On Monday, 16 February 2015 12:01:31 UTC+1, Jeroen Demeyer wrote:
>
> On 2015-02-14 16:40, Bill Hart wrote:
> > Wikipedia claims that it is possible to compute gcd in R[x] for any
> > unique factorisation domain R. But in thinking about it last night, I
> > couldn't see why the algorithm doesn
On 2015-02-14 16:40, Bill Hart wrote:
Wikipedia claims that it is possible to compute gcd in R[x] for any
unique factorisation domain R. But in thinking about it last night, I
couldn't see why the algorithm doesn't work in any GCD domain R.
Perhaps it works for GCD domains, but the proofs are ha
Slightly OT post, more about polynomial GCD than the original question:
The flint fmpz_poly_xgcd function is badly named. It should be
fmpz_poly_bezout. We might even change it in a later release.
The point is, the result of the Euclidean algorithm in the case of Z[x] is
not the gcd, but the re
Le 12/02/2015 16:06, kcrisman a écrit :
As a late postscript, is #8111 related to any of this discussion? I
know this was mostly about Bezout/xgcd stuff, but that ticket looks
like it has languished and could be connected.
This is clearly the same kind of discussions. The point in #8111 is that
As a late postscript, is #8111 related to any of this discussion? I know
this was mostly about Bezout/xgcd stuff, but that ticket looks like it has
languished and could be connected.
- kcrisman
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Le 26/01/2015 18:06, Vincent Delecroix a écrit :
2015-01-26 17:07 UTC+01:00, Jeroen Demeyer :
On 2015-01-26 14:22, Bruno Grenet wrote:
In the special case of univariate polynomials over ZZ, I think there are
two possibilities for xgcd:
- either xgcd(p,q) = (g,u,v) where g = gcd(p,q) = up+vq w
2015-01-26 17:07 UTC+01:00, Jeroen Demeyer :
> On 2015-01-26 14:22, Bruno Grenet wrote:
>> In the special case of univariate polynomials over ZZ, I think there are
>> two possibilities for xgcd:
>>
>> - either xgcd(p,q) = (g,u,v) where g = gcd(p,q) = up+vq with u,v in
>> QQ[x];
>> - or xgcd(p,q) =
On 2015-01-26 14:22, Bruno Grenet wrote:
In the special case of univariate polynomials over ZZ, I think there are
two possibilities for xgcd:
- either xgcd(p,q) = (g,u,v) where g = gcd(p,q) = up+vq with u,v in QQ[x];
- or xgcd(p,q) = (g×r, u, v) where g = gcd(p,q), r = res(p,q), g×r =
up+vq with
Le 26/01/2015 14:41, Vincent Delecroix a écrit :
Hello,
2015-01-26 14:22 UTC+01:00, Bruno Grenet :
In the special case of univariate polynomials over ZZ, I think there are
two possibilities for xgcd:
- either xgcd(p,q) = (g,u,v) where g = gcd(p,q) = up+vq with u,v in QQ[x];
- or xgcd(p,q) = (
Hello,
2015-01-26 14:22 UTC+01:00, Bruno Grenet :
> In the special case of univariate polynomials over ZZ, I think there are
> two possibilities for xgcd:
>
> - either xgcd(p,q) = (g,u,v) where g = gcd(p,q) = up+vq with u,v in QQ[x];
> - or xgcd(p,q) = (g×r, u, v) where g = gcd(p,q), r = res(p,q),
Salut !
I disagree with John and Jeroen that xgcd should not be defined or
should return an error for non-PID UFDs. For now, I checked the
possibilities for the rings of polynomials over a UFD.
In the special case of univariate polynomials over ZZ, I think there are
two possibilities for xgc
Anyways, we should take into account that, even when both gcd and xgcd are
well defined, their definition coincide and everything is happy... there
are still defined only up to product of a unit. So there is yet another
possible source of inconsistency.
El domingo, 25 de enero de 2015, 21:05:00
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