Re: nfrootsof1(bnf) vs. bnf.tu

[Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>]

On Thu, Jan 05, 2017 at 12:19:06PM +0100, Jeroen Demeyer wrote:
> Hello pari-users,
> 
> I somehow always assumed that the unit given by nfrootsof1(bnf) would be the
> same as bnf.tu
> 
> I now realize that this assumption is false. However, it does seem that
> PARI-2.9.1 has made this assumption much more false than it was in
> PARI-2.8.0.
> 
> With PARI-2.9.1 the result of nfrootsof1() seems random (it can be different
> with every call). With PARI-2.8.0, I could only find counter-examples to
> nfrootsof1(bnf) == bnf.tu in degree >= 16.
> 
> Do you consider this a bug? It would be convenient if nfrootsof1(bnf) would
> be the same as bnf.tu, but I don't know how much effort it would take to
> enforce that.

For what purpose would it be convenient ?
If you have a bnf, why not call bnf.tu, and only use nfrootsof1() when
you only have a nf ?

Calling a function nfxxx with a bnf is identical to calling it with
bnf.nf.

It seems to me there is something you do not tell us.

Cheers,
Bill.