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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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