Skip to main content

polynomials - How to express the original ideal elements in the Groebner basis?


Suppose I call


GroebnerBasis[{f1, f2, ...}, {x1,x2, ...}]

The output is a list


{g1,g2,...}


For each $g_j$, there should be an expression $g_j = \sum f_i h_{ij}$ for some polynomials $h_{ij}$. How do I make Mathematica output the $h$'s?




In case there is a better approach, I'll tell you my actual goal. I have polynomials $\{ f_1, f_2, \ldots, \}$ which generate an ideal $I$, and an element $q$ which I know to be invertible in the quotient ring $\mathbb{R}[x_1, x_2, \ldots]/I$. I want to generate an explicit polynomial representative for $q^{-1}$. My plan is to compute the Groebner basis of the ideal $\langle f_1, f_2, \ldots, q \rangle$, which will be $\{ 1 \}$, and find the expression $1 = \sum f_i h_i + pq$; the answer is then $p$. Calling PolynomialReduce[1,{f1, f2, ..., q}] does not have the intended effect because {f1, f2, ..., q} is not a Groebner basis.



Answer



Here is how I would go about finding the reciprocal as a member of that quotient ideal. I'll demonstrate with an example.


We start with polynomials that generate an ideal I in Q[x,y,z].


polys = {x^2 + x*y - z^3 + 3, x*y - y^2 + 2*z - 5, 
58 x - 4*y^2*z - 3*x*z + 7};


Here is the polyniomial we wish to invert in Q/I. We will set it equal to a new variable 'q'.


qpoly = x*y*z^2 + x^2*z^2 - 3*y^2 - 4*x + 2;

We make an explicit inversion relation using a new "reciprocal" variable.


recippoly = q*qrecip - 1;

Now form a Groebner basis, eliminating 'q' but not its reciprocal.


Timing[
gb = GroebnerBasis[
Join[polys, {q - qpoly, recippoly}], {qrecip, x, y, z}, q];]


(* Out[228]= {0.050000, Null} *)

Now filter out members that do not involve that reciprocal.


rpolys = Select[gb, ! FreeQ[#, qrecip] &];

Last we explicitly solve for the reciprocal.


qrecip /. Solve[rpolys == 0, qrecip]

(* Out[237]= \

{(26865976210184208452733700652828855812422384556246644691932501381503\
562817680 -
243502922707328415287593805560731167473204552630142461410610182847\
7051371070 z +
159758899358659297213970169948368802291120932841779568680462060819\
8342054959 z^2 +
117356001085553879314729354875822466004292087013976143694746220636\
08549556104 z^3 +
670871729437581239452745061221877350353798793983114023985403960308\
9597922596 z^4 +

742733530743532277159318817234115610566358359729018365599178560074\
149841571 z^5 -
123017424636603617320966572335457938227737349248956271019315675654\
817148075 z^6 -
120597839923722484422028319621260872787188437259504489268758105960\
490943543 z^7 -
589550060248794720654074609791788306633207078515514201007961400753\
44770617 z^8 -
959376629620096206267937411987608577860579884873406028793609896234\
3719923 z^9 -

161491469959962690666417014268063263295810695614595035192399568987\
4377976 z^10 +
281126429639293832101523727394804818571984663246643057444386184104\
014888 z^11 -
842500954350297074203419395874006533554497667367625627509322289419\
9144 z^12)/
94053848810936873074377725665691572019272504902971426379079192228628\
9247813475} *)

To respond to your original question, which may be relevant if you absolutely require expression the reciprocal in terms of the original idfeal generators, you might have a look at this post: http://forums.wolfram.com/mathgroup/archive/2011/Mar/msg00362.html



Comments

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...