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MathGroup Archive 1998

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Q--PolynomialReduce




Hello,
This is Ted Ersek

I was going over a solution Daniel Lichtblau posted to the Mathgroup a
few  weeks ago.
The lines below are very similar to the solution he gave.

A user may want to let (t=x*u+y*v) and express ( (x*u+y*v)u+(-x*u-y*v)v
)  in terms of (t).  This would give (t*u-t*v).

Daniel pointed out that the last part of what comes from
PolynomialReduce  gives the desired result (when given the right
arguments).

Any of the lines In[1] through In[5] below seem to work.  I was then 
wondering how the third argument of PolynomialReduce effects the
results. 
 For In[6] I let the third argument be something unrelated to the
problem,  and I got a very different result.  So it seems the third
argument is used  in some way, but how?

I had a course in abstract algebra as an undergraduate.  In this course
I  studied Groups, Rings, and Fields.  Also Daniel recently faxed me
his  article on Grobner Bases in Mathematica 3.0.   This limited
background may  help me understand PolynomialReduce, but I still have
very little knowledge  of polynomial algebra.

In[1]:=
PolynomialReduce[(x*u+y*v)u+(-x*u-y*v)v,{x*u+y*v-t},{x}]

Out[1]=
{{u-v},t u-t v}


In[2]:=
PolynomialReduce[(x*u+y*v)u+(-x*u-y*v)v,{x*u+y*v-t},{y}]

Out[2]=
{{u-v},t u-t v}


In[3]:=
PolynomialReduce[(x*u+y*v)u+(-x*u-y*v)v,{x*u+y*v-t},{u}]

Out[3]=
{{u-v},t u-t v}


In[4]:=
PolynomialReduce[(x*u+y*v)u+(-x*u-y*v)v,{x*u+y*v-t},{v}]

Out[4]=
{{u-v},t u-t v}


In[5]:=
PolynomialReduce[(x*u+y*v)u+(-x*u-y*v)v,{x*u+y*v-t},{x,y,u,v}]

Out[5]=
{{u-v},t u-t v}


In[6]:=
PolynomialReduce[(x*u+y*v)u+(-x*u-y*v)v,{x*u+y*v-t},{a}]

Out[6]=
{{0}, u^2*x - u*v*x + u*v*y - v^2*y}




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