Degree of splitting field of $x^5-1$ over $mathbb{Q}$ and the order of its Galois group?
$begingroup$
If I understand the result here correctly, the order of the Galois group is equal to the degree of the extension. That said, I am fairly certain that the splitting field $E$ of $x^5-1$ over $mathbb{Q}$ is
$$E={a_0+a_1omega+a_2omega^2+a_3omega^3+a_4omega^4}$$ where $omega$ is a complex root of unity. $E$, as a $mathbb{Q}$-vector space, has dimension 5.
However, if I understand this correctly, the Galois group for this polynomial is $mathbb{Z}/5mathbb{Z}^*$, which has order 4.
What is it that I'm missing? Am I misinterpreting or miscomputing the dimension of $E$ as a $mathbb{Q}$-vector space, or am I misinterpreting the post and the Galois group is in fact not the aforementioned group of units?
group-theory field-theory galois-theory
asked Dec 10 '18 at 21:24
DanielDaniel
599
$endgroup$
add a comment |
$begingroup$
If I understand the result here correctly, the order of the Galois group is equal to the degree of the extension. That said, I am fairly certain that the splitting field $E$ of $x^5-1$ over $mathbb{Q}$ is
$$E={a_0+a_1omega+a_2omega^2+a_3omega^3+a_4omega^4}$$ where $omega$ is a complex root of unity. $E$, as a $mathbb{Q}$-vector space, has dimension 5.
However, if I understand this correctly, the Galois group for this polynomial is $mathbb{Z}/5mathbb{Z}^*$, which has order 4.
What is it that I'm missing? Am I misinterpreting or miscomputing the dimension of $E$ as a $mathbb{Q}$-vector space, or am I misinterpreting the post and the Galois group is in fact not the aforementioned group of units?
group-theory field-theory galois-theory
asked Dec 10 '18 at 21:24
DanielDaniel
599
$endgroup$
add a comment |
$begingroup$
If I understand the result here correctly, the order of the Galois group is equal to the degree of the extension. That said, I am fairly certain that the splitting field $E$ of $x^5-1$ over $mathbb{Q}$ is
$$E={a_0+a_1omega+a_2omega^2+a_3omega^3+a_4omega^4}$$ where $omega$ is a complex root of unity. $E$, as a $mathbb{Q}$-vector space, has dimension 5.
However, if I understand this correctly, the Galois group for this polynomial is $mathbb{Z}/5mathbb{Z}^*$, which has order 4.
What is it that I'm missing? Am I misinterpreting or miscomputing the dimension of $E$ as a $mathbb{Q}$-vector space, or am I misinterpreting the post and the Galois group is in fact not the aforementioned group of units?
group-theory field-theory galois-theory
asked Dec 10 '18 at 21:24
DanielDaniel
599
$endgroup$
If I understand the result here correctly, the order of the Galois group is equal to the degree of the extension. That said, I am fairly certain that the splitting field $E$ of $x^5-1$ over $mathbb{Q}$ is
$$E={a_0+a_1omega+a_2omega^2+a_3omega^3+a_4omega^4}$$ where $omega$ is a complex root of unity. $E$, as a $mathbb{Q}$-vector space, has dimension 5.
However, if I understand this correctly, the Galois group for this polynomial is $mathbb{Z}/5mathbb{Z}^*$, which has order 4.
What is it that I'm missing? Am I misinterpreting or miscomputing the dimension of $E$ as a $mathbb{Q}$-vector space, or am I misinterpreting the post and the Galois group is in fact not the aforementioned group of units?
group-theory field-theory galois-theory
group-theory field-theory galois-theory
asked Dec 10 '18 at 21:24
DanielDaniel
599
asked Dec 10 '18 at 21:24
DanielDaniel
599
edited Dec 11 '18 at 1:07
Daniel
asked Dec 10 '18 at 21:24
DanielDaniel
599
asked Dec 10 '18 at 21:24
DanielDaniel
599
asked Dec 10 '18 at 21:24
DanielDaniel
599
599
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The polynomial is reducible: $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$.
