Better use of Mathematica's PolynomialReduce[]?

Question

I've been using

scPhiDecomp[expr_]:= PolynomialReduce[expr, {x^2-y^2,2 x y}, {x,y}]

which works great on

scPhiDecomp[a x^2 + 2c x y -a y^2]
>> {{a,c},0}

but doesn't do what I want on

scPhiDecomp[y (a x^2 + 2c x y -a y^2)]
>> {{0,(a x)/2+c y},-a y^3}

How do I use PolynomialReduce[] to produce {{a y, c y}, 0} on the second expression, while still producing the desired result on the 1st?

Thanks, Keith

Answer 1

One way I learned of solving this problem (response from my post on MathGroup) is to use Gröbner bases:

mb = {2*x*y, x^2 - y^2}; (* my basis *)

gb = GroebnerBasis[mb, {x,y}] (* Groebner basis *)
Out = {y^3, x*y, x^2 - y^2}

convGb = {{x/2, -y}, {1/2, 0}, {0, 1}};

bPoly = c*x^2*y + 2*a*x*y^2 - c*y^3;

{bPolyGb, bPolyGbRes} = PolynomialReduce[bPoly, gb, {x,y}]
Out = {{-c, c*x + 2*a*y, 0}, 0}

bPolyGb.convGb // Simplify
Out = {a*y, c*y}

In this simple case, you can derive convGb by solving $\mathrm{gb} = c \cdot \mathrm{mb}$ by hand (where $c$ is convGb).

For the full story on using Gröbner bases you can start at this post.

The central results therein are the functions moduleGroebnerBasis[] and conversionMatrix[].

Answer 2

That comment was probably too cryptic, so:

One way would be to let x -> Cos[t/2] and y -> Sin[t/2]; we then have x^2 - y^2 == Cos[t] and 2 x y == Sin[t].

So, something like the following (broken up into stages to make it easier to follow):

expr = y (a x^2 + 2c x y -a y^2);
f1 = Simplify[expr /. Thread[{x, y} -> Through[{Cos, Sin}[t/2]]]];
f2 = TrigExpand[PolynomialReduce[f1, {Cos[t], Sin[t]}, {Cos[t], Sin[t]}]];
f3 = f2 /. Thread[Through[{Cos, Sin}[t]] -> {Cos[t/2]^2 - Sin[t/2]^2, 2 Sin[t/2] Cos[t/2]}]; (*unneeded here, but needed if f2 ends up containing expressions like Cos[t]*)
f3 /. Thread[Through[{Cos, Sin}[t/2]] -> {x, y}]

should do the trick.