Why isn't the the galois group of a polynomial with n distinct roots isomorphic to Sn?












2














If we consider the polynomial $f(x)=x^3-3x-1 in mathbb{Q}[x]$ which has 3 real roots ${x_1,x_2,x_3}$. I read that its galois group is isomorphic to $A_3$ and not to $S_3$. I don't really understand this.
If I consider the map (Here $E$ is the splitting field of $f$ over $mathbb{Q})$:
begin{align*}
sigma colon E to E\
x_1 mapsto x_2\
x_3 mapsto x_3\
mathbb{Q} mapsto mathbb{Q}
end{align*}



This would correspond to a transposition $(12)$ which is not in the alternating group.

But isn't this a $mathbb{Q}$-automorphism? I actually don't really grasp why the Galois group isn't in general isomorphic to Sn (for n distinct roots).






galois-theory







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  • 1




    What do you mean by $E$?
    – Servaes
    Nov 29 '18 at 15:24










  • $E$ is the splitting field of $f$ over $mathbb{Q}$. Thanks for the comment, I edited.
    – roi_saumon
    Nov 29 '18 at 15:41








  • 3




    There may be hidden relations among the roots. Your cubic is a case in point. If $r=x_1$ is one of the zeros, then the other zeros are $x_2=2-r^2$ and $x_3=r^2-r-2$. So the splitting field is $Bbb{Q}(r)$. If you know $sigma(x_1)=sigma(r)$ then $$sigma(x_2)=sigma(2-r^2)=2-sigma(r)^2$$ is uniquely determined. In other words, you cannot choose $sigma(x_1)$ and $sigma(x_2)$ independently from each other in the case of this polynomial. Looking at the discriminant as in Servaes' answer is a more common constraint, I simply wanted to shed more light to this.
    – Jyrki Lahtonen
    Nov 30 '18 at 4:10












  • @JyrkiLahtonen Oh, I see, but how did you find this hidden relation?
    – roi_saumon
    Nov 30 '18 at 11:09












  • The polynomial is kinda "famous". The zeros are $x_1=-2cos(2pi/9)$, $x_2=-2cos(4pi/9)$ and $x_3=-2cos(8pi/9)$. The implication $r$ is a root $implies$ $2-r^2$ is a root, is simple the double angle cosine formula $cos2alpha=2cos^2alpha-1$ in disguise. More generally, adjoining a single zero $r$ may allow us to factor the polynomial more than just splitting off a factor $x-r$ of $f(x)$, and such factorizations reduce the extension degree of the splitting field (and hence also the number of automorphisms).
    – Jyrki Lahtonen
    Nov 30 '18 at 11:36


















2














If we consider the polynomial $f(x)=x^3-3x-1 in mathbb{Q}[x]$ which has 3 real roots ${x_1,x_2,x_3}$. I read that its galois group is isomorphic to $A_3$ and not to $S_3$. I don't really understand this.
If I consider the map (Here $E$ is the splitting field of $f$ over $mathbb{Q})$:
begin{align*}
sigma colon E to E\
x_1 mapsto x_2\
x_3 mapsto x_3\
mathbb{Q} mapsto mathbb{Q}
end{align*}



This would correspond to a transposition $(12)$ which is not in the alternating group.

But isn't this a $mathbb{Q}$-automorphism? I actually don't really grasp why the Galois group isn't in general isomorphic to Sn (for n distinct roots).






galois-theory







share|cite|improve this question




















  • 1




    What do you mean by $E$?
    – Servaes
    Nov 29 '18 at 15:24










  • $E$ is the splitting field of $f$ over $mathbb{Q}$. Thanks for the comment, I edited.
    – roi_saumon
    Nov 29 '18 at 15:41








