On Thu, Mar 18, 2010 at 8:27 PM, Waldek Hebisch wrote:
> Ralf Hemmecke wrote:
>> On 03/18/2010 11:17 PM, Waldek Hebisch wrote:
>> >
>> > It is a math problem: if base R is a field that does not contain
>> > square root of -1, then Complex(R) is again a field. But we have
>> > no way to check this condition, so need hardcode the choice.
> ...
>> According to the specification...
>>
>> ++ \spadtype {Complex(R)} creates the domain of elements of the form
>> ++ \spad{a + b * i} where \spad{a} and b come from the ring R,
>> ++ and i is a new element such that \spad{i^2 = -1}.
>> ^^^
>>
>> why shouldn't Complex(Complex(Integer)) make sense.
>...
> Yes, asserting Field or IntegralDomain is the problem. Otherwise
> such constructions would be OK.
>
I think that to understand Waldek's point it is import to look at
ComplexCategory.
http://axiom-wiki.newsynthesis.org/Complex
E.g.
if R has IntegralDomain then
IntegralDomain
_exquo : (%,R) -> Union(%,"failed")
++ exquo(x, r) returns the exact quotient of x by r, or
++ "failed" if r does not divide x exactly.
if R has EuclideanDomain then EuclideanDomain
if R has multiplicativeValuation then multiplicativeValuation
if R has additiveValuation then additiveValuation
if R has Field then -- this is a lie; we must know that
Field -- x^2+1 is irreducible in R
...
Regards,
Bill Page.
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