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Prove that $t^4 + t^3 + t^2 + t+1$ is not a primitive polynomial for $mathbb{F}_2$
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I tried showing that $f(t) = t^4 + t^3 + t^2 + t+1$ is reducible over $mathbb{F}_2$, but it turns out that $f(t)$ is irreducible. So I have to check whether or not $t$ is a primitive element of $mathbb{F}_2$, i.e. $t$ generates the cyclic group $mathbb{F}_1 ^* = { 1 }$. However $t^1 ne 1$ and therefore $f(t)$ is not primitive.
Is this correct? Thanks.
field-theory
asked Jan 27 at 12:39
ZacharyZachary
1919
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I tried showing that $f(t) = t^4 + t^3 + t^2 + t+1$ is reducible over $mathbb{F}_2$, but it turns out that $f(t)$ is irreducible. So I have to check whether or not $t$ is a primitive element of $mathbb{F}_2$, i.e. $t$ generates the cyclic group $mathbb{F}_1 ^* = { 1 }$. However $t^1 ne 1$ and therefore $f(t)$ is not primitive.
Is this correct? Thanks.
field-theory
asked Jan 27 at 12:39
ZacharyZachary
1919
$endgroup$
add a comment |
$begingroup$
I tried showing that $f(t) = t^4 + t^3 + t^2 + t+1$ is reducible over $mathbb{F}_2$, but it turns out that $f(t)$ is irreducible. So I have to check whether or not $t$ is a primitive element of $mathbb{F}_2$, i.e. $t$ generates the cyclic group $mathbb{F}_1 ^* = { 1 }$. However $t^1 ne 1$ and therefore $f(t)$ is not primitive.
Is this correct? Thanks.
field-theory
asked Jan 27 at 12:39
ZacharyZachary
1919
$endgroup$
I tried showing that $f(t) = t^4 + t^3 + t^2 + t+1$ is reducible over $mathbb{F}_2$, but it turns out that $f(t)$ is irreducible. So I have to check whether or not $t$ is a primitive element of $mathbb{F}_2$, i.e. $t$ generates the cyclic group $mathbb{F}_1 ^* = { 1 }$. However $t^1 ne 1$ and therefore $f(t)$ is not primitive.
Is this correct? Thanks.
field-theory
field-theory
asked Jan 27 at 12:39
ZacharyZachary
1919
asked Jan 27 at 12:39
ZacharyZachary
1919
asked Jan 27 at 12:39
ZacharyZachary
1919
asked Jan 27 at 12:39
ZacharyZachary
1919
asked Jan 27 at 12:39
ZacharyZachary
1919
1919
add a comment |
add a comment |
2 Answers
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In field theory, a branch of mathematics, a primitive polynomial is the minimal polynomial of a primitive element of the finite extension field GF$(p^m)$.
Definition, straight from Wikipedia. The root doesn't have to be in the base field. It shouldn't be in the base field.
Is the root going to be a primitive element in the larger field's multiplicative group? Well - this is a cyclotomic polynomial. Since $(t-1)f(t)=t^5-1$, any $t$ that is a root is a fifth root of unity. Since $1$ isn't a root, that means it's a primitive fifth root of unity. The only way that can generate the multiplicative group for a field is if that field has exactly five invertible elements - so, then, it has exactly six total elements. There are no such fields. No matter what prime we choose, this isn't a primitive polynomial.
answered Jan 27 at 12:49
jmerryjmerry
16.1k1633
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As $f$ is irreducible, $f$ is primitive iff any of its zeros, $alpha$ say, generates the cyclic group $Bbb F_{2^4}^*$, which has order $15$. But $f(t)mid
(t^5-1)$ as polynomials, so $alpha^5=1$.
answered Jan 27 at 12:50
Lord Shark the UnknownLord Shark the Unknown
107k1162134
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2 Answers
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2 Answers
2
active
oldest
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active
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active
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$begingroup$
In field theory, a branch of mathematics, a primitive polynomial is the minimal polynomial of a primitive element of the finite extension field GF$(p^m)$.
Definition, straight from Wikipedia. The root doesn't have to be in the base field. It shouldn't be in the base field.
Is the root going to be a primitive element in the larger field's multiplicative group? Well - this is a cyclotomic polynomial. Since $(t-1)f(t)=t^5-1$, any $t$ that is a root is a fifth root of unity. Since $1$ isn't a root, that means it's a primitive fifth root of unity. The only way that can generate the multiplicative group for a field is if that field has exactly five invertible elements - so, then, it has exactly six total elements. There are no such fields. No matter what prime we choose, this isn't a primitive polynomial.
answered Jan 27 at 12:49
jmerryjmerry
16.1k1633
$endgroup$
add a comment |
$begingroup$
In field theory, a branch of mathematics, a primitive polynomial is the minimal polynomial of a primitive element of the finite extension field GF$(p^m)$.