You do not need to add the element $1$ to the field (it is already there), only a root of the quaternary polynomial (which generates the rest of the roots). That is why the degree is $4$.
answered Dec 10 '18 at 21:30
A. PongráczA. Pongrácz
5,9631929
$endgroup$
add a comment |
$begingroup$
While your $E$ is the right space, the set ${1,omega,omega^2,omega^3,omega^4}$ is not a basis for it, because that set isn't linearly independent over $mathbb{Q}$. We have the dependence relation $1+omega+omega^2+omega^3+omega^4$ (and nothing else), so any four of those suffice as a basis. As a vector space over $mathbb{Q}$, $E$ has dimension $4$ - so it's not a surprise when the Galois group has order $4$.
answered Dec 10 '18 at 21:53
jmerryjmerry
6,602718
$endgroup$
$begingroup$
Ah! I should've realized that the relation allowed us to generate the rationals using only $mathbb{Q}$-linear sums of the powers of $omega$! Thank you.
$endgroup$
– Daniel
Dec 11 '18 at 1:53
$begingroup$
One additional question if you don't mind: is E going to be equal to $mathbb{Q}(omega)$ or will it be something like $mathbb{Q}(omega)(omega^2)$?
$endgroup$
– Daniel
Dec 11 '18 at 2:06
$begingroup$
I suppose I am wondering how I'm supposed to symbolically represent $E$. My suspicion is that it's $E(omega)(omega^2)$ because elements will end up looking like $${f+f'omega^2 ; | ; f,f'in mathbb{Q}(omega)}={a+bomega + c omega^2 +domega^3 ; | ; a,b,c,d in mathbb{Q}}$$
$endgroup$
– Daniel
Dec 11 '18 at 2:12
$begingroup$
Sorry, $mathbb{Q}(omega)(omega^2)$
$endgroup$
– Daniel
Dec 11 '18 at 2:28
$begingroup$
How you represent it depends on whether you're looking for a vector space basis or an algebraic generator. As a vector space, pick four of the roots to be a basis. If you're allowing multiplication, it's just $mathbb{Q}(omega)$.
$endgroup$
– jmerry
Dec 11 '18 at 2:29
|
show 4 more comments
$begingroup$
Order of Galois extension need not equal to degree of extension it is less than equal to degree of extension example is |Aut (Q(2^{1/3})/Q)| =1 but [ Q(2^{1/3}) : Q ] is 3 and equality hold when extension is splitting field of separable polynomial.
answered Dec 14 '18 at 12:22
user499117user499117
409
$endgroup$
add a comment |
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3 Answers
3
active
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3 Answers
3
active
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$begingroup$
The polynomial is reducible: $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$.
You do not need to add the element $1$ to the field (it is already there), only a root of the quaternary polynomial (which generates the rest of the roots). That is why the degree is $4$.
answered Dec 10 '18 at 21:30
A. PongráczA. Pongrácz
5,9631929
$endgroup$
add a comment |
$begingroup$
The polynomial is reducible: $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$.
You do not need to add the element $1$ to the field (it is already there), only a root of the quaternary polynomial (which generates the rest of the roots). That is why the degree is $4$.
answered Dec 10 '18 at 21:30
A. PongráczA. Pongrácz
5,9631929
$endgroup$
add a comment |
$begingroup$
The polynomial is reducible: $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$.
You do not need to add the element $1$ to the field (it is already there), only a root of the quaternary polynomial (which generates the rest of the roots). That is why the degree is $4$.
answered Dec 10 '18 at 21:30
A. PongráczA. Pongrácz
5,9631929
$endgroup$
The polynomial is reducible: $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$.