  • 3




    There may be hidden relations among the roots. Your cubic is a case in point. If $r=x_1$ is one of the zeros, then the other zeros are $x_2=2-r^2$ and $x_3=r^2-r-2$. So the splitting field is $Bbb{Q}(r)$. If you know $sigma(x_1)=sigma(r)$ then $$sigma(x_2)=sigma(2-r^2)=2-sigma(r)^2$$ is uniquely determined. In other words, you cannot choose $sigma(x_1)$ and $sigma(x_2)$ independently from each other in the case of this polynomial. Looking at the discriminant as in Servaes' answer is a more common constraint, I simply wanted to shed more light to this.
    – Jyrki Lahtonen
    Nov 30 '18 at 4:10












  • @JyrkiLahtonen Oh, I see, but how did you find this hidden relation?
    – roi_saumon
    Nov 30 '18 at 11:09












  • The polynomial is kinda "famous". The zeros are $x_1=-2cos(2pi/9)$, $x_2=-2cos(4pi/9)$ and $x_3=-2cos(8pi/9)$. The implication $r$ is a root $implies$ $2-r^2$ is a root, is simple the double angle cosine formula $cos2alpha=2cos^2alpha-1$ in disguise. More generally, adjoining a single zero $r$ may allow us to factor the polynomial more than just splitting off a factor $x-r$ of $f(x)$, and such factorizations reduce the extension degree of the splitting field (and hence also the number of automorphisms).
    – Jyrki Lahtonen
    Nov 30 '18 at 11:36
















2












2








2







If we consider the polynomial $f(x)=x^3-3x-1 in mathbb{Q}[x]$ which has 3 real roots ${x_1,x_2,x_3}$. I read that its galois group is isomorphic to $A_3$ and not to $S_3$. I don't really understand this.
If I consider the map (Here $E$ is the splitting field of $f$ over $mathbb{Q})$:
begin{align*}
sigma colon E to E\
x_1 mapsto x_2\
x_3 mapsto x_3\
mathbb{Q} mapsto mathbb{Q}
end{align*}



This would correspond to a transposition $(12)$ which is not in the alternating group.

But isn't this a $mathbb{Q}$-automorphism? I actually don't really grasp why the Galois group isn't in general isomorphic to Sn (for n distinct roots).






galois-theory







share|cite|improve this question















If we consider the polynomial $f(x)=x^3-3x-1 in mathbb{Q}[x]$ which has 3 real roots ${x_1,x_2,x_3}$. I read that its galois group is isomorphic to $A_3$ and not to $S_3$. I don't really understand this.
If I consider the map (Here $E$ is the splitting field of $f$ over $mathbb{Q})$:
begin{align*}
sigma colon E to E\
x_1 mapsto x_2\
x_3 mapsto x_3\
mathbb{Q} mapsto mathbb{Q}
end{align*}



This would correspond to a transposition $(12)$ which is not in the alternating group.

But isn't this a $mathbb{Q}$-automorphism? I actually don't really grasp why the Galois group isn't in general isomorphic to Sn (for n distinct roots).






galois-theory




galois-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Nov 29 '18 at 15:40







roi_saumon

















asked Nov 29 '18 at 15:21









roi_saumonroi_saumon

43528




43528








  • 1




    What do you mean by $E$?
    – Servaes
    Nov 29 '18 at 15:24










  • $E$ is the splitting field of $f$ over $mathbb{Q}$. Thanks for the comment, I edited.
    – roi_saumon
    Nov 29 '18 at 15:41








  • 3




    There may be hidden relations among the roots. Your cubic is a case in point. If $r=x_1$ is one of the zeros, then the other zeros are $x_2=2-r^2$ and $x_3=r^2-r-2$. So the splitting field is $Bbb{Q}(r)$. If you know $sigma(x_1)=sigma(r)$ then $$sigma(x_2)=sigma(2-r^2)=2-sigma(r)^2$$ is uniquely determined. In other words, you cannot choose $sigma(x_1)$ and $sigma(x_2)$ independently from each other in the case of this polynomial. Looking at the discriminant as in Servaes' answer is a more common constraint, I simply wanted to shed more light to this.
    – Jyrki Lahtonen
    Nov 30 '18 at 4:10












  • @JyrkiLahtonen Oh, I see, but how did you find this hidden relation?
    – roi_saumon
    Nov 30 '18 at 11:09












