Galois group of $x^3-x^2-4$
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In determining the Galois group of the polynomial $p(x) = x^3-x^2-4,$ I concluded that is must be the Klein-$4$ group as follows. First, $p(x) = (x-2)(x^2+x+2)$ and the roots of the irreducible quadratic $x^2+x+2$ are:
$$x_{1,2} = dfrac{-1+sqrt{-7}}{2}.$$ Therefore, the splitting field of $p(x)$ is
$mathbb{Q}(sqrt{7}, i).$ Since this is a biquadratic extension and none of $i, sqrt{7}$ and $sqrt{7}i$ are squares, the Galois group is then Klein-$4$ group.
However, I found two different answers that disagree with mine. First is from the Dummit and Foote. Specifically, on page 612 it states that:
If the cubic polynomial is reducible and it splits to a linear factor and an irreducible quadratic, it's Galois group is group of order $2.$
The second source is here, where it proceeds to conclude that the polynomial is irreducible and also its Galois group is $S_3,$ on page $5.$
What is the correct answer here?
abstract-algebra field-theory galois-theory
asked Jan 3 at 1:42
dezdichadodezdichado
6,4091929
$endgroup$
add a comment |
$begingroup$
In determining the Galois group of the polynomial $p(x) = x^3-x^2-4,$ I concluded that is must be the Klein-$4$ group as follows. First, $p(x) = (x-2)(x^2+x+2)$ and the roots of the irreducible quadratic $x^2+x+2$ are:
$$x_{1,2} = dfrac{-1+sqrt{-7}}{2}.$$ Therefore, the splitting field of $p(x)$ is
$mathbb{Q}(sqrt{7}, i).$ Since this is a biquadratic extension and none of $i, sqrt{7}$ and $sqrt{7}i$ are squares, the Galois group is then Klein-$4$ group.
However, I found two different answers that disagree with mine. First is from the Dummit and Foote. Specifically, on page 612 it states that:
If the cubic polynomial is reducible and it splits to a linear factor and an irreducible quadratic, it's Galois group is group of order $2.$
The second source is here, where it proceeds to conclude that the polynomial is irreducible and also its Galois group is $S_3,$ on page $5.$
What is the correct answer here?
abstract-algebra field-theory galois-theory
asked Jan 3 at 1:42
dezdichadodezdichado
6,4091929
$endgroup$
4
$begingroup$
Is the splitting field not $mathbb Q(sqrt{7}i)$?
$endgroup$
– Cheerful Parsnip
Jan 3 at 1:59
1
$begingroup$
Well, I agree 2 is a root. Therefore Dummit and Foote are right. The problem with your answer is that you've got the splitting field wrong. See Cheerful Parsnip's comment above.
$endgroup$
– jgon
Jan 3 at 2:00
1
$begingroup$
@CheerfulParsnip, actually you are right. I don't know what I was thinking. $p(x)$ is the minimal polynomial of its roots and so the group must be of order $2$ too.
$endgroup$
– dezdichado
Jan 3 at 2:03
$begingroup$
It's all good. That's how we learn.
$endgroup$
– Cheerful Parsnip
Jan 3 at 2:24
$begingroup$
@CheerfulParsnip maybe you should write an answer.
$endgroup$
– Kenny Lau
Jan 3 at 18:14
add a comment |
$begingroup$
In determining the Galois group of the polynomial $p(x) = x^3-x^2-4,$ I concluded that is must be the Klein-$4$ group as follows. First, $p(x) = (x-2)(x^2+x+2)$ and the roots of the irreducible quadratic $x^2+x+2$ are:
$$x_{1,2} = dfrac{-1+sqrt{-7}}{2}.$$ Therefore, the splitting field of $p(x)$ is
$mathbb{Q}(sqrt{7}, i).$ Since this is a biquadratic extension and none of $i, sqrt{7}$ and $sqrt{7}i$ are squares, the Galois group is then Klein-$4$ group.
