Smart way to solve octics like $x^8+5992704x-304129728=0$












6












$begingroup$


Question: How to quickly find explicit formula for roots of polynomials (solvable by radicals) like $f(x)=x^8+5992704x-304129728$?





My current approach is just a brute force - not very fast and convenient:




  • observe that Galois group of above $f(x)$ is $C_2 wr S_4$, so one needs to find quartic polynomial $g(x)=x^4+cx^2+dx+e$ such that $f(x)$ splits into quadratic (and sextic) polynomial over it

  • iterate over small integer triples $c,d,e$ until correct $g(x)$ is found

  • to speed up the process I limited $g(x)$ to such that $Delta(g(x))$ divides $Delta(f(x))$ - here $Delta$ means discriminant


More explicitly, using GAP system:



x:=Indeterminate(Rationals, "x");
f:=x^8+5992704*x-304129728;
discrF:=Discriminant(f);

# define some arbitrary range of the c,d,e coefficients
for c in [-40..40] do
for d in [1..6000] do
for e in [-30000..30000] do
# small perf trick, calculate discriminant
# without actual construction of g(x)
discrG:=256*e^3-128*c^2*e^2+144*c*d^2*e-27*d^4+16*c^4*e-4*c^3*d^2;
if (discrG <> 0 and discrF mod discrG = 0) then
g:=x^4+c*x^2+d*x+e;
if IsIrreducible(g) then
e:=AlgebraicExtension(Rationals,g);
factors:=FactorsPolynomialAlgExt(e, f);
if Size(factors) > 1 then
Print("Success: g(x)=", g," : factors=", factors,"n");
fi;
fi;
fi;
od;
od;
od;


Couple hours later...:




  • $g(x)=x^4-34x^2+1632x-7871$

  • $f(x)={ x^2+(-a^3/56-a^2/56+89a/56-1207/56)x+(3a^3/28+6a^2/7+153a/28+459/7) }times{x^6+(a^3/56+a^2/56-89a/56+1207/56)x^5+(3a^3/28+33a^2/14+153a/28+561/14)x^4+(-27a^3/14-153a^2/14+2907a/14-4743/14)x^3+(-153a^2-3060a+4437)x^2+(-153a^3/7+8415a^2/7+13617a/7-2078199/7)x+(-2754a^3-5508a^2+109242a-2465748) }$


where $g(a) = 0$



Now one can find (very long and fancy) explicit expression for roots of $f(x)$, using only $+$, $-$, $times$, $div$, and $sqrt{}$ operations.





Loosely related question: On solvable octic trinomials like $x^8-5x-5=0$










share|cite|improve this question









$endgroup$












  • $begingroup$
    If you're going to resort to a computer algebra system, there's probably one that solves solvable octics quickly. Maybe Maple, maybe Mathematica.
    $endgroup$
    – Gerry Myerson
    Dec 21 '18 at 19:41










  • $begingroup$
    @N.S. As far as I understand degree of extension is determined by Galois group of roots of $f(x)$. Quadratic $g(x)$ is not possible for $C_2 wr S_4$. On the contrary, quadratic $g(x)$ is required e.g. for another group $S_4 wr C_2$.
    $endgroup$
    – klajok
    Dec 21 '18 at 19:56










  • $begingroup$
    @GerryMyerson I tried RadiRoot package in GAP system, but $C_2 wr S_4$ group is too big for it and calculations finish with out of memory error. Maybe one day I will get Maple or Mathematica...
    $endgroup$
    – klajok
    Dec 21 '18 at 20:01
















6












$begingroup$


Question: How to quickly find explicit formula for roots of polynomials (solvable by radicals) like $f(x)=x^8+5992704x-304129728$?





My current approach is just a brute force - not very fast and convenient:




  • observe that Galois group of above $f(x)$ is $C_2 wr S_4$, so one needs to find quartic polynomial $g(x)=x^4+cx^2+dx+e$ such that $f(x)$ splits into quadratic (and sextic) polynomial over it

  • iterate over small integer triples $c,d,e$ until correct $g(x)$ is found

  • to speed up the process I limited $g(x)$ to such that $Delta(g(x))$ divides $Delta(f(x))$ - here $Delta$ means discriminant


More explicitly, using GAP system:



x:=Indeterminate(Rationals, "x");
f:=x^8+5992704*x-304129728;
discrF:=Discriminant(f);

