This also works fine:
sage: GF(5)(-1).sqrt()
2
as does this
sage: a=GF(5)(2).sqrt()
sage: a
sqrt2
sage: a^2
2
but not GF(5)(sqrt(2)). I can see that coercing simple symbolic
expressions like sqrt(integer) into various fields would not be hard,
but handling arbitrary symbolic expressions would be very hard
(probably impossible); if what you want to a solution to an equation
in a certain field then it would be better (surely) to construct it
that way instead of coercing a symbolic expression.
John
2008/5/16 William Stein <[EMAIL PROTECTED]>:
>
> On Thu, May 15, 2008 at 4:00 PM, John H Palmieri <[EMAIL PROTECTED]> wrote:
>>
>> The fact that i is not in CC doesn't bother me too much, since i is a
>> "formal square root of -1":
>>
>> sage: i in CC
>> False
>>
>> I can force it to be in CC by doing this:
>>
>> sage: CC(i)
>> 1.00000000000000*I
>>
>> But then I think I should be able to do this:
>>
>> sage: GF(5)(i)
>>
>> Someone who knows more number theory than I do can come up with a
>> better solution, but coercion code like this ought to work: replace
>> the end ("raise TypeError...") of the _coerc_impl() method for
>> Finite_Field_prime_modn (in sage.rings.finite_field_prime_modn.py)
>> with this:
>>
>> from sage.functions.constants import I
>> from sage.rings.arith import primitive_root
>> p = self.__char
>> if x == i and p % 4 == 1:
>> return self(primitive_root(p))**((p-1)/4)
>> raise TypeError, "no canonical coercion of x"
>>
>> But this isn't the right place to patch. When I do this,
>> GF(5)._coerce_(i) works, but GF(5)(i) doesn't.
>>
>> Perhaps IntegerMod in the file integer_mod.pyx should be patched, but
>> I don't want to mess around with Cython. Perhaps some other file
>> should be patched. Can anyone help?
>
> You should put the code in __call__. _coerce_ is only
> for canonical coercions, which the above definitely isn't.
>
> By the way, it would be (asymptotically) a lot faster to
> choose a random element of GF(p) and raise it to the
> power (p-1)/4 and check the resulting power is not +/-1.
> Finding a primitive root requires factoring p-1, which is
> potentially very slow.
>
> -- William
>
> >
>
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