Cyclotomic cosets and minimal polynomials
Let $mathbb{F}_{p^m}$ be a field and let $alpha in mathbb{F}_{p^m}$. Let $M^{(i)}$ be the minimal polynomial of $alpha^i$. Then I know that $M^{(i)}(x) = prod_{j in C_s} (x - alpha^j)$, where $i$ is in the cyclotomic coset $C_s$.
Further, from this, we get that $x^{p^m - 1} - 1 = prod_{s} M^{s}(x)$, where $s$ runs through the coset representatives mod $p^m - 1$
How do we get this last equation? Is $alpha$ assumed to be primitive here?
field-theory finite-fields minimal-polynomials
asked Nov 26 '18 at 20:03
the man
697715
add a comment |
Let $mathbb{F}_{p^m}$ be a field and let $alpha in mathbb{F}_{p^m}$. Let $M^{(i)}$ be the minimal polynomial of $alpha^i$. Then I know that $M^{(i)}(x) = prod_{j in C_s} (x - alpha^j)$, where $i$ is in the cyclotomic coset $C_s$.
Further, from this, we get that $x^{p^m - 1} - 1 = prod_{s} M^{s}(x)$, where $s$ runs through the coset representatives mod $p^m - 1$
How do we get this last equation? Is $alpha$ assumed to be primitive here?
field-theory finite-fields minimal-polynomials
asked Nov 26 '18 at 20:03
the man
697715
Yes, it sure looks like $alpha$ is assumed to be primitive so that $alpha^j$ ranges over the zeros of $x^{p^m-1}-1$ as $j$ ranges over $0,1,ldots,p^m-2$.
– Jyrki Lahtonen
Nov 27 '18 at 3:08
add a comment |
Let $mathbb{F}_{p^m}$ be a field and let $alpha in mathbb{F}_{p^m}$. Let $M^{(i)}$ be the minimal polynomial of $alpha^i$. Then I know that $M^{(i)}(x) = prod_{j in C_s} (x - alpha^j)$, where $i$ is in the cyclotomic coset $C_s$.
Further, from this, we get that $x^{p^m - 1} - 1 = prod_{s} M^{s}(x)$, where $s$ runs through the coset representatives mod $p^m - 1$
How do we get this last equation? Is $alpha$ assumed to be primitive here?
field-theory finite-fields minimal-polynomials
asked Nov 26 '18 at 20:03
the man
697715
Let $mathbb{F}_{p^m}$ be a field and let $alpha in mathbb{F}_{p^m}$. Let $M^{(i)}$ be the minimal polynomial of $alpha^i$. Then I know that $M^{(i)}(x) = prod_{j in C_s} (x - alpha^j)$, where $i$ is in the cyclotomic coset $C_s$.
Further, from this, we get that $x^{p^m - 1} - 1 = prod_{s} M^{s}(x)$, where $s$ runs through the coset representatives mod $p^m - 1$
How do we get this last equation? Is $alpha$ assumed to be primitive here?
field-theory finite-fields minimal-polynomials
field-theory finite-fields minimal-polynomials
asked Nov 26 '18 at 20:03
the man
697715
asked Nov 26 '18 at 20:03
the man
697715
asked Nov 26 '18 at 20:03
the man
697715
asked Nov 26 '18 at 20:03
the man
697715
asked Nov 26 '18 at 20:03
the man
697715
697715
Yes, it sure looks like $alpha$ is assumed to be primitive so that $alpha^j$ ranges over the zeros of $x^{p^m-1}-1$ as $j$ ranges over $0,1,ldots,p^m-2$.
– Jyrki Lahtonen
Nov 27 '18 at 3:08
add a comment |
Yes, it sure looks like $alpha$ is assumed to be primitive so that $alpha^j$ ranges over the zeros of $x^{p^m-1}-1$ as $j$ ranges over $0,1,ldots,p^m-2$.
– Jyrki Lahtonen
Nov 27 '18 at 3:08
Yes, it sure looks like $alpha$ is assumed to be primitive so that $alpha^j$ ranges over the zeros of $x^{p^m-1}-1$ as $j$ ranges over $0,1,ldots,p^m-2$.
– Jyrki Lahtonen
Nov 27 '18 at 3:08
Yes, it sure looks like $alpha$ is assumed to be primitive so that $alpha^j$ ranges over the zeros of $x^{p^m-1}-1$ as $j$ ranges over $0,1,ldots,p^m-2$.
– Jyrki Lahtonen
Nov 27 '18 at 3:08
add a comment |
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Yes, it sure looks like $alpha$ is assumed to be primitive so that $alpha^j$ ranges over the zeros of $x^{p^m-1}-1$ as $j$ ranges over $0,1,ldots,p^m-2$.
– Jyrki Lahtonen
Nov 27 '18 at 3:08