# define some arbitrary range of the c,d,e coefficients
for c in [-40..40] do
for d in [1..6000] do
for e in [-30000..30000] do
# small perf trick, calculate discriminant
# without actual construction of g(x)
discrG:=256*e^3-128*c^2*e^2+144*c*d^2*e-27*d^4+16*c^4*e-4*c^3*d^2;
if (discrG <> 0 and discrF mod discrG = 0) then
g:=x^4+c*x^2+d*x+e;
if IsIrreducible(g) then
e:=AlgebraicExtension(Rationals,g);
factors:=FactorsPolynomialAlgExt(e, f);
if Size(factors) > 1 then
Print("Success: g(x)=", g," : factors=", factors,"n");
fi;
fi;
fi;
od;
od;
od;


Couple hours later...:




  • $g(x)=x^4-34x^2+1632x-7871$

  • $f(x)={ x^2+(-a^3/56-a^2/56+89a/56-1207/56)x+(3a^3/28+6a^2/7+153a/28+459/7) }times{x^6+(a^3/56+a^2/56-89a/56+1207/56)x^5+(3a^3/28+33a^2/14+153a/28+561/14)x^4+(-27a^3/14-153a^2/14+2907a/14-4743/14)x^3+(-153a^2-3060a+4437)x^2+(-153a^3/7+8415a^2/7+13617a/7-2078199/7)x+(-2754a^3-5508a^2+109242a-2465748) }$


where $g(a) = 0$



Now one can find (very long and fancy) explicit expression for roots of $f(x)$, using only $+$, $-$, $times$, $div$, and $sqrt{}$ operations.





Loosely related question: On solvable octic trinomials like $x^8-5x-5=0$










share|cite|improve this question









$endgroup$












  • $begingroup$
    If you're going to resort to a computer algebra system, there's probably one that solves solvable octics quickly. Maybe Maple, maybe Mathematica.
    $endgroup$
    – Gerry Myerson
    Dec 21 '18 at 19:41










  • $begingroup$
    @N.S. As far as I understand degree of extension is determined by Galois group of roots of $f(x)$. Quadratic $g(x)$ is not possible for $C_2 wr S_4$. On the contrary, quadratic $g(x)$ is required e.g. for another group $S_4 wr C_2$.
    $endgroup$
    – klajok
    Dec 21 '18 at 19:56










  • $begingroup$
    @GerryMyerson I tried RadiRoot package in GAP system, but $C_2 wr S_4$ group is too big for it and calculations finish with out of memory error. Maybe one day I will get Maple or Mathematica...
    $endgroup$
    – klajok
    Dec 21 '18 at 20:01














6












6








6


1



$begingroup$


Question: How to quickly find explicit formula for roots of polynomials (solvable by radicals) like $f(x)=x^8+5992704x-304129728$?





My current approach is just a brute force - not very fast and convenient:




  • observe that Galois group of above $f(x)$ is $C_2 wr S_4$, so one needs to find quartic polynomial $g(x)=x^4+cx^2+dx+e$ such that $f(x)$ splits into quadratic (and sextic) polynomial over it

  • iterate over small integer triples $c,d,e$ until correct $g(x)$ is found

  • to speed up the process I limited $g(x)$ to such that $Delta(g(x))$ divides $Delta(f(x))$ - here $Delta$ means discriminant


More explicitly, using GAP system:



x:=Indeterminate(Rationals, "x");
f:=x^8+5992704*x-304129728;
discrF:=Discriminant(f);

# define some arbitrary range of the c,d,e coefficients
for c in [-40..40] do
for d in [1..6000] do
for e in [-30000..30000] do
# small perf trick, calculate discriminant
# without actual construction of g(x)
discrG:=256*e^3-128*c^2*e^2+144*c*d^2*e-27*d^4+16*c^4*e-4*c^3*d^2;
if (discrG <> 0 and discrF mod discrG = 0) then
g:=x^4+c*x^2+d*x+e;
if IsIrreducible(g) then
e:=AlgebraicExtension(Rationals,g);
factors:=FactorsPolynomialAlgExt(e, f);
if Size(factors) > 1 then
Print("Success: g(x)=", g," : factors=", factors,"n");
fi;
fi;
fi;
od;
od;
od;


Couple hours later...:




