Galois extension - minimum polynomial
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Let $K$ be a Galois extension of $F$ and let $ain K$. Let $n=[K:F]$, $r=[F(a):F]$, $G=text{Gal}(K/F)$ and $H=text{Gal}(K/F(a))$.
We symbolize with $tau_1, ldots , tau_r$ the representatives of the left cosets of $H$ in $G$.
Show that $displaystyle{min (F,a)=prod_{i=1}^rleft (x-tau_i(a)right )}$.
Show that $displaystyle{prod_{sigma in G}left (x- tau_i(a)right )=min (F,a)^{n/r}}$, where $tau_i$ is the unique representative such that $sigma in left[ tau_i right]$.
$$$$
Could you give me a hint how we could show these two points? I don't really have an idea.
abstract-algebra field-theory galois-theory
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Let $K$ be a Galois extension of $F$ and let $ain K$. Let $n=[K:F]$, $r=[F(a):F]$, $G=text{Gal}(K/F)$ and $H=text{Gal}(K/F(a))$.
We symbolize with $tau_1, ldots , tau_r$ the representatives of the left cosets of $H$ in $G$.
Show that $displaystyle{min (F,a)=prod_{i=1}^rleft (x-tau_i(a)right )}$.
Show that $displaystyle{prod_{sigma in G}left (x- tau_i(a)right )=min (F,a)^{n/r}}$, where $tau_i$ is the unique representative such that $sigma in left[ tau_i right]$.
$$$$
Could you give me a hint how we could show these two points? I don't really have an idea.
abstract-algebra field-theory galois-theory
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $K$ be a Galois extension of $F$ and let $ain K$. Let $n=[K:F]$, $r=[F(a):F]$, $G=text{Gal}(K/F)$ and $H=text{Gal}(K/F(a))$.
We symbolize with $tau_1, ldots , tau_r$ the representatives of the left cosets of $H$ in $G$.
Show that $displaystyle{min (F,a)=prod_{i=1}^rleft (x-tau_i(a)right )}$.
Show that $displaystyle{prod_{sigma in G}left (x- tau_i(a)right )=min (F,a)^{n/r}}$, where $tau_i$ is the unique representative such that $sigma in left[ tau_i right]$.
$$$$
Could you give me a hint how we could show these two points? I don't really have an idea.
abstract-algebra field-theory galois-theory
Let $K$ be a Galois extension of $F$ and let $ain K$. Let $n=[K:F]$, $r=[F(a):F]$, $G=text{Gal}(K/F)$ and $H=text{Gal}(K/F(a))$.
We symbolize with $tau_1, ldots , tau_r$ the representatives of the left cosets of $H$ in $G$.
Show that $displaystyle{min (F,a)=prod_{i=1}^rleft (x-tau_i(a)right )}$.
Show that $displaystyle{prod_{sigma in G}left (x- tau_i(a)right )=min (F,a)^{n/r}}$, where $tau_i$ is the unique representative such that $sigma in left[ tau_i right]$.
$$$$
Could you give me a hint how we could show these two points? I don't really have an idea.
abstract-algebra field-theory galois-theory
abstract-algebra field-theory galois-theory
edited Nov 29 at 14:46
Bilo
1089
1089
asked Nov 19 at 23:09
Mary Star
2,92082266
2,92082266
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1 Answer
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For the first point, try to show that if $i neq j$, then $tau_i(a) neq tau_j(a)$.
For the second one, try to show that if $sigma, tau$ belong to the same class modulo $H$, then $sigma(a) = tau(a)$
1
Do I have to write a full proof?
– Bilo
Nov 23 at 20:39
For the first part: Do they have to be different because the number of the left costes is equal to the extension degree $[F(a):F]$ ?
– Mary Star
Nov 24 at 8:38
1
Yes, since $tau_i$ maps a root of a polynomial into a root of the same polynomial. With this claim you can show that $tau_i(a)$ define all' the roots of $min(F,a)$, since it's degree is exactly $r$, so It must have the form you have described above
– Bilo
Nov 24 at 14:57
Ah ok!! Could you explain to me also further the second point?
– Mary Star
Nov 24 at 16:41
1
Yes, for sure. First show that if $sigma,tau$ belong to the same class modulo $H$, then $sigma(a)=tau(a)$; Since $vert G:H vert = r$, each coset contains exacly $frac{n}{r}$ elements. Denote with $C_1,...,C_r$ the distinct cosets of $G$ modulo $H$., then begin{align} prod_{sigma in G}(x-tau_i(a)) &= prod_{sigma in C_1}(x-tau_1(a)) cdots prod_{sigma in C_r}(x-tau_r(a)) \ &= (x-tau_1(a))^{n/r} cdots (x-tau_r(a))^{n/r} \ &= (prod_{1}^r(x-tau_i(a)) )^{n/r} \ &= min(F,a)^{n/r} end{align} Hope it's clear.