You do not need to add the element $1$ to the field (it is already there), only a root of the quaternary polynomial (which generates the rest of the roots). That is why the degree is $4$.
answered Dec 10 '18 at 21:30
A. PongráczA. Pongrácz
5,9631929
answered Dec 10 '18 at 21:30
A. PongráczA. Pongrácz
5,9631929
answered Dec 10 '18 at 21:30
A. PongráczA. Pongrácz
5,9631929
answered Dec 10 '18 at 21:30
A. PongráczA. Pongrácz
5,9631929
5,9631929
add a comment |
add a comment |
$begingroup$
While your $E$ is the right space, the set ${1,omega,omega^2,omega^3,omega^4}$ is not a basis for it, because that set isn't linearly independent over $mathbb{Q}$. We have the dependence relation $1+omega+omega^2+omega^3+omega^4$ (and nothing else), so any four of those suffice as a basis. As a vector space over $mathbb{Q}$, $E$ has dimension $4$ - so it's not a surprise when the Galois group has order $4$.
answered Dec 10 '18 at 21:53
jmerryjmerry
6,602718
$endgroup$
$begingroup$
Ah! I should've realized that the relation allowed us to generate the rationals using only $mathbb{Q}$-linear sums of the powers of $omega$! Thank you.
$endgroup$
– Daniel
Dec 11 '18 at 1:53
$begingroup$
One additional question if you don't mind: is E going to be equal to $mathbb{Q}(omega)$ or will it be something like $mathbb{Q}(omega)(omega^2)$?
$endgroup$
– Daniel
Dec 11 '18 at 2:06
$begingroup$
I suppose I am wondering how I'm supposed to symbolically represent $E$. My suspicion is that it's $E(omega)(omega^2)$ because elements will end up looking like $${f+f'omega^2 ; | ; f,f'in mathbb{Q}(omega)}={a+bomega + c omega^2 +domega^3 ; | ; a,b,c,d in mathbb{Q}}$$
$endgroup$
– Daniel
Dec 11 '18 at 2:12
$begingroup$
Sorry, $mathbb{Q}(omega)(omega^2)$
$endgroup$
– Daniel
Dec 11 '18 at 2:28
$begingroup$
How you represent it depends on whether you're looking for a vector space basis or an algebraic generator. As a vector space, pick four of the roots to be a basis. If you're allowing multiplication, it's just $mathbb{Q}(omega)$.
$endgroup$
– jmerry
Dec 11 '18 at 2:29
|
show 4 more comments
$begingroup$
While your $E$ is the right space, the set ${1,omega,omega^2,omega^3,omega^4}$ is not a basis for it, because that set isn't linearly independent over $mathbb{Q}$. We have the dependence relation $1+omega+omega^2+omega^3+omega^4$ (and nothing else), so any four of those suffice as a basis. As a vector space over $mathbb{Q}$, $E$ has dimension $4$ - so it's not a surprise when the Galois group has order $4$.
answered Dec 10 '18 at 21:53
jmerryjmerry
6,602718
$endgroup$
$begingroup$
Ah! I should've realized that the relation allowed us to generate the rationals using only $mathbb{Q}$-linear sums of the powers of $omega$! Thank you.
$endgroup$
– Daniel
Dec 11 '18 at 1:53
$begingroup$
One additional question if you don't mind: is E going to be equal to $mathbb{Q}(omega)$ or will it be something like $mathbb{Q}(omega)(omega^2)$?
$endgroup$
– Daniel
Dec 11 '18 at 2:06
$begingroup$
I suppose I am wondering how I'm supposed to symbolically represent $E$. My suspicion is that it's $E(omega)(omega^2)$ because elements will end up looking like $${f+f'omega^2 ; | ; f,f'in mathbb{Q}(omega)}={a+bomega + c omega^2 +domega^3 ; | ; a,b,c,d in mathbb{Q}}$$
$endgroup$
– Daniel
Dec 11 '18 at 2:12
$begingroup$
Sorry, $mathbb{Q}(omega)(omega^2)$
$endgroup$
– Daniel
Dec 11 '18 at 2:28
$begingroup$
How you represent it depends on whether you're looking for a vector space basis or an algebraic generator. As a vector space, pick four of the roots to be a basis. If you're allowing multiplication, it's just $mathbb{Q}(omega)$.