  • $g(x)=x^4-34x^2+1632x-7871$

  • $f(x)={ x^2+(-a^3/56-a^2/56+89a/56-1207/56)x+(3a^3/28+6a^2/7+153a/28+459/7) }times{x^6+(a^3/56+a^2/56-89a/56+1207/56)x^5+(3a^3/28+33a^2/14+153a/28+561/14)x^4+(-27a^3/14-153a^2/14+2907a/14-4743/14)x^3+(-153a^2-3060a+4437)x^2+(-153a^3/7+8415a^2/7+13617a/7-2078199/7)x+(-2754a^3-5508a^2+109242a-2465748) }$


where $g(a) = 0$



Now one can find (very long and fancy) explicit expression for roots of $f(x)$, using only $+$, $-$, $times$, $div$, and $sqrt{}$ operations.





Loosely related question: On solvable octic trinomials like $x^8-5x-5=0$










share|cite|improve this question









$endgroup$




Question: How to quickly find explicit formula for roots of polynomials (solvable by radicals) like $f(x)=x^8+5992704x-304129728$?





My current approach is just a brute force - not very fast and convenient:




  • observe that Galois group of above $f(x)$ is $C_2 wr S_4$, so one needs to find quartic polynomial $g(x)=x^4+cx^2+dx+e$ such that $f(x)$ splits into quadratic (and sextic) polynomial over it

  • iterate over small integer triples $c,d,e$ until correct $g(x)$ is found

  • to speed up the process I limited $g(x)$ to such that $Delta(g(x))$ divides $Delta(f(x))$ - here $Delta$ means discriminant


More explicitly, using GAP system:



x:=Indeterminate(Rationals, "x");
f:=x^8+5992704*x-304129728;
discrF:=Discriminant(f);

# define some arbitrary range of the c,d,e coefficients
for c in [-40..40] do
for d in [1..6000] do
for e in [-30000..30000] do
# small perf trick, calculate discriminant
# without actual construction of g(x)
discrG:=256*e^3-128*c^2*e^2+144*c*d^2*e-27*d^4+16*c^4*e-4*c^3*d^2;
if (discrG <> 0 and discrF mod discrG = 0) then
g:=x^4+c*x^2+d*x+e;
if IsIrreducible(g) then
e:=AlgebraicExtension(Rationals,g);
factors:=FactorsPolynomialAlgExt(e, f);
if Size(factors) > 1 then
Print("Success: g(x)=", g," : factors=", factors,"n");
fi;
fi;
fi;
od;
od;
od;


Couple hours later...:




  • $g(x)=x^4-34x^2+1632x-7871$

  • $f(x)={ x^2+(-a^3/56-a^2/56+89a/56-1207/56)x+(3a^3/28+6a^2/7+153a/28+459/7) }times{x^6+(a^3/56+a^2/56-89a/56+1207/56)x^5+(3a^3/28+33a^2/14+153a/28+561/14)x^4+(-27a^3/14-153a^2/14+2907a/14-4743/14)x^3+(-153a^2-3060a+4437)x^2+(-153a^3/7+8415a^2/7+13617a/7-2078199/7)x+(-2754a^3-5508a^2+109242a-2465748) }$


where $g(a) = 0$



Now one can find (very long and fancy) explicit expression for roots of $f(x)$, using only $+$, $-$, $times$, $div$, and $sqrt{}$ operations.





Loosely related question: On solvable octic trinomials like $x^8-5x-5=0$







galois-theory radicals gap






share|cite|improve this question













share|cite|improve this question











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asked Dec 21 '18 at 18:27









klajokklajok

37519




37519












  • $begingroup$
    If you're going to resort to a computer algebra system, there's probably one that solves solvable octics quickly. Maybe Maple, maybe Mathematica.
    $endgroup$
    – Gerry Myerson
    Dec 21 '18 at 19:41










  • $begingroup$
    @N.S. As far as I understand degree of extension is determined by Galois group of roots of $f(x)$. Quadratic $g(x)$ is not possible for $C_2 wr S_4$. On the contrary, quadratic $g(x)$ is required e.g. for another group $S_4 wr C_2$.
    $endgroup$
    – klajok
    Dec 21 '18 at 19:56










  • $begingroup$
    @GerryMyerson I tried RadiRoot package in GAP system, but $C_2 wr S_4$ group is too big for it and calculations finish with out of memory error. Maybe one day I will get Maple or Mathematica...
    $endgroup$
    – klajok
    Dec 21 '18 at 20:01


















