There is an intermediate extension of degree $p$
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Let $K$ be a Galois extension of $F$ and $p$ be a prime factor of the degree $[K:F]$.
I want to show that there is an intermediate extension $Fsubseteq Lsubseteq K$ with $[K:L]=p$.
$$$$
Do we us for that the fact that the intermediate fields are in one to one correspondence with the subgroups of the Galois group?
Then since $p$ is a prime factor of the degree $[K:F]$ there must be an intermediate field with that degree according to Lagrange theorem.
Is that correct?
abstract-algebra field-theory galois-theory galois-extensions
asked Nov 19 at 21:59
Mary Star
2,90582266
|
show 1 more comment
up vote
2
down vote
favorite
Let $K$ be a Galois extension of $F$ and $p$ be a prime factor of the degree $[K:F]$.
I want to show that there is an intermediate extension $Fsubseteq Lsubseteq K$ with $[K:L]=p$.
$$$$
Do we us for that the fact that the intermediate fields are in one to one correspondence with the subgroups of the Galois group?
Then since $p$ is a prime factor of the degree $[K:F]$ there must be an intermediate field with that degree according to Lagrange theorem.
Is that correct?
abstract-algebra field-theory galois-theory galois-extensions
asked Nov 19 at 21:59
Mary Star
2,90582266
3
So, it is a matter of finding a subgroup of the Galois group, of order $p$. This is done by Cauchy's theorem.
– i707107
Nov 19 at 22:24
1
Here is Cauchy'theorem I am referring to: en.wikipedia.org/wiki/Cauchy%27s_theorem_(group_theory)
– i707107
Nov 19 at 22:26
Ah ok! So since $pmid [K:F]=|text{Gal}(K/F)|$ the Galois group $text{Gal}(K/F)$ has an element of order $p$, so there is a subgroup of the Galois group of order $p$, and so there is an intermediate field $L$ with $[K:L]=p$. Is everything correct? $$$$ Do we use the same argument also for the following: Let $K$ be a finite and normal extension of $F$ and suppose that the extension $K/F$ has no proper intermediate extension. Show that the degree $[K:F]$ is a prime number. $$$$ Do we use here again the one correspondence with the subgroups? But we don't have here a Galois group. @i707107
– Mary Star
Nov 19 at 22:54
1
@MaryStar With $G = Aut(K/F)$ then $K/F$ is Galois iff $F = K^G$ (the subfield fixed by $G$). If $K/F$ is normal : if $G$ is non-trivial then look at $K^G$ and $K^H,H le G$. If $G$ is trivial then show the minimal polynomial of any $ain K$ is of the form $f(t) = (t-a)^n$ with $f$ irreducible thus $gcd(f(t),f'(t)) = 1$ thus $n = {p^m}, p= char(F)$.
– reuns
Nov 20 at 1:24
Could you explain that further to me? I got stuck right now. @reuns
– Mary Star
Nov 24 at 16:44
|
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $K$ be a Galois extension of $F$ and $p$ be a prime factor of the degree $[K:F]$.
I want to show that there is an intermediate extension $Fsubseteq Lsubseteq K$ with $[K:L]=p$.
$$$$
Do we us for that the fact that the intermediate fields are in one to one correspondence with the subgroups of the Galois group?
Then since $p$ is a prime factor of the degree $[K:F]$ there must be an intermediate field with that degree according to Lagrange theorem.
Is that correct?
abstract-algebra field-theory galois-theory galois-extensions
asked Nov 19 at 21:59
Mary Star
2,90582266
Let $K$ be a Galois extension of $F$ and $p$ be a prime factor of the degree $[K:F]$.
I want to show that there is an intermediate extension $Fsubseteq Lsubseteq K$ with $[K:L]=p$.
$$$$
Do we us for that the fact that the intermediate fields are in one to one correspondence with the subgroups of the Galois group?
Then since $p$ is a prime factor of the degree $[K:F]$ there must be an intermediate field with that degree according to Lagrange theorem.