  • The polynomial is kinda "famous". The zeros are $x_1=-2cos(2pi/9)$, $x_2=-2cos(4pi/9)$ and $x_3=-2cos(8pi/9)$. The implication $r$ is a root $implies$ $2-r^2$ is a root, is simple the double angle cosine formula $cos2alpha=2cos^2alpha-1$ in disguise. More generally, adjoining a single zero $r$ may allow us to factor the polynomial more than just splitting off a factor $x-r$ of $f(x)$, and such factorizations reduce the extension degree of the splitting field (and hence also the number of automorphisms).
    – Jyrki Lahtonen
    Nov 30 '18 at 11:36
















  • 1




    What do you mean by $E$?
    – Servaes
    Nov 29 '18 at 15:24










  • $E$ is the splitting field of $f$ over $mathbb{Q}$. Thanks for the comment, I edited.
    – roi_saumon
    Nov 29 '18 at 15:41








  • 3




    There may be hidden relations among the roots. Your cubic is a case in point. If $r=x_1$ is one of the zeros, then the other zeros are $x_2=2-r^2$ and $x_3=r^2-r-2$. So the splitting field is $Bbb{Q}(r)$. If you know $sigma(x_1)=sigma(r)$ then $$sigma(x_2)=sigma(2-r^2)=2-sigma(r)^2$$ is uniquely determined. In other words, you cannot choose $sigma(x_1)$ and $sigma(x_2)$ independently from each other in the case of this polynomial. Looking at the discriminant as in Servaes' answer is a more common constraint, I simply wanted to shed more light to this.
    – Jyrki Lahtonen
    Nov 30 '18 at 4:10












  • @JyrkiLahtonen Oh, I see, but how did you find this hidden relation?
    – roi_saumon
    Nov 30 '18 at 11:09












  • The polynomial is kinda "famous". The zeros are $x_1=-2cos(2pi/9)$, $x_2=-2cos(4pi/9)$ and $x_3=-2cos(8pi/9)$. The implication $r$ is a root $implies$ $2-r^2$ is a root, is simple the double angle cosine formula $cos2alpha=2cos^2alpha-1$ in disguise. More generally, adjoining a single zero $r$ may allow us to factor the polynomial more than just splitting off a factor $x-r$ of $f(x)$, and such factorizations reduce the extension degree of the splitting field (and hence also the number of automorphisms).
    – Jyrki Lahtonen
    Nov 30 '18 at 11:36










1




1




What do you mean by $E$?
– Servaes
Nov 29 '18 at 15:24




What do you mean by $E$?
– Servaes
Nov 29 '18 at 15:24












$E$ is the splitting field of $f$ over $mathbb{Q}$. Thanks for the comment, I edited.
– roi_saumon
Nov 29 '18 at 15:41






$E$ is the splitting field of $f$ over $mathbb{Q}$. Thanks for the comment, I edited.
– roi_saumon
Nov 29 '18 at 15:41






3




3




There may be hidden relations among the roots. Your cubic is a case in point. If $r=x_1$ is one of the zeros, then the other zeros are $x_2=2-r^2$ and $x_3=r^2-r-2$. So the splitting field is $Bbb{Q}(r)$. If you know $sigma(x_1)=sigma(r)$ then $$sigma(x_2)=sigma(2-r^2)=2-sigma(r)^2$$ is uniquely determined. In other words, you cannot choose $sigma(x_1)$ and $sigma(x_2)$ independently from each other in the case of this polynomial. Looking at the discriminant as in Servaes' answer is a more common constraint, I simply wanted to shed more light to this.
– Jyrki Lahtonen
Nov 30 '18 at 4:10






There may be hidden relations among the roots. Your cubic is a case in point. If $r=x_1$ is one of the zeros, then the other zeros are $x_2=2-r^2$ and $x_3=r^2-r-2$. So the splitting field is $Bbb{Q}(r)$. If you know $sigma(x_1)=sigma(r)$ then $$sigma(x_2)=sigma(2-r^2)=2-sigma(r)^2$$ is uniquely determined. In other words, you cannot choose $sigma(x_1)$ and $sigma(x_2)$ independently from each other in the case of this polynomial. Looking at the discriminant as in Servaes' answer is a more common constraint, I simply wanted to shed more light to this.
– Jyrki Lahtonen
Nov 30 '18 at 4:10