Definition, straight from Wikipedia. The root doesn't have to be in the base field. It shouldn't be in the base field.
Is the root going to be a primitive element in the larger field's multiplicative group? Well - this is a cyclotomic polynomial. Since $(t-1)f(t)=t^5-1$, any $t$ that is a root is a fifth root of unity. Since $1$ isn't a root, that means it's a primitive fifth root of unity. The only way that can generate the multiplicative group for a field is if that field has exactly five invertible elements - so, then, it has exactly six total elements. There are no such fields. No matter what prime we choose, this isn't a primitive polynomial.
answered Jan 27 at 12:49
jmerryjmerry
16.1k1633
$endgroup$
add a comment |
$begingroup$
In field theory, a branch of mathematics, a primitive polynomial is the minimal polynomial of a primitive element of the finite extension field GF$(p^m)$.
Definition, straight from Wikipedia. The root doesn't have to be in the base field. It shouldn't be in the base field.
Is the root going to be a primitive element in the larger field's multiplicative group? Well - this is a cyclotomic polynomial. Since $(t-1)f(t)=t^5-1$, any $t$ that is a root is a fifth root of unity. Since $1$ isn't a root, that means it's a primitive fifth root of unity. The only way that can generate the multiplicative group for a field is if that field has exactly five invertible elements - so, then, it has exactly six total elements. There are no such fields. No matter what prime we choose, this isn't a primitive polynomial.
answered Jan 27 at 12:49
jmerryjmerry
16.1k1633
$endgroup$
In field theory, a branch of mathematics, a primitive polynomial is the minimal polynomial of a primitive element of the finite extension field GF$(p^m)$.
Definition, straight from Wikipedia. The root doesn't have to be in the base field. It shouldn't be in the base field.
Is the root going to be a primitive element in the larger field's multiplicative group? Well - this is a cyclotomic polynomial. Since $(t-1)f(t)=t^5-1$, any $t$ that is a root is a fifth root of unity. Since $1$ isn't a root, that means it's a primitive fifth root of unity. The only way that can generate the multiplicative group for a field is if that field has exactly five invertible elements - so, then, it has exactly six total elements. There are no such fields. No matter what prime we choose, this isn't a primitive polynomial.
answered Jan 27 at 12:49
jmerryjmerry
16.1k1633
answered Jan 27 at 12:49
jmerryjmerry
16.1k1633
answered Jan 27 at 12:49
jmerryjmerry
16.1k1633
answered Jan 27 at 12:49
jmerryjmerry
16.1k1633
16.1k1633
add a comment |
add a comment |
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As $f$ is irreducible, $f$ is primitive iff any of its zeros, $alpha$ say, generates the cyclic group $Bbb F_{2^4}^*$, which has order $15$. But $f(t)mid
(t^5-1)$ as polynomials, so $alpha^5=1$.
answered Jan 27 at 12:50
Lord Shark the UnknownLord Shark the Unknown
107k1162134
$endgroup$
add a comment |
$begingroup$
As $f$ is irreducible, $f$ is primitive iff any of its zeros, $alpha$ say, generates the cyclic group $Bbb F_{2^4}^*$, which has order $15$. But $f(t)mid
(t^5-1)$ as polynomials, so $alpha^5=1$.
answered Jan 27 at 12:50
Lord Shark the UnknownLord Shark the Unknown
107k1162134
$endgroup$
add a comment |
$begingroup$
As $f$ is irreducible, $f$ is primitive iff any of its zeros, $alpha$ say, generates the cyclic group $Bbb F_{2^4}^*$, which has order $15$. But $f(t)mid
(t^5-1)$ as polynomials, so $alpha^5=1$.
answered Jan 27 at 12:50
Lord Shark the UnknownLord Shark the Unknown
107k1162134
$endgroup$
As $f$ is irreducible, $f$ is primitive iff any of its zeros, $alpha$ say, generates the cyclic group $Bbb F_{2^4}^*$, which has order $15$. But $f(t)mid
(t^5-1)$ as polynomials, so $alpha^5=1$.
answered Jan 27 at 12:50
Lord Shark the UnknownLord Shark the Unknown
107k1162134
answered Jan 27 at 12:50
Lord Shark the UnknownLord Shark the Unknown
107k1162134
answered Jan 27 at 12:50
Lord Shark the UnknownLord Shark the Unknown
107k1162134
answered Jan 27 at 12:50
Lord Shark the UnknownLord Shark the Unknown
107k1162134
107k1162134
add a comment |
add a comment |
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