  • $begingroup$
    If you're going to resort to a computer algebra system, there's probably one that solves solvable octics quickly. Maybe Maple, maybe Mathematica.
    $endgroup$
    – Gerry Myerson
    Dec 21 '18 at 19:41










  • $begingroup$
    @N.S. As far as I understand degree of extension is determined by Galois group of roots of $f(x)$. Quadratic $g(x)$ is not possible for $C_2 wr S_4$. On the contrary, quadratic $g(x)$ is required e.g. for another group $S_4 wr C_2$.
    $endgroup$
    – klajok
    Dec 21 '18 at 19:56










  • $begingroup$
    @GerryMyerson I tried RadiRoot package in GAP system, but $C_2 wr S_4$ group is too big for it and calculations finish with out of memory error. Maybe one day I will get Maple or Mathematica...
    $endgroup$
    – klajok
    Dec 21 '18 at 20:01
















$begingroup$
If you're going to resort to a computer algebra system, there's probably one that solves solvable octics quickly. Maybe Maple, maybe Mathematica.
$endgroup$
– Gerry Myerson
Dec 21 '18 at 19:41




$begingroup$
If you're going to resort to a computer algebra system, there's probably one that solves solvable octics quickly. Maybe Maple, maybe Mathematica.
$endgroup$
– Gerry Myerson
Dec 21 '18 at 19:41












$begingroup$
@N.S. As far as I understand degree of extension is determined by Galois group of roots of $f(x)$. Quadratic $g(x)$ is not possible for $C_2 wr S_4$. On the contrary, quadratic $g(x)$ is required e.g. for another group $S_4 wr C_2$.
$endgroup$
– klajok
Dec 21 '18 at 19:56




$begingroup$
@N.S. As far as I understand degree of extension is determined by Galois group of roots of $f(x)$. Quadratic $g(x)$ is not possible for $C_2 wr S_4$. On the contrary, quadratic $g(x)$ is required e.g. for another group $S_4 wr C_2$.
$endgroup$
– klajok
Dec 21 '18 at 19:56












$begingroup$
@GerryMyerson I tried RadiRoot package in GAP system, but $C_2 wr S_4$ group is too big for it and calculations finish with out of memory error. Maybe one day I will get Maple or Mathematica...
$endgroup$
– klajok
Dec 21 '18 at 20:01




$begingroup$
@GerryMyerson I tried RadiRoot package in GAP system, but $C_2 wr S_4$ group is too big for it and calculations finish with out of memory error. Maybe one day I will get Maple or Mathematica...
$endgroup$
– klajok
Dec 21 '18 at 20:01










1 Answer
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$begingroup$

If I understand your question correctly, the abstract translation of your observation is that the Galois group is imprimitive, that is that the field of degree 8 defined by $f$ has a subfield of degree $2$ (over which it has degree $2$), that is over this subfield $f$ will have a quadratic factor.



Such subfields correspond to polynomial decompositions, that is pairs of polynomials $g,h$ such that $f(x)$ divides $g(h(x))$. The polynomial $g$ defines the subfield you are seeking, the polynomial $h$ expresses a root of $g$ in terms of a root of $f$.



In GAP, you can find such decompositions using the function DecomPoly. It takes a polynomial $f$ and returns a list of all possible decompositions (up to equality of the intermediate fields.



In your example, we indeed get the degree 2 factor you were seeking (and this is much faster than the hours you quote):



gap> DecomPoly(f);
[ [ x^4+273770356608*x^3+4408835773957146427392*x^2- 7845555523405311768791676626141184*x-986470018747508750165370725578689408242024448,
3663*x^7-22492*x^6-364752*x^5-5179288*x^4-28801128*x^3+150835968*x^2+121269024*x-49235223744 ] ]
gap> g:=last[1][1];
x^4+273770356608*x^3+4408835773957146427392*x^2- 7845555523405311768791676626141184*x-986470018747508750165370725578689408242024448
gap> e:=AlgebraicExtension(Rationals,g);
<algebraic extension over the Rationals of degree 4>
gap> fe:=Value(f,X(e)); # write f over e
x_1^8+!5992704*x_1+(!-304129728)
gap> Factors(fe);
[ x_1^2+(-1/514753642938357559490856611217408*a^3-1/1141573548613903985664*a^2+1/233510010048*a+18)*x_1+1/1342011552*a,
x_1^6+(1/514753642938357559490856611217408*a^3+1/1141573548613903985664*a^2-1/233510010048*a-18)*x_1^5+(-1/12256039117579941892639443124224*a^3-1/104457710330684024832*a^2+47/19459167504*a+408)*x_1^4+(5/14298712303843265541412683644928*a^3-1/26114427582671006208*a^2-9/926627024*a+1224)*x_1^3+(1/420550361877743104159196577792*a^3+1/932658127952535936*a^2-15/95388076*a-30600)*x_1^2+(-1/14501736616473900143420571648*a^3-1/112562187856340544*a^2+9/6578488*a+105264)*x_1+(-1/2416956102745650023903428608*a^3-1/8828406890693376*a^2-3/1644622*a+3246048) ]