$endgroup$
– jmerry
Dec 11 '18 at 2:29
|
show 4 more comments
$begingroup$
While your $E$ is the right space, the set ${1,omega,omega^2,omega^3,omega^4}$ is not a basis for it, because that set isn't linearly independent over $mathbb{Q}$. We have the dependence relation $1+omega+omega^2+omega^3+omega^4$ (and nothing else), so any four of those suffice as a basis. As a vector space over $mathbb{Q}$, $E$ has dimension $4$ - so it's not a surprise when the Galois group has order $4$.
answered Dec 10 '18 at 21:53
jmerryjmerry
6,602718
$endgroup$
While your $E$ is the right space, the set ${1,omega,omega^2,omega^3,omega^4}$ is not a basis for it, because that set isn't linearly independent over $mathbb{Q}$. We have the dependence relation $1+omega+omega^2+omega^3+omega^4$ (and nothing else), so any four of those suffice as a basis. As a vector space over $mathbb{Q}$, $E$ has dimension $4$ - so it's not a surprise when the Galois group has order $4$.
answered Dec 10 '18 at 21:53
jmerryjmerry
6,602718
answered Dec 10 '18 at 21:53
jmerryjmerry
6,602718
answered Dec 10 '18 at 21:53
jmerryjmerry
6,602718
answered Dec 10 '18 at 21:53
jmerryjmerry
6,602718
6,602718
$begingroup$
Ah! I should've realized that the relation allowed us to generate the rationals using only $mathbb{Q}$-linear sums of the powers of $omega$! Thank you.
$endgroup$
– Daniel
Dec 11 '18 at 1:53
$begingroup$
One additional question if you don't mind: is E going to be equal to $mathbb{Q}(omega)$ or will it be something like $mathbb{Q}(omega)(omega^2)$?
$endgroup$
– Daniel
Dec 11 '18 at 2:06
$begingroup$
I suppose I am wondering how I'm supposed to symbolically represent $E$. My suspicion is that it's $E(omega)(omega^2)$ because elements will end up looking like $${f+f'omega^2 ; | ; f,f'in mathbb{Q}(omega)}={a+bomega + c omega^2 +domega^3 ; | ; a,b,c,d in mathbb{Q}}$$
$endgroup$
– Daniel
Dec 11 '18 at 2:12
$begingroup$
Sorry, $mathbb{Q}(omega)(omega^2)$
$endgroup$
– Daniel
Dec 11 '18 at 2:28
$begingroup$
How you represent it depends on whether you're looking for a vector space basis or an algebraic generator. As a vector space, pick four of the roots to be a basis. If you're allowing multiplication, it's just $mathbb{Q}(omega)$.
$endgroup$
– jmerry
Dec 11 '18 at 2:29
|
show 4 more comments
$begingroup$
Ah! I should've realized that the relation allowed us to generate the rationals using only $mathbb{Q}$-linear sums of the powers of $omega$! Thank you.
$endgroup$
– Daniel
Dec 11 '18 at 1:53
$begingroup$
One additional question if you don't mind: is E going to be equal to $mathbb{Q}(omega)$ or will it be something like $mathbb{Q}(omega)(omega^2)$?
$endgroup$
– Daniel
Dec 11 '18 at 2:06
$begingroup$
I suppose I am wondering how I'm supposed to symbolically represent $E$. My suspicion is that it's $E(omega)(omega^2)$ because elements will end up looking like $${f+f'omega^2 ; | ; f,f'in mathbb{Q}(omega)}={a+bomega + c omega^2 +domega^3 ; | ; a,b,c,d in mathbb{Q}}$$
$endgroup$
– Daniel
Dec 11 '18 at 2:12
$begingroup$
Sorry, $mathbb{Q}(omega)(omega^2)$
$endgroup$
– Daniel
Dec 11 '18 at 2:28
$begingroup$
How you represent it depends on whether you're looking for a vector space basis or an algebraic generator. As a vector space, pick four of the roots to be a basis. If you're allowing multiplication, it's just $mathbb{Q}(omega)$.
$endgroup$
– jmerry
Dec 11 '18 at 2:29
$begingroup$
Ah! I should've realized that the relation allowed us to generate the rationals using only $mathbb{Q}$-linear sums of the powers of $omega$! Thank you.