@JyrkiLahtonen Oh, I see, but how did you find this hidden relation?
– roi_saumon
Nov 30 '18 at 11:09






@JyrkiLahtonen Oh, I see, but how did you find this hidden relation?
– roi_saumon
Nov 30 '18 at 11:09














The polynomial is kinda "famous". The zeros are $x_1=-2cos(2pi/9)$, $x_2=-2cos(4pi/9)$ and $x_3=-2cos(8pi/9)$. The implication $r$ is a root $implies$ $2-r^2$ is a root, is simple the double angle cosine formula $cos2alpha=2cos^2alpha-1$ in disguise. More generally, adjoining a single zero $r$ may allow us to factor the polynomial more than just splitting off a factor $x-r$ of $f(x)$, and such factorizations reduce the extension degree of the splitting field (and hence also the number of automorphisms).
– Jyrki Lahtonen
Nov 30 '18 at 11:36






The polynomial is kinda "famous". The zeros are $x_1=-2cos(2pi/9)$, $x_2=-2cos(4pi/9)$ and $x_3=-2cos(8pi/9)$. The implication $r$ is a root $implies$ $2-r^2$ is a root, is simple the double angle cosine formula $cos2alpha=2cos^2alpha-1$ in disguise. More generally, adjoining a single zero $r$ may allow us to factor the polynomial more than just splitting off a factor $x-r$ of $f(x)$, and such factorizations reduce the extension degree of the splitting field (and hence also the number of automorphisms).
– Jyrki Lahtonen
Nov 30 '18 at 11:36












1 Answer
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TL;DR The element $(x_1-x_2)(x_1-x_3)(x_2-x_3)$ is contained in $Bbb{Q}$ but not fixed by $sigma$.




The following theorem tells us when the Galois group of a polynomial over $Bbb{Q}$ is contained in $A_n$:




Let $finBbb{Q}[x]$ be irreducible of degree $n$, and let $E$ be a splitting field of $f$. Identify $operatorname{Gal}(E/Bbb{Q})$ with a subgroup of $S_n$ by enumerating the roots of $f$ in $E$. If $Delta(f)inBbb{Q}$ is a square in $Bbb{Q}$ then $operatorname{Gal}(E/Bbb{Q})$ is contained in $A_n$.




The proof shows in particular why your map is not a field automorphism of $E$:



Let $x_1,ldots,x_nin E$ be the roots of $f$ and let $S_n$ act on $E$ by its action on the indices of the roots, i.e. $sigma(x_i):=x_{sigma(i)}$ for all $i$. Consider the element
$$delta:=prod_{1leq i<jleq n}(x_i-x_j)in E.$$
Note that for all $sigmain S_n$ we have $sigma(delta)=operatorname{sgn}(sigma)delta$ and that $delta^2=Delta(f)$.



If $Delta(f)$ is a square in $Bbb{Q}$ then it $deltainBbb{Q}$. It follows that $sigma(delta)=delta$ for all $sigmainoperatorname{Gal}(EBbb{Q})$ and hence that $operatorname{sgn}(sigma)=1$ for all $sigmainoperatorname{Gal}(EBbb{Q})$. This means precisely that $operatorname{Gal}(EBbb{Q})$ is contained in $A_n$.




In this particular case we see that $Delta(x^3-3x-1)=81$, so the element
$$delta=(x_1-x_2)(x_1-x_3)(x_2-x_3)=pm9,$$
is not fixed by your map $sigma$, so $sigma$ cannot be a field automorphism of $E$.






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    1 Answer
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    1 Answer
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    active

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    3














    TL;DR The element $(x_1-x_2)(x_1-x_3)(x_2-x_3)$ is contained in $Bbb{Q}$ but not fixed by $sigma$.




