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Some of my colleges are interested in finding common zeros of the following four polynomials:

{-1 + 12*t^4 + 24*t^2*w^2 + 12*w^4 + 4*x^4 + 12*y^4 + 24*y^2*z^2 + 12*z^4,

-1 + 8*t^2 - 16*t^4 + 8*w^2 - 32*t^2*w^2 - 16*w^4 + 16*t^2*w*x + 16*w^3*x + 16*w*y + 64*w^2*x*y + 8*y^2 - 24*t^2*y^2 + 40*w^2*y^2 + 64*w*x*y^2 + 16*x*y^3 - 16*y^4 - 16*t*z + 96*t^3*z + 96*t*w^2*z + 64*t*w*x*z + 128*t*w*y*z + 64*t*x*y*z + 96*t*y^2*z + 8*z^2 + 40*t^2*z^2 - 24*w^2*z^2 + 16*x*y*z^2 - 32*y^2*z^2 + 96*t*z^3 - 16*z^4,

-2 + 3*t^2 + 3*w^2 + x^2 + 3*y^2 + 3*z^2,

-(t^3*x) - t*w^2*x + t*y - 2*t^3*y - 2*t*w^2*y - 4*t*w*x*y + 4*t*x*y^2 - 4*t*y^3 + w*z - 4*t^2*w*z - 4*w^3*z + 4*w^2*x*z - 4*w*x*y*z - 2*w*y^2*z - x*y^2*z - 4*t*y*z^2 - 2*w*z^3 - x*z^3}

which are of degree $4$, in the five variables $x$, $y$, $z$, $w$, and $t$.

I have tried GroebnerBasis, (also in Macaulay2), to simplify this a bit, before feeding it to Reduce, but no luck, it takes all the memory of my computer and never finishes. I am looking for a simpler way to represent the ideal generated by the four polynomials above. So, if anyone have some special trick to attack very specific systems like this, I would appreciate it.

EDIT: I am not so familiar with the details, but this is a simplification of a system of the form $p1=p2=p4=p4=1/4$ for four homogeneous degree-4 polynomials, so this I think could help.

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Could you please add some context to the problem? The intention is that the people trying to help you doesn't feel you just dropped some random integers and powers :) –  belisarius Nov 24 '13 at 23:16

2 Answers 2

I know this isn't what you want, but perhaps it can be twisted to approach it:

ns = NSolve[And @@ Thread[Equal[1.0 list, 0.]], {x, y, z, w, t}, Complexes]

Retuns:

NSolve::infsolns: Infinite solution set has dimension at least 1. Returning intersection of solutions with (153196 t)/195501+(185938 w)/195501+(38650 x)/65167-(41688 y)/65167-(153968 z)/195501 == 1. >>

{{x -> -1.0224 - 0.00777854 I, y -> 1.37461 + 1.39918 I, z -> -1.4993 + 1.41945 I, w -> 1.84474 + 0.660893 I, t -> -0.573711 + 1.77259 I}, {x -> -1.0224 + 0.00777854 I, y -> 1.37461 - 1.39918 I, z -> -1.4993 - 1.41945 I, w -> 1.84474 - 0.660893 I, t -> -0.573711 - 1.77259 I}, {x -> 2.2041 + 1.21154 I, y -> 0.415665 + 0.834055 I, z -> -0.0729957 + 1.43551 I, w -> 1.05233 + 0.203707 I, t -> -1.40334 + 0.95941 I}, {x -> 2.2041 - 1.21154 I, y -> 0.415665 - 0.834055 I, z -> -0.0729957 - 1.43551 I, w -> 1.05233 - 0.203707 I, t -> -1.40334 - 0.95941 I}, .....

ListPlot[Transpose[{x, y, z, w, t} /. ns /. Complex[a_, b_] -> {a, b}], PlotRange -> All]

enter image description here

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Not sure if it helps but since you have five variables and four equations you might eliminate three variables and get a bivariate.

