Re: separable elements
William Stein
> Just a quick status report. I have been thinking about the problem of
> generating a separating element of a function field. In principal I know
> how to find one if I can take pth roots. However, there's the additional
> problem that not only do I need to find an element x in my function field
> F=k(C) so that F/k(x) is finite and separable, but I need to explicitly
> construct (a copy of) F as an extension of k(x). Is anyone there working on
> pth roots or factorization over F?
In some cases, if R is a domain such that factorization in R[x] is
possible, and S is a finite algebraic extension of R, then factoring
over S[x] is also possible, as long as we have GCD over S[x]. This
works by taking norms, factoring over R[x], then taking gcd's. More
precisely, we factor f(x) in S[x] by:
(1) Compute g(x)=Norm(f(x)) in R[x]
(2) Factor g(x) as a product of prime powers p^e
(3) Recover the factorization of f(x) from the gcd(f, p).
This isn't implemented as much as it should be in Sage. Jen B. and I
used this "by hand" the other day to factor over extensions of Q_p.
> A second problem that arises with inseparable extensions is that we need to
> convert divisors between two representations of F. The best way to
> represent a divisor for a finite extension F/k(t) is to look at the integral
> closure of k[t] and k[1/t] in F and to consider fractional ideals for the
> respective rings. (The ideals must give the same ideal over the integral
> closure of k[t,1/t] in F.) The problem is that when we switch to a
> different representation F/k(x) of F, we also switch the affine cover of C
> (the curve of F) and so it's not trivial to convert the divisor. Any
> thoughts?
Not yet.
> I'll try to return to Hess's algorithm using the assumption that F is
> separable over its base rational function field k(t), i.e. we can take x=t.
> The algorithm will take a divisor represented as described in the previous
> paragraph. If we can solve the aforementioned problems (e.g. finding x,
> representing F/k(x), and converting divisors), then it should be easy to
> apply the algorithm to a general function field.
Let me know what happens.
>
>
>
--
William Stein
Professor of Mathematics
University of Washington
http://wstein.org
John Cremona
> On Thu, Jul 8, 2010 at 10:47 PM, Chris Hall <Christop...@uwyo.edu> wrote:
>> Just a quick status report. I have been thinking about the problem of
>> generating a separating element of a function field. In principal I know
>> how to find one if I can take pth roots. However, there's the additional
>> problem that not only do I need to find an element x in my function field
>> F=k(C) so that F/k(x) is finite and separable, but I need to explicitly
>> construct (a copy of) F as an extension of k(x). Is anyone there working on
>> pth roots or factorization over F?
>
> In some cases, if R is a domain such that factorization in R[x] is
> possible, and S is a finite algebraic extension of R, then factoring
> over S[x] is also possible, as long as we have GCD over S[x]. This
> works by taking norms, factoring over R[x], then taking gcd's. More
> precisely, we factor f(x) in S[x] by:
>
> (1) Compute g(x)=Norm(f(x)) in R[x]
> (2) Factor g(x) as a product of prime powers p^e
> (3) Recover the factorization of f(x) from the gcd(f, p).
Be careful. I think this only works if the norm is square-free.
(Special case where it certainly does not work: if f is already in
R[x] and irreducible there, then g is a power of f and you get no
info.) In this case one does a random shift, replacing x by x+c where
c is in S (not in R!) and then takes the norm.
This is in Cohen's book in the number field context.
John
William Stein
> On 11 July 2010 13:50, William Stein <wst...@gmail.com> wrote:
>> On Thu, Jul 8, 2010 at 10:47 PM, Chris Hall <Christop...@uwyo.edu> wrote:
>>> Just a quick status report. I have been thinking about the problem of
>>> generating a separating element of a function field. In principal I know
>>> how to find one if I can take pth roots. However, there's the additional
>>> problem that not only do I need to find an element x in my function field
>>> F=k(C) so that F/k(x) is finite and separable, but I need to explicitly
>>> construct (a copy of) F as an extension of k(x). Is anyone there working on
>>> pth roots or factorization over F?
>>
>> In some cases, if R is a domain such that factorization in R[x] is
>> possible, and S is a finite algebraic extension of R, then factoring
>> over S[x] is also possible, as long as we have GCD over S[x]. This
>> works by taking norms, factoring over R[x], then taking gcd's. More
>> precisely, we factor f(x) in S[x] by:
>>
>> (1) Compute g(x)=Norm(f(x)) in R[x]
>> (2) Factor g(x) as a product of prime powers p^e
>> (3) Recover the factorization of f(x) from the gcd(f, p).
>
> Be careful. I think this only works if the norm is square-free.
I was careful to be vague about (3) -- I didn't say how one recovers
the factorization.
Definitely the above will work in characteristic zero even if the norm
is not square free. Just take valuations at each prime divisor.
> (Special case where it certainly does not work: if f is already in
> R[x] and irreducible there, then g is a power of f and you get no
> info.)
I think you get complete info -- you get that f is irreducible. ?? Or
am I confused?
> In this case one does a random shift, replacing x by x+c where
> c is in S (not in R!) and then takes the norm.
>
> This is in Cohen's book in the number field context.
Yes, that's where I learned about this.
Resultants are relevant...
John Cremona
One of us is. If f is irreducible over R but not over S then your
gcd will be f again which does not help you factor over S.
Basically what one needs is that the conjugates of f (whose product is
the norm) are coprime.
?
John
William Stein
Yes, that's a good point. This won't work without some further trick
like you suggest.
And the tricks to make this general are in Cohen's book, I think, as you say.
William Stein
You're vindicated -- see the embarrassing (to me)
http://trac.sagemath.org/sage_trac/ticket/9498
William