*To*: mathgroup@smc.vnet.net*Subject*: [mg11936] Q--PolynomialReduce*From*: Ersek_Ted%PAX1A@mr.nawcad.navy.mil*Date*: Thu, 9 Apr 1998 00:33:42 -0400

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}

**Follow-Ups**:**Re: Q--PolynomialReduce***From:*Daniel Lichtblau <danl@wolfram.com>