Is it true that he polynomial $frac{x^p- 1}{x-1}$ ($p$ is prime) is irreducible in $mathbb{F}_2[x]$ iff $p$...
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Is it true that the polynomial $frac{x^p- 1}{x-1}$ ($p$ is prime) is irreducible in $mathbb{F}_2[x]$ iff $p$ is prime?
I know it will be true in $mathbb Q[x]$. Can anyone please help me to understand what happens in $mathbb{F}_2[x]$?
abstract-algebra polynomials ring-theory finite-fields irreducible-polynomials
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show 5 more comments
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Is it true that the polynomial $frac{x^p- 1}{x-1}$ ($p$ is prime) is irreducible in $mathbb{F}_2[x]$ iff $p$ is prime?
I know it will be true in $mathbb Q[x]$. Can anyone please help me to understand what happens in $mathbb{F}_2[x]$?
abstract-algebra polynomials ring-theory finite-fields irreducible-polynomials
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Are you sure that "($p$ is prime)" is supposed to be there? Should it, perhaps, be "($p$ is a natural number)"?
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– Arthur
Dec 14 '18 at 15:30
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en.wikipedia.org/wiki/Cyclotomic_polynomial
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– vadim123
Dec 14 '18 at 15:33
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Is the statement true for any natural number?@Arthur
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– cmi
Dec 14 '18 at 15:35
2
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This is false when $p=7$. For prime $p> 2$, it is true if and only if $2$ is a generator mod $p$.
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– KCd
Dec 14 '18 at 15:43
$begingroup$
"2 is a generator mod p" - means?@KCd
$endgroup$
– cmi
Dec 14 '18 at 15:45
|
show 5 more comments
$begingroup$
Is it true that the polynomial $frac{x^p- 1}{x-1}$ ($p$ is prime) is irreducible in $mathbb{F}_2[x]$ iff $p$ is prime?
I know it will be true in $mathbb Q[x]$. Can anyone please help me to understand what happens in $mathbb{F}_2[x]$?
abstract-algebra polynomials ring-theory finite-fields irreducible-polynomials
$endgroup$
Is it true that the polynomial $frac{x^p- 1}{x-1}$ ($p$ is prime) is irreducible in $mathbb{F}_2[x]$ iff $p$ is prime?
I know it will be true in $mathbb Q[x]$. Can anyone please help me to understand what happens in $mathbb{F}_2[x]$?
abstract-algebra polynomials ring-theory finite-fields irreducible-polynomials
abstract-algebra polynomials ring-theory finite-fields irreducible-polynomials
edited Dec 14 '18 at 21:27
Batominovski
33.1k33293
33.1k33293
asked Dec 14 '18 at 15:28
cmicmi
1,121312
1,121312
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Are you sure that "($p$ is prime)" is supposed to be there? Should it, perhaps, be "($p$ is a natural number)"?
$endgroup$
– Arthur
Dec 14 '18 at 15:30
$begingroup$
en.wikipedia.org/wiki/Cyclotomic_polynomial
$endgroup$
– vadim123
Dec 14 '18 at 15:33
$begingroup$
Is the statement true for any natural number?@Arthur
$endgroup$
– cmi
Dec 14 '18 at 15:35
2
$begingroup$
This is false when $p=7$. For prime $p> 2$, it is true if and only if $2$ is a generator mod $p$.
$endgroup$
– KCd
Dec 14 '18 at 15:43
$begingroup$
"2 is a generator mod p" - means?@KCd
$endgroup$
– cmi
Dec 14 '18 at 15:45
|
show 5 more comments
$begingroup$
Are you sure that "($p$ is prime)" is supposed to be there? Should it, perhaps, be "($p$ is a natural number)"?
$endgroup$
– Arthur
Dec 14 '18 at 15:30
$begingroup$
en.wikipedia.org/wiki/Cyclotomic_polynomial
$endgroup$
– vadim123
Dec 14 '18 at 15:33
$begingroup$
Is the statement true for any natural number?@Arthur
$endgroup$
– cmi
Dec 14 '18 at 15:35
2
$begingroup$
This is false when $p=7$. For prime $p> 2$, it is true if and only if $2$ is a generator mod $p$.
$endgroup$
– KCd
Dec 14 '18 at 15:43
$begingroup$
"2 is a generator mod p" - means?@KCd
$endgroup$
– cmi
Dec 14 '18 at 15:45
$begingroup$
Are you sure that "($p$ is prime)" is supposed to be there? Should it, perhaps, be "($p$ is a natural number)"?
$endgroup$
– Arthur
Dec 14 '18 at 15:30
$begingroup$
Are you sure that "($p$ is prime)" is supposed to be there? Should it, perhaps, be "($p$ is a natural number)"?
$endgroup$
– Arthur
Dec 14 '18 at 15:30
$begingroup$
en.wikipedia.org/wiki/Cyclotomic_polynomial
$endgroup$
– vadim123
Dec 14 '18 at 15:33
$begingroup$
en.wikipedia.org/wiki/Cyclotomic_polynomial
$endgroup$
– vadim123
Dec 14 '18 at 15:33
$begingroup$
Is the statement true for any natural number?@Arthur
$endgroup$
– cmi
Dec 14 '18 at 15:35
$begingroup$
Is the statement true for any natural number?@Arthur
$endgroup$
– cmi
Dec 14 '18 at 15:35
2
2
$begingroup$
This is false when $p=7$. For prime $p> 2$, it is true if and only if $2$ is a generator mod $p$.
$endgroup$
– KCd
Dec 14 '18 at 15:43
$begingroup$
This is false when $p=7$. For prime $p> 2$, it is true if and only if $2$ is a generator mod $p$.
$endgroup$
– KCd
Dec 14 '18 at 15:43
$begingroup$
"2 is a generator mod p" - means?@KCd
$endgroup$
– cmi
Dec 14 '18 at 15:45
$begingroup$
"2 is a generator mod p" - means?@KCd
$endgroup$
– cmi
Dec 14 '18 at 15:45
|
show 5 more comments
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$begingroup$
Are you sure that "($p$ is prime)" is supposed to be there? Should it, perhaps, be "($p$ is a natural number)"?
$endgroup$
– Arthur
Dec 14 '18 at 15:30
$begingroup$
en.wikipedia.org/wiki/Cyclotomic_polynomial
$endgroup$
– vadim123
Dec 14 '18 at 15:33
$begingroup$
Is the statement true for any natural number?@Arthur
$endgroup$
– cmi
Dec 14 '18 at 15:35
2
$begingroup$
This is false when $p=7$. For prime $p> 2$, it is true if and only if $2$ is a generator mod $p$.
$endgroup$
– KCd
Dec 14 '18 at 15:43
$begingroup$
"2 is a generator mod p" - means?@KCd
$endgroup$
– cmi
Dec 14 '18 at 15:45