The algorithm used by GAP is described (disclaimer: self-promotion) in



Hulpke, Alexander
Block systems of a Galois group, Experiment. Math. 4 (1995), no. 1, 1–9.



but there are by now better algorithms with



van Hoeij, Mark; Klüners, Jürgen; Novocin, Andrew
Generating subfields,
J. Symbolic Comput. 52 (2013), 17–34.



as far as I know the state of the art.






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    $begingroup$

    If I understand your question correctly, the abstract translation of your observation is that the Galois group is imprimitive, that is that the field of degree 8 defined by $f$ has a subfield of degree $2$ (over which it has degree $2$), that is over this subfield $f$ will have a quadratic factor.



    Such subfields correspond to polynomial decompositions, that is pairs of polynomials $g,h$ such that $f(x)$ divides $g(h(x))$. The polynomial $g$ defines the subfield you are seeking, the polynomial $h$ expresses a root of $g$ in terms of a root of $f$.



    In GAP, you can find such decompositions using the function DecomPoly. It takes a polynomial $f$ and returns a list of all possible decompositions (up to equality of the intermediate fields.



    In your example, we indeed get the degree 2 factor you were seeking (and this is much faster than the hours you quote):



    gap> DecomPoly(f);
    [ [ x^4+273770356608*x^3+4408835773957146427392*x^2- 7845555523405311768791676626141184*x-986470018747508750165370725578689408242024448,
    3663*x^7-22492*x^6-364752*x^5-5179288*x^4-28801128*x^3+150835968*x^2+121269024*x-49235223744 ] ]
    gap> g:=last[1][1];
    x^4+273770356608*x^3+4408835773957146427392*x^2- 7845555523405311768791676626141184*x-986470018747508750165370725578689408242024448
    gap> e:=AlgebraicExtension(Rationals,g);
    <algebraic extension over the Rationals of degree 4>
    gap> fe:=Value(f,X(e)); # write f over e
    x_1^8+!5992704*x_1+(!-304129728)
    gap> Factors(fe);
    [ x_1^2+(-1/514753642938357559490856611217408*a^3-1/1141573548613903985664*a^2+1/233510010048*a+18)*x_1+1/1342011552*a,
    x_1^6+(1/514753642938357559490856611217408*a^3+1/1141573548613903985664*a^2-1/233510010048*a-18)*x_1^5+(-1/12256039117579941892639443124224*a^3-1/104457710330684024832*a^2+47/19459167504*a+408)*x_1^4+(5/14298712303843265541412683644928*a^3-1/26114427582671006208*a^2-9/926627024*a+1224)*x_1^3+(1/420550361877743104159196577792*a^3+1/932658127952535936*a^2-15/95388076*a-30600)*x_1^2+(-1/14501736616473900143420571648*a^3-1/112562187856340544*a^2+9/6578488*a+105264)*x_1+(-1/2416956102745650023903428608*a^3-1/8828406890693376*a^2-3/1644622*a+3246048) ]


    The algorithm used by GAP is described (disclaimer: self-promotion) in



    Hulpke, Alexander
    Block systems of a Galois group, Experiment. Math. 4 (1995), no. 1, 1–9.



    but there are by now better algorithms with



    van Hoeij, Mark; Klüners, Jürgen; Novocin, Andrew
    Generating subfields,
    J. Symbolic Comput. 52 (2013), 17–34.



    as far as I know the state of the art.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      If I understand your question correctly, the abstract translation of your observation is that the Galois group is imprimitive, that is that the field of degree 8 defined by $f$ has a subfield of degree $2$ (over which it has degree $2$), that is over this subfield $f$ will have a quadratic factor.