Is that correct?
abstract-algebra field-theory galois-theory galois-extensions
abstract-algebra field-theory galois-theory galois-extensions
asked Nov 19 at 21:59
Mary Star
2,90582266
asked Nov 19 at 21:59
Mary Star
2,90582266
asked Nov 19 at 21:59
Mary Star
2,90582266
asked Nov 19 at 21:59
Mary Star
2,90582266
asked Nov 19 at 21:59
Mary Star
2,90582266
2,90582266
3
So, it is a matter of finding a subgroup of the Galois group, of order $p$. This is done by Cauchy's theorem.
– i707107
Nov 19 at 22:24
1
Here is Cauchy'theorem I am referring to: en.wikipedia.org/wiki/Cauchy%27s_theorem_(group_theory)
– i707107
Nov 19 at 22:26
Ah ok! So since $pmid [K:F]=|text{Gal}(K/F)|$ the Galois group $text{Gal}(K/F)$ has an element of order $p$, so there is a subgroup of the Galois group of order $p$, and so there is an intermediate field $L$ with $[K:L]=p$. Is everything correct? $$$$ Do we use the same argument also for the following: Let $K$ be a finite and normal extension of $F$ and suppose that the extension $K/F$ has no proper intermediate extension. Show that the degree $[K:F]$ is a prime number. $$$$ Do we use here again the one correspondence with the subgroups? But we don't have here a Galois group. @i707107
– Mary Star
Nov 19 at 22:54
1
@MaryStar With $G = Aut(K/F)$ then $K/F$ is Galois iff $F = K^G$ (the subfield fixed by $G$). If $K/F$ is normal : if $G$ is non-trivial then look at $K^G$ and $K^H,H le G$. If $G$ is trivial then show the minimal polynomial of any $ain K$ is of the form $f(t) = (t-a)^n$ with $f$ irreducible thus $gcd(f(t),f'(t)) = 1$ thus $n = {p^m}, p= char(F)$.
– reuns
Nov 20 at 1:24
Could you explain that further to me? I got stuck right now. @reuns
– Mary Star
Nov 24 at 16:44
|
show 1 more comment
3
So, it is a matter of finding a subgroup of the Galois group, of order $p$. This is done by Cauchy's theorem.
– i707107
Nov 19 at 22:24
1
Here is Cauchy'theorem I am referring to: en.wikipedia.org/wiki/Cauchy%27s_theorem_(group_theory)
– i707107
Nov 19 at 22:26
Ah ok! So since $pmid [K:F]=|text{Gal}(K/F)|$ the Galois group $text{Gal}(K/F)$ has an element of order $p$, so there is a subgroup of the Galois group of order $p$, and so there is an intermediate field $L$ with $[K:L]=p$. Is everything correct? $$$$ Do we use the same argument also for the following: Let $K$ be a finite and normal extension of $F$ and suppose that the extension $K/F$ has no proper intermediate extension. Show that the degree $[K:F]$ is a prime number. $$$$ Do we use here again the one correspondence with the subgroups? But we don't have here a Galois group. @i707107
– Mary Star
Nov 19 at 22:54
1
@MaryStar With $G = Aut(K/F)$ then $K/F$ is Galois iff $F = K^G$ (the subfield fixed by $G$). If $K/F$ is normal : if $G$ is non-trivial then look at $K^G$ and $K^H,H le G$. If $G$ is trivial then show the minimal polynomial of any $ain K$ is of the form $f(t) = (t-a)^n$ with $f$ irreducible thus $gcd(f(t),f'(t)) = 1$ thus $n = {p^m}, p= char(F)$.
– reuns
Nov 20 at 1:24
Could you explain that further to me? I got stuck right now. @reuns
– Mary Star
Nov 24 at 16:44
3
3
So, it is a matter of finding a subgroup of the Galois group, of order $p$. This is done by Cauchy's theorem.
– i707107
Nov 19 at 22:24
So, it is a matter of finding a subgroup of the Galois group, of order $p$. This is done by Cauchy's theorem.