    The following theorem tells us when the Galois group of a polynomial over $Bbb{Q}$ is contained in $A_n$:




    Let $finBbb{Q}[x]$ be irreducible of degree $n$, and let $E$ be a splitting field of $f$. Identify $operatorname{Gal}(E/Bbb{Q})$ with a subgroup of $S_n$ by enumerating the roots of $f$ in $E$. If $Delta(f)inBbb{Q}$ is a square in $Bbb{Q}$ then $operatorname{Gal}(E/Bbb{Q})$ is contained in $A_n$.




    The proof shows in particular why your map is not a field automorphism of $E$:



    Let $x_1,ldots,x_nin E$ be the roots of $f$ and let $S_n$ act on $E$ by its action on the indices of the roots, i.e. $sigma(x_i):=x_{sigma(i)}$ for all $i$. Consider the element
    $$delta:=prod_{1leq i<jleq n}(x_i-x_j)in E.$$
    Note that for all $sigmain S_n$ we have $sigma(delta)=operatorname{sgn}(sigma)delta$ and that $delta^2=Delta(f)$.



    If $Delta(f)$ is a square in $Bbb{Q}$ then it $deltainBbb{Q}$. It follows that $sigma(delta)=delta$ for all $sigmainoperatorname{Gal}(EBbb{Q})$ and hence that $operatorname{sgn}(sigma)=1$ for all $sigmainoperatorname{Gal}(EBbb{Q})$. This means precisely that $operatorname{Gal}(EBbb{Q})$ is contained in $A_n$.




    In this particular case we see that $Delta(x^3-3x-1)=81$, so the element
    $$delta=(x_1-x_2)(x_1-x_3)(x_2-x_3)=pm9,$$
    is not fixed by your map $sigma$, so $sigma$ cannot be a field automorphism of $E$.






    share|cite|improve this answer




























      3














      TL;DR The element $(x_1-x_2)(x_1-x_3)(x_2-x_3)$ is contained in $Bbb{Q}$ but not fixed by $sigma$.




      The following theorem tells us when the Galois group of a polynomial over $Bbb{Q}$ is contained in $A_n$:




      Let $finBbb{Q}[x]$ be irreducible of degree $n$, and let $E$ be a splitting field of $f$. Identify $operatorname{Gal}(E/Bbb{Q})$ with a subgroup of $S_n$ by enumerating the roots of $f$ in $E$. If $Delta(f)inBbb{Q}$ is a square in $Bbb{Q}$ then $operatorname{Gal}(E/Bbb{Q})$ is contained in $A_n$.




      The proof shows in particular why your map is not a field automorphism of $E$:



      Let $x_1,ldots,x_nin E$ be the roots of $f$ and let $S_n$ act on $E$ by its action on the indices of the roots, i.e. $sigma(x_i):=x_{sigma(i)}$ for all $i$. Consider the element
      $$delta:=prod_{1leq i<jleq n}(x_i-x_j)in E.$$
      Note that for all $sigmain S_n$ we have $sigma(delta)=operatorname{sgn}(sigma)delta$ and that $delta^2=Delta(f)$.



      If $Delta(f)$ is a square in $Bbb{Q}$ then it $deltainBbb{Q}$. It follows that $sigma(delta)=delta$ for all $sigmainoperatorname{Gal}(EBbb{Q})$ and hence that $operatorname{sgn}(sigma)=1$ for all $sigmainoperatorname{Gal}(EBbb{Q})$. This means precisely that $operatorname{Gal}(EBbb{Q})$ is contained in $A_n$.




      In this particular case we see that $Delta(x^3-3x-1)=81$, so the element
      $$delta=(x_1-x_2)(x_1-x_3)(x_2-x_3)=pm9,$$
      is not fixed by your map $sigma$, so $sigma$ cannot be a field automorphism of $E$.






      share|cite|improve this answer


























        3












        3








        3






        TL;DR The element $(x_1-x_2)(x_1-x_3)(x_2-x_3)$ is contained in $Bbb{Q}$ but not fixed by $sigma$.




