On Monday, March 14, 2011 11:29:20 AM UTC-7, avishek wrote:
>
> I fully agree with you. But it will not certainly give all prime ideals!
>
>
> On Sun, Mar 13, 2011 at 9:30 PM, John Cremona <john.c...@gmail.com> wrote:
>
>> This is not a computational question at all but an easy exercise.
>> There is one prime ideal for each prime factor p of n, namely pZ/nZ.
>> See any book on elementary ring theory!
>>
>> John Cremona
>>
>> On Mar 12, 11:41 pm, Avishek Adhikari <avish...@gmail.com> wrote:
>> > Hello,
>> > I shall be glad, if you kindly send the help towards finding the
>> solution
>> > of the following problem using sage:
>> >
>> > In Z_n (the ring of integers modulo n), find all prime ideals.
>>
>
As far as I can tell, there are two obstacles to doing this in Sage: first,
we have no way of dealing with the ring Z/nZ for a variable n, only one n at
a time (IntegerModRing(6) works, but var('n'); IntegerModRing(n) doesn't).
Is there any reasonable way of doing this? Second, we have no way of
producing all of the ideals of a ring, nor all of its prime ideals.
Something like the following *ought* to work, but the method "is_maximal"
looks pretty broken to me. (The method "is_prime" isn't implemented for
ideals in Z/nZ.)
sage: R = IntegerModRing(6)
sage: [R.principal_ideal(i) for i in R if
R.principal_ideal(i).is_maximal()]
This produces a list including the zero ideal, which is not maximal in Z/6Z,
last time I checked. I just created
<http://trac.sagemath.org/sage_trac/ticket/10934> to track this.
--
John
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