Re: A question about algebraic numbers using Mathematica
- To: mathgroup at smc.vnet.net
- Subject: [mg62795] Re: [mg62762] A question about algebraic numbers using Mathematica
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Mon, 5 Dec 2005 13:40:58 -0500 (EST)
- References: <200512050837.DAA08323@smc.vnet.net> <DA57D525-3C39-4D6F-95B4-D74C1973B5FF@mimuw.edu.pl>
- Sender: owner-wri-mathgroup at wolfram.com
On 5 Dec 2005, at 23:24, Andrzej Kozlowski wrote: > > On 5 Dec 2005, at 17:37, Kent Holing wrote: > >> I want to find the inverse of 2 - r in Q[r] where r is a root of >> the equation >> x^4 - 2c x^3 + (c^2 - 2a^2) x^2 + 2a^2 c x - a^2 c^2 = 0 for a, b >> and c integers. >> >> Can this be done for general a, b and c? (I know how to do it for >> specific given numerical values of a, b and c.) >> >> Kent Holing >> > > > I think this can be done using Groebner basis. Suppose r satisfies > the above equation and let v be 2-r. Let us first find the > algebraic equation satisfied by v. > > In[1]:= > poly1 = First[GroebnerBasis[{r^4 - 2*c*r^3 + (c^2 - 2*a^2)*r^2 + > 2*a^2*c*r - a^2*c^2, v - (2 - r)}, {a, b, c, v}, > {r}]] > > Out[1]= > v^4 + 2*c*v^3 - 8*v^3 - 2*a^2*v^2 + c^2*v^2 - 12*c*v^2 + 24*v^2 + > 8*a^2*v - 4*c^2*v - 2*a^2*c*v + 24*c*v - 32*v - > 8*a^2 - a^2*c^2 + 4*c^2 + 4*a^2*c - 16*c + 16 > > From this we can easily see what the inverse of v is as a > polynomial in v with rational coefficients, and by substituting 2-r > for v we can also get a polynomial in r. We proceed as follows. > First we find the free term (not involving v) in poly1: > > In[2]:= > free = poly1 /. v -> 0 > > Out[2]= > (-c^2)*a^2 + 4*c*a^2 - 8*a^2 + 4*c^2 - 16*c + 16 > > Next we subtract the free term from poly1 and divide by v: > > In[3]:= > poly2 = Cancel[(poly1 - free)/v] > > Out[3]= > v^3 + 2*c*v^2 - 8*v^2 - 2*a^2*v + c^2*v - 12*c*v + 24*v + 8*a^2 - > 4*c^2 - 2*a^2*c + 24*c - 32 > > Now substitute 2-r for v: > > In[4]:= > q = Expand[poly2 /. v -> 2 - r] > > Out[4]= > -r^3 + 2*c*r^2 - 2*r^2 + 2*a^2*r - c^2*r + 4*c*r - 4*r + 4*a^2 - > 2*c^2 - 2*a^2*c + 8*c - 8 > > The inverse ought to be the polynomial poly3 given by: > > In[5]:= > poly3 = Expand[(q/free)*(r - 2)] > > Out[5]= > -(r^4/((-c^2)*a^2 + 4*c*a^2 - 8*a^2 + 4*c^2 - 16*c + 16)) + > (2*c*r^3)/((-c^2)*a^2 + 4*c*a^2 - 8*a^2 + 4*c^2 - 16*c + 16) + > (2*a^2*r^2)/((-c^2)*a^2 + 4*c*a^2 - 8*a^2 + 4*c^2 - 16*c + 16) - > (c^2*r^2)/((-c^2)*a^2 + 4*c*a^2 - 8*a^2 + 4*c^2 - 16*c + 16) - > (2*a^2*c*r)/((-c^2)*a^2 + 4*c*a^2 - 8*a^2 + 4*c^2 - 16*c + 16) - > (8*a^2)/((-c^2)*a^2 + 4*c*a^2 - 8*a^2 + 4*c^2 - 16*c + 16) + > (4*c^2)/((-c^2)*a^2 + 4*c*a^2 - 8*a^2 + 4*c^2 - 16*c + 16) + > (4*a^2*c)/((-c^2)*a^2 + 4*c*a^2 - 8*a^2 + 4*c^2 - 16*c + 16) - > (16*c)/((-c^2)*a^2 + 4*c*a^2 - 8*a^2 + 4*c^2 - 16*c + 16) + 16/((- > c^2)*a^2 + 4*c*a^2 - 8*a^2 + 4*c^2 - 16*c + 16) > > Finally we check that this is indeed the inverse of r-2 modulo the > original polynomial p: > > In[6]:= > FullSimplify[PolynomialReduce[poly3, r^4 - 2*c*r^3 + (c^2 - 2*a^2) > *r^2 + 2*a^2*c*r - a^2*c^2, r]] > > Out[6]= > {{1/(a^2*((c - 4)*c + 8) - 4*(c - 2)^2)}, 1} > > We get the remainder 1, as we should. > > Andrzej Kozlowski Sorry, I got confused near the end. The inverse of (r-2) is of course the polynomial inv=q/free above. The last part of the argument shows that this polynomial ( inv) is the inverse of (r-2) modulo the original polynomial in r.
- References:
- A question about algebraic numbers using Mathematica
- From: Kent Holing <KHO@statoil.com>
- A question about algebraic numbers using Mathematica