On May 14, 2010, at 11:13 AM, Simon King wrote:
Hi Robert!
On 14 Mai, 18:34, Robert Bradshaw <rober...@math.washington.edu>
wrote:
1. Do you agree this is a bug?
The p-adic fields are of capped precision, not set precision, but
each
element remembers its own actual precision, so this is why the
coercion goes in that direction, and I don't think its a bug.
What does capped or set precision mean?
Is it like this?
- The precision of the elements of RR is determined by RR, and so the
precision must be non-increasing along a coercion.
Yep, otherwise elements have spuriously high precision.
- The precision of the elements of one p-adic ring can be different,
but there is an upper bound for the precision. This upper bound must
be non-decreasing under coercion.
Yep, otherwise we loose information.
If we're going to change this, it should be on its own ticket as it's
quite a big change.
I thought it is just changing one byte: '>' to '<'.
:). There are many places I could just change '>' to '<' that would
wreck havoc all over, the above would be a big semantic change.
Anyway, if you say we should keep it then I need to do a change in a
different place: The Completion construction functor allows merging,
and the rule is that two Completion functors with respect to the same
prime (resp. +Infinity, for the completion functor yielding RR) are
merged so that the *smaller* precision results.
By consequence, if you have two parent structures based on p-adic
rings of different precision, then it could be that there is no
coercion into their pushout.
Solution: In the merge method, I can distinguish the case "completion
at +Infinity" and "completion at some prime", so that the smaller
precision is chosen in the first case and the bigger precision in the
second case.
Yes, we'd have to special case this.
On May 14, 2010, at 11:39 AM, Simon King wrote:
Would the following be what you want?
sage: R1.<a> = Zp(5,prec=20)[]
sage: R2 = Qp(5,prec=40)
sage: R2(1)+a
(1 + O(5^20))*a + (1 + O(5^40))
This results when one changes the merge method (and makes fraction
field functor and completion functor commute).
Yep, that looks right to me. (Sometimes, though, I wonder if the
generator of a inexact ring should be considered to have infinite
precision, but that's totally separate.)
- Robert
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