$endgroup$
– Daniel
Dec 11 '18 at 1:53
$begingroup$
Ah! I should've realized that the relation allowed us to generate the rationals using only $mathbb{Q}$-linear sums of the powers of $omega$! Thank you.
$endgroup$
– Daniel
Dec 11 '18 at 1:53
$begingroup$
One additional question if you don't mind: is E going to be equal to $mathbb{Q}(omega)$ or will it be something like $mathbb{Q}(omega)(omega^2)$?
$endgroup$
– Daniel
Dec 11 '18 at 2:06
$begingroup$
One additional question if you don't mind: is E going to be equal to $mathbb{Q}(omega)$ or will it be something like $mathbb{Q}(omega)(omega^2)$?
$endgroup$
– Daniel
Dec 11 '18 at 2:06
$begingroup$
I suppose I am wondering how I'm supposed to symbolically represent $E$. My suspicion is that it's $E(omega)(omega^2)$ because elements will end up looking like $${f+f'omega^2 ; | ; f,f'in mathbb{Q}(omega)}={a+bomega + c omega^2 +domega^3 ; | ; a,b,c,d in mathbb{Q}}$$
$endgroup$
– Daniel
Dec 11 '18 at 2:12
$begingroup$
I suppose I am wondering how I'm supposed to symbolically represent $E$. My suspicion is that it's $E(omega)(omega^2)$ because elements will end up looking like $${f+f'omega^2 ; | ; f,f'in mathbb{Q}(omega)}={a+bomega + c omega^2 +domega^3 ; | ; a,b,c,d in mathbb{Q}}$$
$endgroup$
– Daniel
Dec 11 '18 at 2:12
$begingroup$
Sorry, $mathbb{Q}(omega)(omega^2)$
$endgroup$
– Daniel
Dec 11 '18 at 2:28
$begingroup$
Sorry, $mathbb{Q}(omega)(omega^2)$
$endgroup$
– Daniel
Dec 11 '18 at 2:28
$begingroup$
How you represent it depends on whether you're looking for a vector space basis or an algebraic generator. As a vector space, pick four of the roots to be a basis. If you're allowing multiplication, it's just $mathbb{Q}(omega)$.
$endgroup$
– jmerry
Dec 11 '18 at 2:29
$begingroup$
How you represent it depends on whether you're looking for a vector space basis or an algebraic generator. As a vector space, pick four of the roots to be a basis. If you're allowing multiplication, it's just $mathbb{Q}(omega)$.
$endgroup$
– jmerry
Dec 11 '18 at 2:29
|
show 4 more comments
$begingroup$
Order of Galois extension need not equal to degree of extension it is less than equal to degree of extension example is |Aut (Q(2^{1/3})/Q)| =1 but [ Q(2^{1/3}) : Q ] is 3 and equality hold when extension is splitting field of separable polynomial.
answered Dec 14 '18 at 12:22
user499117user499117
409
$endgroup$
add a comment |
$begingroup$
Order of Galois extension need not equal to degree of extension it is less than equal to degree of extension example is |Aut (Q(2^{1/3})/Q)| =1 but [ Q(2^{1/3}) : Q ] is 3 and equality hold when extension is splitting field of separable polynomial.
answered Dec 14 '18 at 12:22
user499117user499117
409
$endgroup$
add a comment |
$begingroup$
Order of Galois extension need not equal to degree of extension it is less than equal to degree of extension example is |Aut (Q(2^{1/3})/Q)| =1 but [ Q(2^{1/3}) : Q ] is 3 and equality hold when extension is splitting field of separable polynomial.
answered Dec 14 '18 at 12:22
user499117user499117
409
$endgroup$
Order of Galois extension need not equal to degree of extension it is less than equal to degree of extension example is |Aut (Q(2^{1/3})/Q)| =1 but [ Q(2^{1/3}) : Q ] is 3 and equality hold when extension is splitting field of separable polynomial.
answered Dec 14 '18 at 12:22
user499117user499117
409
answered Dec 14 '18 at 12:22
user499117user499117
409
answered Dec 14 '18 at 12:22
user499117user499117
409
answered Dec 14 '18 at 12:22
user499117user499117
409
409
add a comment |
add a comment |
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