      Such subfields correspond to polynomial decompositions, that is pairs of polynomials $g,h$ such that $f(x)$ divides $g(h(x))$. The polynomial $g$ defines the subfield you are seeking, the polynomial $h$ expresses a root of $g$ in terms of a root of $f$.



      In GAP, you can find such decompositions using the function DecomPoly. It takes a polynomial $f$ and returns a list of all possible decompositions (up to equality of the intermediate fields.



      In your example, we indeed get the degree 2 factor you were seeking (and this is much faster than the hours you quote):



      gap> DecomPoly(f);
      [ [ x^4+273770356608*x^3+4408835773957146427392*x^2- 7845555523405311768791676626141184*x-986470018747508750165370725578689408242024448,
      3663*x^7-22492*x^6-364752*x^5-5179288*x^4-28801128*x^3+150835968*x^2+121269024*x-49235223744 ] ]
      gap> g:=last[1][1];
      x^4+273770356608*x^3+4408835773957146427392*x^2- 7845555523405311768791676626141184*x-986470018747508750165370725578689408242024448
      gap> e:=AlgebraicExtension(Rationals,g);
      <algebraic extension over the Rationals of degree 4>
      gap> fe:=Value(f,X(e)); # write f over e
      x_1^8+!5992704*x_1+(!-304129728)
      gap> Factors(fe);
      [ x_1^2+(-1/514753642938357559490856611217408*a^3-1/1141573548613903985664*a^2+1/233510010048*a+18)*x_1+1/1342011552*a,
      x_1^6+(1/514753642938357559490856611217408*a^3+1/1141573548613903985664*a^2-1/233510010048*a-18)*x_1^5+(-1/12256039117579941892639443124224*a^3-1/104457710330684024832*a^2+47/19459167504*a+408)*x_1^4+(5/14298712303843265541412683644928*a^3-1/26114427582671006208*a^2-9/926627024*a+1224)*x_1^3+(1/420550361877743104159196577792*a^3+1/932658127952535936*a^2-15/95388076*a-30600)*x_1^2+(-1/14501736616473900143420571648*a^3-1/112562187856340544*a^2+9/6578488*a+105264)*x_1+(-1/2416956102745650023903428608*a^3-1/8828406890693376*a^2-3/1644622*a+3246048) ]


      The algorithm used by GAP is described (disclaimer: self-promotion) in



      Hulpke, Alexander
      Block systems of a Galois group, Experiment. Math. 4 (1995), no. 1, 1–9.



      but there are by now better algorithms with



      van Hoeij, Mark; Klüners, Jürgen; Novocin, Andrew
      Generating subfields,
      J. Symbolic Comput. 52 (2013), 17–34.



      as far as I know the state of the art.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        If I understand your question correctly, the abstract translation of your observation is that the Galois group is imprimitive, that is that the field of degree 8 defined by $f$ has a subfield of degree $2$ (over which it has degree $2$), that is over this subfield $f$ will have a quadratic factor.



        Such subfields correspond to polynomial decompositions, that is pairs of polynomials $g,h$ such that $f(x)$ divides $g(h(x))$. The polynomial $g$ defines the subfield you are seeking, the polynomial $h$ expresses a root of $g$ in terms of a root of $f$.



        In GAP, you can find such decompositions using the function DecomPoly. It takes a polynomial $f$ and returns a list of all possible decompositions (up to equality of the intermediate fields.



        In your example, we indeed get the degree 2 factor you were seeking (and this is much faster than the hours you quote):



        gap> DecomPoly(f);
        [ [ x^4+273770356608*x^3+4408835773957146427392*x^2- 7845555523405311768791676626141184*x-986470018747508750165370725578689408242024448,
        3663*x^7-22492*x^6-364752*x^5-5179288*x^4-28801128*x^3+150835968*x^2+121269024*x-49235223744 ] ]
        gap> g:=last[1][1];
        x^4+273770356608*x^3+4408835773957146427392*x^2- 7845555523405311768791676626141184*x-986470018747508750165370725578689408242024448
        gap> e:=AlgebraicExtension(Rationals,g);
        <algebraic extension over the Rationals of degree 4>
        gap> fe:=Value(f,X(e)); # write f over e
        x_1^8+!5992704*x_1+(!-304129728)
        gap> Factors(fe);
        [ x_1^2+(-1/514753642938357559490856611217408*a^3-1/1141573548613903985664*a^2+1/233510010048*a+18)*x_1+1/1342011552*a,
        x_1^6+(1/514753642938357559490856611217408*a^3+1/1141573548613903985664*a^2-1/233510010048*a-18)*x_1^5+(-1/12256039117579941892639443124224*a^3-1/104457710330684024832*a^2+47/19459167504*a+408)*x_1^4+(5/14298712303843265541412683644928*a^3-1/26114427582671006208*a^2-9/926627024*a+1224)*x_1^3+(1/420550361877743104159196577792*a^3+1/932658127952535936*a^2-15/95388076*a-30600)*x_1^2+(-1/14501736616473900143420571648*a^3-1/112562187856340544*a^2+9/6578488*a+105264)*x_1+(-1/2416956102745650023903428608*a^3-1/8828406890693376*a^2-3/1644622*a+3246048) ]