However, I found two different answers that disagree with mine. First is from the Dummit and Foote. Specifically, on page 612 it states that:
If the cubic polynomial is reducible and it splits to a linear factor and an irreducible quadratic, it's Galois group is group of order $2.$
The second source is here, where it proceeds to conclude that the polynomial is irreducible and also its Galois group is $S_3,$ on page $5.$
What is the correct answer here?
abstract-algebra field-theory galois-theory
asked Jan 3 at 1:42
dezdichadodezdichado
6,4091929
$endgroup$
In determining the Galois group of the polynomial $p(x) = x^3-x^2-4,$ I concluded that is must be the Klein-$4$ group as follows. First, $p(x) = (x-2)(x^2+x+2)$ and the roots of the irreducible quadratic $x^2+x+2$ are:
$$x_{1,2} = dfrac{-1+sqrt{-7}}{2}.$$ Therefore, the splitting field of $p(x)$ is
$mathbb{Q}(sqrt{7}, i).$ Since this is a biquadratic extension and none of $i, sqrt{7}$ and $sqrt{7}i$ are squares, the Galois group is then Klein-$4$ group.
However, I found two different answers that disagree with mine. First is from the Dummit and Foote. Specifically, on page 612 it states that:
If the cubic polynomial is reducible and it splits to a linear factor and an irreducible quadratic, it's Galois group is group of order $2.$
The second source is here, where it proceeds to conclude that the polynomial is irreducible and also its Galois group is $S_3,$ on page $5.$
What is the correct answer here?
abstract-algebra field-theory galois-theory
abstract-algebra field-theory galois-theory
asked Jan 3 at 1:42
dezdichadodezdichado
6,4091929
asked Jan 3 at 1:42
dezdichadodezdichado
6,4091929
asked Jan 3 at 1:42
dezdichadodezdichado
6,4091929
asked Jan 3 at 1:42
dezdichadodezdichado
6,4091929
asked Jan 3 at 1:42
dezdichadodezdichado
6,4091929
6,4091929
4
$begingroup$
Is the splitting field not $mathbb Q(sqrt{7}i)$?
$endgroup$
– Cheerful Parsnip
Jan 3 at 1:59
1
$begingroup$
Well, I agree 2 is a root. Therefore Dummit and Foote are right. The problem with your answer is that you've got the splitting field wrong. See Cheerful Parsnip's comment above.
$endgroup$
– jgon
Jan 3 at 2:00
1
$begingroup$
@CheerfulParsnip, actually you are right. I don't know what I was thinking. $p(x)$ is the minimal polynomial of its roots and so the group must be of order $2$ too.
$endgroup$
– dezdichado
Jan 3 at 2:03
$begingroup$
It's all good. That's how we learn.
$endgroup$
– Cheerful Parsnip
Jan 3 at 2:24
$begingroup$
@CheerfulParsnip maybe you should write an answer.
$endgroup$
– Kenny Lau
Jan 3 at 18:14
add a comment |
4
$begingroup$
Is the splitting field not $mathbb Q(sqrt{7}i)$?
$endgroup$
– Cheerful Parsnip
Jan 3 at 1:59
1
$begingroup$
Well, I agree 2 is a root. Therefore Dummit and Foote are right. The problem with your answer is that you've got the splitting field wrong. See Cheerful Parsnip's comment above.
$endgroup$
– jgon
Jan 3 at 2:00
1
$begingroup$
@CheerfulParsnip, actually you are right. I don't know what I was thinking. $p(x)$ is the minimal polynomial of its roots and so the group must be of order $2$ too.
$endgroup$
– dezdichado
Jan 3 at 2:03
$begingroup$
It's all good. That's how we learn.
$endgroup$
– Cheerful Parsnip
Jan 3 at 2:24
$begingroup$
@CheerfulParsnip maybe you should write an answer.
$endgroup$
– Kenny Lau
Jan 3 at 18:14
4
4
$begingroup$
Is the splitting field not $mathbb Q(sqrt{7}i)$?
$endgroup$
– Cheerful Parsnip
Jan 3 at 1:59
$begingroup$
Is the splitting field not $mathbb Q(sqrt{7}i)$?
$endgroup$
– Cheerful Parsnip
Jan 3 at 1:59
1
1
$begingroup$
Well, I agree 2 is a root. Therefore Dummit and Foote are right. The problem with your answer is that you've got the splitting field wrong. See Cheerful Parsnip's comment above.
$endgroup$
– jgon
Jan 3 at 2:00
$begingroup$
Well, I agree 2 is a root. Therefore Dummit and Foote are right. The problem with your answer is that you've got the splitting field wrong. See Cheerful Parsnip's comment above.
