Galois group of $x^3-x^2-4$












4














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







share|cite|improve this question


















  • 4




    Is the splitting field not $mathbb Q(sqrt{7}i)$?
    – Cheerful Parsnip
    Jan 3 at 1:59






  • 1




    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.
    – jgon
    Jan 3 at 2:00








  • 1




    @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.
    – dezdichado
    Jan 3 at 2:03










  • It's all good. That's how we learn.
    – Cheerful Parsnip
    Jan 3 at 2:24










  • @CheerfulParsnip maybe you should write an answer.
    – Kenny Lau
    2 days ago
















4














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







share|cite|improve this question


















  • 4




    Is the splitting field not $mathbb Q(sqrt{7}i)$?
    – Cheerful Parsnip
    Jan 3 at 1:59






  • 1




    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.
    – jgon
    Jan 3 at 2:00








  • 1




    @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.
    – dezdichado
    Jan 3 at 2:03










  • It's all good. That's how we learn.
    – Cheerful Parsnip
    Jan 3 at 2:24










  • @CheerfulParsnip maybe you should write an answer.
    – Kenny Lau
    2 days ago














4












4








4







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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 3 at 1:42









dezdichado

6,2551929




6,2551929








  • 4




    Is the splitting field not $mathbb Q(sqrt{7}i)$?
    – Cheerful Parsnip
    Jan 3 at 1:59






  • 1




    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.
    – jgon
    Jan 3 at 2:00








  • 1




    @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.
    – dezdichado
    Jan 3 at 2:03










  • It's all good. That's how we learn.
    – Cheerful Parsnip
    Jan 3 at 2:24










  • @CheerfulParsnip maybe you should write an answer.
    – Kenny Lau
    2 days ago














  • 4




    Is the splitting field not $mathbb Q(sqrt{7}i)$?
    – Cheerful Parsnip
    Jan 3 at 1:59






  • 1




    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.
    – jgon
    Jan 3 at 2:00








  • 1




    @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.
    – dezdichado
    Jan 3 at 2:03










  • It's all good. That's how we learn.
    – Cheerful Parsnip
    Jan 3 at 2:24










  • @CheerfulParsnip maybe you should write an answer.
    – Kenny Lau
    2 days ago








4




4




Is the splitting field not $mathbb Q(sqrt{7}i)$?
– Cheerful Parsnip
Jan 3 at 1:59




Is the splitting field not $mathbb Q(sqrt{7}i)$?
– Cheerful Parsnip
Jan 3 at 1:59




1




1




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.
– jgon
Jan 3 at 2:00






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.
– jgon
Jan 3 at 2:00






1




1




@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.
– dezdichado
Jan 3 at 2:03




@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.
– dezdichado
Jan 3 at 2:03












It's all good. That's how we learn.
– Cheerful Parsnip
Jan 3 at 2:24




It's all good. That's how we learn.
– Cheerful Parsnip
Jan 3 at 2:24












@CheerfulParsnip maybe you should write an answer.
– Kenny Lau
2 days ago




@CheerfulParsnip maybe you should write an answer.
– Kenny Lau
2 days ago










1 Answer
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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.






share|cite|improve this answer





















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    1 Answer
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    active

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    active

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    2














    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.






    share|cite|improve this answer


























      2














      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.






      share|cite|improve this answer
























        2












        2








        2






        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        Cheerful Parsnip

        20.9k23396




        20.9k23396






























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