# 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.

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

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

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Nice! And I also learned about FreeQ, which I really should have learned sooner. (I was using Exponent[expr, form]!=0 , which was ugly and no doubt buggy, for this task.) –  David Speyer Aug 14 '12 at 20:13
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