– i707107
Nov 19 at 22:24
1
1
Here is Cauchy'theorem I am referring to: en.wikipedia.org/wiki/Cauchy%27s_theorem_(group_theory)
– i707107
Nov 19 at 22:26
Here is Cauchy'theorem I am referring to: en.wikipedia.org/wiki/Cauchy%27s_theorem_(group_theory)
– i707107
Nov 19 at 22:26
Ah ok! So since $pmid [K:F]=|text{Gal}(K/F)|$ the Galois group $text{Gal}(K/F)$ has an element of order $p$, so there is a subgroup of the Galois group of order $p$, and so there is an intermediate field $L$ with $[K:L]=p$. Is everything correct? $$$$ Do we use the same argument also for the following: Let $K$ be a finite and normal extension of $F$ and suppose that the extension $K/F$ has no proper intermediate extension. Show that the degree $[K:F]$ is a prime number. $$$$ Do we use here again the one correspondence with the subgroups? But we don't have here a Galois group. @i707107
– Mary Star
Nov 19 at 22:54
Ah ok! So since $pmid [K:F]=|text{Gal}(K/F)|$ the Galois group $text{Gal}(K/F)$ has an element of order $p$, so there is a subgroup of the Galois group of order $p$, and so there is an intermediate field $L$ with $[K:L]=p$. Is everything correct? $$$$ Do we use the same argument also for the following: Let $K$ be a finite and normal extension of $F$ and suppose that the extension $K/F$ has no proper intermediate extension. Show that the degree $[K:F]$ is a prime number. $$$$ Do we use here again the one correspondence with the subgroups? But we don't have here a Galois group. @i707107
– Mary Star
Nov 19 at 22:54
1
1
@MaryStar With $G = Aut(K/F)$ then $K/F$ is Galois iff $F = K^G$ (the subfield fixed by $G$). If $K/F$ is normal : if $G$ is non-trivial then look at $K^G$ and $K^H,H le G$. If $G$ is trivial then show the minimal polynomial of any $ain K$ is of the form $f(t) = (t-a)^n$ with $f$ irreducible thus $gcd(f(t),f'(t)) = 1$ thus $n = {p^m}, p= char(F)$.
– reuns
Nov 20 at 1:24
@MaryStar With $G = Aut(K/F)$ then $K/F$ is Galois iff $F = K^G$ (the subfield fixed by $G$). If $K/F$ is normal : if $G$ is non-trivial then look at $K^G$ and $K^H,H le G$. If $G$ is trivial then show the minimal polynomial of any $ain K$ is of the form $f(t) = (t-a)^n$ with $f$ irreducible thus $gcd(f(t),f'(t)) = 1$ thus $n = {p^m}, p= char(F)$.
– reuns
Nov 20 at 1:24
Could you explain that further to me? I got stuck right now. @reuns
– Mary Star
Nov 24 at 16:44
Could you explain that further to me? I got stuck right now. @reuns
– Mary Star
Nov 24 at 16:44
|
show 1 more comment
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3
So, it is a matter of finding a subgroup of the Galois group, of order $p$. This is done by Cauchy's theorem.
– i707107
Nov 19 at 22:24
1
Here is Cauchy'theorem I am referring to: en.wikipedia.org/wiki/Cauchy%27s_theorem_(group_theory)
– i707107
Nov 19 at 22:26
Ah ok! So since $pmid [K:F]=|text{Gal}(K/F)|$ the Galois group $text{Gal}(K/F)$ has an element of order $p$, so there is a subgroup of the Galois group of order $p$, and so there is an intermediate field $L$ with $[K:L]=p$. Is everything correct? $$$$ Do we use the same argument also for the following: Let $K$ be a finite and normal extension of $F$ and suppose that the extension $K/F$ has no proper intermediate extension. Show that the degree $[K:F]$ is a prime number. $$$$ Do we use here again the one correspondence with the subgroups? But we don't have here a Galois group. @i707107
– Mary Star
Nov 19 at 22:54
1
@MaryStar With $G = Aut(K/F)$ then $K/F$ is Galois iff $F = K^G$ (the subfield fixed by $G$). If $K/F$ is normal : if $G$ is non-trivial then look at $K^G$ and $K^H,H le G$. If $G$ is trivial then show the minimal polynomial of any $ain K$ is of the form $f(t) = (t-a)^n$ with $f$ irreducible thus $gcd(f(t),f'(t)) = 1$ thus $n = {p^m}, p= char(F)$.
– reuns
Nov 20 at 1:24
Could you explain that further to me? I got stuck right now. @reuns
– Mary Star
Nov 24 at 16:44