        The following theorem tells us when the Galois group of a polynomial over $Bbb{Q}$ is contained in $A_n$:




        Let $finBbb{Q}[x]$ be irreducible of degree $n$, and let $E$ be a splitting field of $f$. Identify $operatorname{Gal}(E/Bbb{Q})$ with a subgroup of $S_n$ by enumerating the roots of $f$ in $E$. If $Delta(f)inBbb{Q}$ is a square in $Bbb{Q}$ then $operatorname{Gal}(E/Bbb{Q})$ is contained in $A_n$.




        The proof shows in particular why your map is not a field automorphism of $E$:



        Let $x_1,ldots,x_nin E$ be the roots of $f$ and let $S_n$ act on $E$ by its action on the indices of the roots, i.e. $sigma(x_i):=x_{sigma(i)}$ for all $i$. Consider the element
        $$delta:=prod_{1leq i<jleq n}(x_i-x_j)in E.$$
        Note that for all $sigmain S_n$ we have $sigma(delta)=operatorname{sgn}(sigma)delta$ and that $delta^2=Delta(f)$.



        If $Delta(f)$ is a square in $Bbb{Q}$ then it $deltainBbb{Q}$. It follows that $sigma(delta)=delta$ for all $sigmainoperatorname{Gal}(EBbb{Q})$ and hence that $operatorname{sgn}(sigma)=1$ for all $sigmainoperatorname{Gal}(EBbb{Q})$. This means precisely that $operatorname{Gal}(EBbb{Q})$ is contained in $A_n$.




        In this particular case we see that $Delta(x^3-3x-1)=81$, so the element
        $$delta=(x_1-x_2)(x_1-x_3)(x_2-x_3)=pm9,$$
        is not fixed by your map $sigma$, so $sigma$ cannot be a field automorphism of $E$.






        share|cite|improve this answer














        TL;DR The element $(x_1-x_2)(x_1-x_3)(x_2-x_3)$ is contained in $Bbb{Q}$ but not fixed by $sigma$.




        The following theorem tells us when the Galois group of a polynomial over $Bbb{Q}$ is contained in $A_n$:




        Let $finBbb{Q}[x]$ be irreducible of degree $n$, and let $E$ be a splitting field of $f$. Identify $operatorname{Gal}(E/Bbb{Q})$ with a subgroup of $S_n$ by enumerating the roots of $f$ in $E$. If $Delta(f)inBbb{Q}$ is a square in $Bbb{Q}$ then $operatorname{Gal}(E/Bbb{Q})$ is contained in $A_n$.




        The proof shows in particular why your map is not a field automorphism of $E$:



        Let $x_1,ldots,x_nin E$ be the roots of $f$ and let $S_n$ act on $E$ by its action on the indices of the roots, i.e. $sigma(x_i):=x_{sigma(i)}$ for all $i$. Consider the element
        $$delta:=prod_{1leq i<jleq n}(x_i-x_j)in E.$$
        Note that for all $sigmain S_n$ we have $sigma(delta)=operatorname{sgn}(sigma)delta$ and that $delta^2=Delta(f)$.



        If $Delta(f)$ is a square in $Bbb{Q}$ then it $deltainBbb{Q}$. It follows that $sigma(delta)=delta$ for all $sigmainoperatorname{Gal}(EBbb{Q})$ and hence that $operatorname{sgn}(sigma)=1$ for all $sigmainoperatorname{Gal}(EBbb{Q})$. This means precisely that $operatorname{Gal}(EBbb{Q})$ is contained in $A_n$.




        In this particular case we see that $Delta(x^3-3x-1)=81$, so the element
        $$delta=(x_1-x_2)(x_1-x_3)(x_2-x_3)=pm9,$$
        is not fixed by your map $sigma$, so $sigma$ cannot be a field automorphism of $E$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 29 '18 at 16:17

























        answered Nov 29 '18 at 15:56









        ServaesServaes

        22.4k33793




        22.4k33793






























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