this is a very small field, you can check the 12 non-zero elements by hand, just multiply and reduce mod 13 as you go. A primitive root of unity for $mathbb Z/13mathbb Z$ is one where its powers give you all 12 non-zero elements. It's enough to raise it to the 6th power, if none of those are one then it must be primitive.
$endgroup$
– Gregory Grant
Mar 2 '15 at 6:33
$begingroup$
if $(i,phi(n))=1$and g is a primitive roots $g^i$is primitive roots and they are all primitve roots
$endgroup$
– ali
Mar 2 '15 at 18:14
$begingroup$
The command bmod (or pmod, depending on your preferences) renders the mod notation nicely.
$endgroup$
– Travis
Mar 2 '15 at 6:14
$begingroup$
The command bmod (or pmod, depending on your preferences) renders the mod notation nicely.
$endgroup$
– Travis
Mar 2 '15 at 6:14
$begingroup$
Thank you. I've edited the post appropriately.
$endgroup$
– aldnoah.Algebra
Mar 2 '15 at 6:21
$begingroup$
Thank you. I've edited the post appropriately.
$endgroup$
– aldnoah.Algebra
Mar 2 '15 at 6:21
$begingroup$
Note that if $g$ is a primitive root then $g^3$ and its powers are 4th roots of unity.
$endgroup$
– Gerry Myerson
Mar 2 '15 at 6:27
$begingroup$
Note that if $g$ is a primitive root then $g^3$ and its powers are 4th roots of unity.
$endgroup$
– Gerry Myerson
Mar 2 '15 at 6:27
$begingroup$
Yes, this is a very small field, you can check the 12 non-zero elements by hand, just multiply and reduce mod 13 as you go. A primitive root of unity for $mathbb Z/13mathbb Z$ is one where its powers give you all 12 non-zero elements. It's enough to raise it to the 6th power, if none of those are one then it must be primitive.
$endgroup$
– Gregory Grant
Mar 2 '15 at 6:33
$begingroup$
Yes, this is a very small field, you can check the 12 non-zero elements by hand, just multiply and reduce mod 13 as you go. A primitive root of unity for $mathbb Z/13mathbb Z$ is one where its powers give you all 12 non-zero elements. It's enough to raise it to the 6th power, if none of those are one then it must be primitive.
$endgroup$
– Gregory Grant
Mar 2 '15 at 6:33
$begingroup$
if $(i,phi(n))=1$and g is a primitive roots $g^i$is primitive roots and they are all primitve roots
$endgroup$
– ali
Mar 2 '15 at 18:14
$begingroup$
if $(i,phi(n))=1$and g is a primitive roots $g^i$is primitive roots and they are all primitve roots
$endgroup$
– ali
Mar 2 '15 at 18:14
add a comment |
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$begingroup$
The command bmod (or pmod, depending on your preferences) renders the mod notation nicely.
$endgroup$
– Travis
Mar 2 '15 at 6:14
$begingroup$
Thank you. I've edited the post appropriately.
$endgroup$
– aldnoah.Algebra
Mar 2 '15 at 6:21
$begingroup$
Note that if $g$ is a primitive root then $g^3$ and its powers are 4th roots of unity.
$endgroup$
– Gerry Myerson
Mar 2 '15 at 6:27
$begingroup$
Yes, this is a very small field, you can check the 12 non-zero elements by hand, just multiply and reduce mod 13 as you go. A primitive root of unity for $mathbb Z/13mathbb Z$ is one where its powers give you all 12 non-zero elements. It's enough to raise it to the 6th power, if none of those are one then it must be primitive.
$endgroup$
– Gregory Grant
Mar 2 '15 at 6:33
$begingroup$
if $(i,phi(n))=1$and g is a primitive roots $g^i$is primitive roots and they are all primitve roots
$endgroup$
– ali
Mar 2 '15 at 18:14