       # 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 through In below seem to work.  I was then
wondering how the third argument of PolynomialReduce effects the
results.
For In 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:=
PolynomialReduce[(x*u+y*v)u+(-x*u-y*v)v,{x*u+y*v-t},{x}]

Out=
{{u-v},t u-t v}

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

Out=
{{u-v},t u-t v}

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

Out=
{{u-v},t u-t v}

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

Out=
{{u-v},t u-t v}

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

Out=
{{u-v},t u-t v}

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

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

```

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