Timing[
 gbelim = GroebnerBasis[polys, {x, y}, {z, t, w}, 
    MonomialOrder -> EliminationOrder];]

(* t[41]= {359.164000, Null} *)

The result is not all that large actually.

gbelim

(* Out[46]= {6115962008563732 + 83974195238436880 x^2 - 
  1855144677799197876 x^4 - 5333259112295668568 x^6 + 
  272688937685506335637 x^8 - 2495862660863566832976 x^10 + 
  13159656903363735414738 x^12 - 48437638593030585155328 x^14 + 
  135066040808147089369164 x^16 - 298285593409584372532096 x^18 + 
  536105262590487154429184 x^20 - 798003574719003785780736 x^22 + 
  995374445402970297750272 x^24 - 1048653474833206115627008 x^26 + 
  938173467457848425508864 x^28 - 715085620186802468683776 x^30 + 
  464675329662790939705344 x^32 - 256368287742756290297856 x^34 + 
  118502904257204262535168 x^36 - 44544791348207069167616 x^38 + 
  12847710284044409241600 x^40 - 2538510800076023005184 x^42 + 
  262167606402979201024 x^44 + 273822354127843872 x y + 
  1728012212939615616 x^3 y - 49487524513343872704 x^5 y - 
  22872117271385872128 x^7 y + 4512723568057797067512 x^9 y - 
  39589496603845140541824 x^11 y + 193728389315612745670848 x^13 y - 
  656635049872198551220464 x^15 y + 
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  98453512446046126120894464 x^22 y^18 + 
  32114981483400004934565888 x^24 y^18 - 
  3080639140597793342619648 x^26 y^18 + 
  531614718952716550275072 x^5 y^19 + 
  9180577261914220425904128 x^7 y^19 - 
  84015572314488934810779648 x^9 y^19 + 
  289873148868948558971142144 x^11 y^19 - 
  556498377143695628801409024 x^13 y^19 + 
  655174247669303708479389696 x^15 y^19 - 
  461257555571320486832898048 x^17 y^19 + 
  147604871389217722170605568 x^19 y^19 + 
  31140354498653357925728256 x^21 y^19 - 
  41874882477506288267821056 x^23 y^19 + 
  10141573100021054189862912 x^25 y^19 + 
  12758753254865197206601728 x^8 y^20 - 
  92255600458256041340043264 x^10 y^20 + 
  301302865326431964802056192 x^12 y^20 - 
  584939764607665964241125376 x^14 y^20 + 
  740989131340247991614177280 x^16 y^20 - 
  626160352046461216754761728 x^18 y^20 + 
  342523452765227217315692544 x^20 y^20 - 
  109921566503454006703030272 x^22 y^20 + 
  15703080929064858100432896 x^24 y^20} *)

--- edit ---

To determine the lowest order polynomial containing the solution set, one can compute a degree-based Groebner basis and then check the first member. I do so below. I will have the heuristic variable sorting because we don't care about variable ordering in this case, and it makes the computation go significantly faster.

Timing[gbdrlsort =                                                      
           GroebnerBasis[polys, {x, y, z, t, w},                                
            MonomialOrder -> DegreeReverseLexicographic, Sort -> True];]        

(* Out[2]= {2.855566, Null} *)

gbdrlsort[[1]] // InputForm                                             

(* Out[4]//InputForm= -2 + 3*t^2 + 3*w^2 + x^2 + 3*y^2 + 3*z^2 *)

So the solution set lies on an ellipsoid centered at the origin and aligned with the axes. But we knew that since it was one of the original equations. In any case, it does not lie on a plane.

--- end edit ---

share|improve this answer
    
Ah, this is nice! –  Paxinum Nov 26 '13 at 9:29
    
Is there some meaning to the variables that might distinguish one as a parameter, say, rather than being a variable on equal footing to the others? That is to say, is the interest in finding a parametrized solution set over C(t) or C(x) or some such? –  Daniel Lichtblau Nov 26 '13 at 15:17
    
I do not think so; the person I got the problem from did not state any such think, but he thought that one can find A,B,C,D,E,F such that Ax+By+Cz+Dw+Et=F, for the solutions, but this does not feel right. –  Paxinum Nov 27 '13 at 9:23
    
Straightforward to check that. Use a degree-based term order in GroebnerBasis. The hypothesized relation will exist iff the first polynomial is linear. See edit. –  Daniel Lichtblau Nov 27 '13 at 17:06

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