– Bilo
Nov 24 at 17:19
|
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
For the first point, try to show that if $i neq j$, then $tau_i(a) neq tau_j(a)$.
For the second one, try to show that if $sigma, tau$ belong to the same class modulo $H$, then $sigma(a) = tau(a)$
1
Do I have to write a full proof?
– Bilo
Nov 23 at 20:39
For the first part: Do they have to be different because the number of the left costes is equal to the extension degree $[F(a):F]$ ?
– Mary Star
Nov 24 at 8:38
1
Yes, since $tau_i$ maps a root of a polynomial into a root of the same polynomial. With this claim you can show that $tau_i(a)$ define all' the roots of $min(F,a)$, since it's degree is exactly $r$, so It must have the form you have described above
– Bilo
Nov 24 at 14:57
Ah ok!! Could you explain to me also further the second point?
– Mary Star
Nov 24 at 16:41
1
Yes, for sure. First show that if $sigma,tau$ belong to the same class modulo $H$, then $sigma(a)=tau(a)$; Since $vert G:H vert = r$, each coset contains exacly $frac{n}{r}$ elements. Denote with $C_1,...,C_r$ the distinct cosets of $G$ modulo $H$., then begin{align} prod_{sigma in G}(x-tau_i(a)) &= prod_{sigma in C_1}(x-tau_1(a)) cdots prod_{sigma in C_r}(x-tau_r(a)) \ &= (x-tau_1(a))^{n/r} cdots (x-tau_r(a))^{n/r} \ &= (prod_{1}^r(x-tau_i(a)) )^{n/r} \ &= min(F,a)^{n/r} end{align} Hope it's clear.
– Bilo
Nov 24 at 17:19
|
show 1 more comment
up vote
2
down vote
accepted
For the first point, try to show that if $i neq j$, then $tau_i(a) neq tau_j(a)$.
For the second one, try to show that if $sigma, tau$ belong to the same class modulo $H$, then $sigma(a) = tau(a)$
1
Do I have to write a full proof?
– Bilo
Nov 23 at 20:39
For the first part: Do they have to be different because the number of the left costes is equal to the extension degree $[F(a):F]$ ?
– Mary Star
Nov 24 at 8:38
1
Yes, since $tau_i$ maps a root of a polynomial into a root of the same polynomial. With this claim you can show that $tau_i(a)$ define all' the roots of $min(F,a)$, since it's degree is exactly $r$, so It must have the form you have described above
– Bilo
Nov 24 at 14:57
Ah ok!! Could you explain to me also further the second point?
– Mary Star
Nov 24 at 16:41
1
Yes, for sure. First show that if $sigma,tau$ belong to the same class modulo $H$, then $sigma(a)=tau(a)$; Since $vert G:H vert = r$, each coset contains exacly $frac{n}{r}$ elements. Denote with $C_1,...,C_r$ the distinct cosets of $G$ modulo $H$., then begin{align} prod_{sigma in G}(x-tau_i(a)) &= prod_{sigma in C_1}(x-tau_1(a)) cdots prod_{sigma in C_r}(x-tau_r(a)) \ &= (x-tau_1(a))^{n/r} cdots (x-tau_r(a))^{n/r} \ &= (prod_{1}^r(x-tau_i(a)) )^{n/r} \ &= min(F,a)^{n/r} end{align} Hope it's clear.
– Bilo
Nov 24 at 17:19
|
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
For the first point, try to show that if $i neq j$, then $tau_i(a) neq tau_j(a)$.
For the second one, try to show that if $sigma, tau$ belong to the same class modulo $H$, then $sigma(a) = tau(a)$
For the first point, try to show that if $i neq j$, then $tau_i(a) neq tau_j(a)$.
For the second one, try to show that if $sigma, tau$ belong to the same class modulo $H$, then $sigma(a) = tau(a)$
answered Nov 23 at 12:52
Bilo
1089
1089
1
Do I have to write a full proof?
– Bilo
Nov 23 at 20:39
For the first part: Do they have to be different because the number of the left costes is equal to the extension degree $[F(a):F]$ ?
– Mary Star
Nov 24 at 8:38
1
Yes, since $tau_i$ maps a root of a polynomial into a root of the same polynomial. With this claim you can show that $tau_i(a)$ define all' the roots of $min(F,a)$, since it's degree is exactly $r$, so It must have the form you have described above
– Bilo
Nov 24 at 14:57
Ah ok!! Could you explain to me also further the second point?