$endgroup$
– jgon
Jan 3 at 2:00
1
1
$begingroup$
@CheerfulParsnip, actually you are right. I don't know what I was thinking. $p(x)$ is the minimal polynomial of its roots and so the group must be of order $2$ too.
$endgroup$
– dezdichado
Jan 3 at 2:03
$begingroup$
@CheerfulParsnip, actually you are right. I don't know what I was thinking. $p(x)$ is the minimal polynomial of its roots and so the group must be of order $2$ too.
$endgroup$
– dezdichado
Jan 3 at 2:03
$begingroup$
It's all good. That's how we learn.
$endgroup$
– Cheerful Parsnip
Jan 3 at 2:24
$begingroup$
It's all good. That's how we learn.
$endgroup$
– Cheerful Parsnip
Jan 3 at 2:24
$begingroup$
@CheerfulParsnip maybe you should write an answer.
$endgroup$
– Kenny Lau
Jan 3 at 18:14
$begingroup$
@CheerfulParsnip maybe you should write an answer.
$endgroup$
– Kenny Lau
Jan 3 at 18:14
add a comment |
1 Answer
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$begingroup$
In the interests of moving this question off the unanswered queue, I'm converting my comment to an answer:
The splitting field is actually $mathbb Q(sqrt{7}i)$. Also, the second source is clearly wrong, since $2$ is a root.
answered Jan 3 at 18:58
Cheerful ParsnipCheerful Parsnip
21.1k23598
$endgroup$
add a comment |
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$begingroup$
In the interests of moving this question off the unanswered queue, I'm converting my comment to an answer:
The splitting field is actually $mathbb Q(sqrt{7}i)$. Also, the second source is clearly wrong, since $2$ is a root.
answered Jan 3 at 18:58
Cheerful ParsnipCheerful Parsnip
21.1k23598
$endgroup$
add a comment |
$begingroup$
In the interests of moving this question off the unanswered queue, I'm converting my comment to an answer:
The splitting field is actually $mathbb Q(sqrt{7}i)$. Also, the second source is clearly wrong, since $2$ is a root.
answered Jan 3 at 18:58
Cheerful ParsnipCheerful Parsnip
21.1k23598
$endgroup$
add a comment |
$begingroup$
In the interests of moving this question off the unanswered queue, I'm converting my comment to an answer:
The splitting field is actually $mathbb Q(sqrt{7}i)$. Also, the second source is clearly wrong, since $2$ is a root.
answered Jan 3 at 18:58
Cheerful ParsnipCheerful Parsnip
21.1k23598
$endgroup$
In the interests of moving this question off the unanswered queue, I'm converting my comment to an answer:
The splitting field is actually $mathbb Q(sqrt{7}i)$. Also, the second source is clearly wrong, since $2$ is a root.
answered Jan 3 at 18:58
Cheerful ParsnipCheerful Parsnip
21.1k23598
answered Jan 3 at 18:58
Cheerful ParsnipCheerful Parsnip
21.1k23598
answered Jan 3 at 18:58
Cheerful ParsnipCheerful Parsnip
21.1k23598
answered Jan 3 at 18:58
Cheerful ParsnipCheerful Parsnip
21.1k23598
21.1k23598
add a comment |
add a comment |
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4
$begingroup$
Is the splitting field not $mathbb Q(sqrt{7}i)$?
$endgroup$
– Cheerful Parsnip
Jan 3 at 1:59
1
$begingroup$
Well, I agree 2 is a root. Therefore Dummit and Foote are right. The problem with your answer is that you've got the splitting field wrong. See Cheerful Parsnip's comment above.
$endgroup$
– jgon
Jan 3 at 2:00
1
$begingroup$
@CheerfulParsnip, actually you are right. I don't know what I was thinking. $p(x)$ is the minimal polynomial of its roots and so the group must be of order $2$ too.
$endgroup$
– dezdichado
Jan 3 at 2:03
$begingroup$
It's all good. That's how we learn.
$endgroup$
– Cheerful Parsnip
Jan 3 at 2:24
$begingroup$
@CheerfulParsnip maybe you should write an answer.
$endgroup$
– Kenny Lau
Jan 3 at 18:14