        The algorithm used by GAP is described (disclaimer: self-promotion) in



        Hulpke, Alexander
        Block systems of a Galois group, Experiment. Math. 4 (1995), no. 1, 1–9.



        but there are by now better algorithms with



        van Hoeij, Mark; Klüners, Jürgen; Novocin, Andrew
        Generating subfields,
        J. Symbolic Comput. 52 (2013), 17–34.



        as far as I know the state of the art.






        share|cite|improve this answer









        $endgroup$



        If I understand your question correctly, the abstract translation of your observation is that the Galois group is imprimitive, that is that the field of degree 8 defined by $f$ has a subfield of degree $2$ (over which it has degree $2$), that is over this subfield $f$ will have a quadratic factor.



        Such subfields correspond to polynomial decompositions, that is pairs of polynomials $g,h$ such that $f(x)$ divides $g(h(x))$. The polynomial $g$ defines the subfield you are seeking, the polynomial $h$ expresses a root of $g$ in terms of a root of $f$.



        In GAP, you can find such decompositions using the function DecomPoly. It takes a polynomial $f$ and returns a list of all possible decompositions (up to equality of the intermediate fields.



        In your example, we indeed get the degree 2 factor you were seeking (and this is much faster than the hours you quote):



        gap> DecomPoly(f);
        [ [ x^4+273770356608*x^3+4408835773957146427392*x^2- 7845555523405311768791676626141184*x-986470018747508750165370725578689408242024448,
        3663*x^7-22492*x^6-364752*x^5-5179288*x^4-28801128*x^3+150835968*x^2+121269024*x-49235223744 ] ]
        gap> g:=last[1][1];
        x^4+273770356608*x^3+4408835773957146427392*x^2- 7845555523405311768791676626141184*x-986470018747508750165370725578689408242024448
        gap> e:=AlgebraicExtension(Rationals,g);
        <algebraic extension over the Rationals of degree 4>
        gap> fe:=Value(f,X(e)); # write f over e
        x_1^8+!5992704*x_1+(!-304129728)
        gap> Factors(fe);
        [ x_1^2+(-1/514753642938357559490856611217408*a^3-1/1141573548613903985664*a^2+1/233510010048*a+18)*x_1+1/1342011552*a,
        x_1^6+(1/514753642938357559490856611217408*a^3+1/1141573548613903985664*a^2-1/233510010048*a-18)*x_1^5+(-1/12256039117579941892639443124224*a^3-1/104457710330684024832*a^2+47/19459167504*a+408)*x_1^4+(5/14298712303843265541412683644928*a^3-1/26114427582671006208*a^2-9/926627024*a+1224)*x_1^3+(1/420550361877743104159196577792*a^3+1/932658127952535936*a^2-15/95388076*a-30600)*x_1^2+(-1/14501736616473900143420571648*a^3-1/112562187856340544*a^2+9/6578488*a+105264)*x_1+(-1/2416956102745650023903428608*a^3-1/8828406890693376*a^2-3/1644622*a+3246048) ]


        The algorithm used by GAP is described (disclaimer: self-promotion) in



        Hulpke, Alexander
        Block systems of a Galois group, Experiment. Math. 4 (1995), no. 1, 1–9.



        but there are by now better algorithms with



        van Hoeij, Mark; Klüners, Jürgen; Novocin, Andrew
        Generating subfields,
        J. Symbolic Comput. 52 (2013), 17–34.



        as far as I know the state of the art.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 23 '18 at 15:48









        ahulpkeahulpke

        7,2421026




        7,2421026






























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