Awesome! I'll explore that. Thanks! On Monday, July 10, 2017 at 3:38:27 PM UTC+2, John Cremona wrote: > > Sage does have a function is_norm() for number field elements so the > underlying algebraic problem should be solvable. > > Example (p=3): > > sage: Q3.<z> = CyclotomicField(3) > sage: a=2+3*z > sage: b=3+4*z > sage: x=polygen(Q3) > sage: L.<a3>=Q3.extension(x^3-a) > sage: b.is_norm(L) > False > > On 10 July 2017 at 14:23, Pierre <pierre....@gmail.com <javascript:>> > wrote: > > Hi all ! > > > > I wanted to know whether Sagemath had any support for cyclic algebras. > From > > the manual, I strongly suspect the answer is "no", but you never know. > > > > Let me be more concrete. For a prime p, let K be QQ with the p-th roots > of > > unity adjoined. For a, b in K, there is a cyclic algebra (a,b) over K > > (technically, this depends on a choice of primitive root in K); for p=2, > > this is the quaternion algebra (a, b) over QQ. > > > > I would like to be able to answer questions such as: is (a,b) trivial? > For > > p=2, Sage does this, essentially with hilbert_conductor(a,b). > > > > Also, (a,b) is trivial if and only if b is a norm from K[a^(1/p)]. > Finding > > explicitly an element from this field whose norm is b would be awesome. > When > > p=2 and so K=QQ, it's a matter of finding x, y in QQ such that x^2 - > ay^2 = > > b, and trying random values for x and y (essentially...) works fine. > Over > > more complicated fields, PARI has functions accessible through sage to > find > > points on conics. > > > > If any of the above can be facilitated by Sage for p odd, it would be > great. > > > > Thanks! > > Pierre > > > > -- > > You received this message because you are subscribed to the Google > Groups > > "sage-support" group. > > To unsubscribe from this group and stop receiving emails from it, send > an > > email to sage-support...@googlegroups.com <javascript:>. > > To post to this group, send email to sage-s...@googlegroups.com > <javascript:>. > > Visit this group at https://groups.google.com/group/sage-support. > > For more options, visit https://groups.google.com/d/optout. >
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