– Mary Star
Nov 24 at 16:41
1
Yes, for sure. First show that if $sigma,tau$ belong to the same class modulo $H$, then $sigma(a)=tau(a)$; Since $vert G:H vert = r$, each coset contains exacly $frac{n}{r}$ elements. Denote with $C_1,...,C_r$ the distinct cosets of $G$ modulo $H$., then begin{align} prod_{sigma in G}(x-tau_i(a)) &= prod_{sigma in C_1}(x-tau_1(a)) cdots prod_{sigma in C_r}(x-tau_r(a)) \ &= (x-tau_1(a))^{n/r} cdots (x-tau_r(a))^{n/r} \ &= (prod_{1}^r(x-tau_i(a)) )^{n/r} \ &= min(F,a)^{n/r} end{align} Hope it's clear.
– Bilo
Nov 24 at 17:19
|
show 1 more comment
1
Do I have to write a full proof?
– Bilo
Nov 23 at 20:39
For the first part: Do they have to be different because the number of the left costes is equal to the extension degree $[F(a):F]$ ?
– Mary Star
Nov 24 at 8:38
1
Yes, since $tau_i$ maps a root of a polynomial into a root of the same polynomial. With this claim you can show that $tau_i(a)$ define all' the roots of $min(F,a)$, since it's degree is exactly $r$, so It must have the form you have described above
– Bilo
Nov 24 at 14:57
Ah ok!! Could you explain to me also further the second point?
– Mary Star
Nov 24 at 16:41
1
Yes, for sure. First show that if $sigma,tau$ belong to the same class modulo $H$, then $sigma(a)=tau(a)$; Since $vert G:H vert = r$, each coset contains exacly $frac{n}{r}$ elements. Denote with $C_1,...,C_r$ the distinct cosets of $G$ modulo $H$., then begin{align} prod_{sigma in G}(x-tau_i(a)) &= prod_{sigma in C_1}(x-tau_1(a)) cdots prod_{sigma in C_r}(x-tau_r(a)) \ &= (x-tau_1(a))^{n/r} cdots (x-tau_r(a))^{n/r} \ &= (prod_{1}^r(x-tau_i(a)) )^{n/r} \ &= min(F,a)^{n/r} end{align} Hope it's clear.
– Bilo
Nov 24 at 17:19
1
1
Do I have to write a full proof?
– Bilo
Nov 23 at 20:39
Do I have to write a full proof?
– Bilo
Nov 23 at 20:39
For the first part: Do they have to be different because the number of the left costes is equal to the extension degree $[F(a):F]$ ?
– Mary Star
Nov 24 at 8:38
For the first part: Do they have to be different because the number of the left costes is equal to the extension degree $[F(a):F]$ ?
– Mary Star
Nov 24 at 8:38
1
1
Yes, since $tau_i$ maps a root of a polynomial into a root of the same polynomial. With this claim you can show that $tau_i(a)$ define all' the roots of $min(F,a)$, since it's degree is exactly $r$, so It must have the form you have described above
– Bilo
Nov 24 at 14:57
Yes, since $tau_i$ maps a root of a polynomial into a root of the same polynomial. With this claim you can show that $tau_i(a)$ define all' the roots of $min(F,a)$, since it's degree is exactly $r$, so It must have the form you have described above
– Bilo
Nov 24 at 14:57
Ah ok!! Could you explain to me also further the second point?
– Mary Star
Nov 24 at 16:41
Ah ok!! Could you explain to me also further the second point?
– Mary Star
Nov 24 at 16:41
1
1
Yes, for sure. First show that if $sigma,tau$ belong to the same class modulo $H$, then $sigma(a)=tau(a)$; Since $vert G:H vert = r$, each coset contains exacly $frac{n}{r}$ elements. Denote with $C_1,...,C_r$ the distinct cosets of $G$ modulo $H$., then begin{align} prod_{sigma in G}(x-tau_i(a)) &= prod_{sigma in C_1}(x-tau_1(a)) cdots prod_{sigma in C_r}(x-tau_r(a)) \ &= (x-tau_1(a))^{n/r} cdots (x-tau_r(a))^{n/r} \ &= (prod_{1}^r(x-tau_i(a)) )^{n/r} \ &= min(F,a)^{n/r} end{align} Hope it's clear.
– Bilo
Nov 24 at 17:19
Yes, for sure. First show that if $sigma,tau$ belong to the same class modulo $H$, then $sigma(a)=tau(a)$; Since $vert G:H vert = r$, each coset contains exacly $frac{n}{r}$ elements. Denote with $C_1,...,C_r$ the distinct cosets of $G$ modulo $H$., then begin{align} prod_{sigma in G}(x-tau_i(a)) &= prod_{sigma in C_1}(x-tau_1(a)) cdots prod_{sigma in C_r}(x-tau_r(a)) \ &= (x-tau_1(a))^{n/r} cdots (x-tau_r(a))^{n/r} \ &= (prod_{1}^r(x-tau_i(a)) )^{n/r} \ &= min(F,a)^{n/r} end{align} Hope it's clear.
– Bilo
Nov 24 at 17:19
|
show 1 more comment
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