prime ideals in Z_n

[Avishek Adhikari]

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.



Waiting for your reply.
Best regards,
Avishek

[John Cremona]

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

[Avishek Adhikari]

I fully agree with you. But it will not certainly give all prime ideals!

 

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Dr. Avishek Adhikari
website : http://imbic.org/avishek.html
Secretary, IMBIC
http://www.imbic.org/forthcoming.html
Faculty Member, Dept. Of Pure Mathematics,
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[John H Palmieri]

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

[John Cremona]

On Mar 14, 10:29 am, Avishek Adhikari <avishek....@gmail.com> wrote:
> I fully agree with you. But it will not certainly give all prime ideals!

Yes it will. The number of prime ideals of Z/nZ is exactly the number
of prime divisors of n.

In general the prime ideals of R/I (where R is a commutative ring and
I an ideal) are of the form P/I where P is a prime ideal of R which
contains I, and P <--> P/I is a bijection. In this case, R=Z, I=nZ so
P has to be pZ where (1) p is prime, so that P is a non-zero prime
ideal and (2) p divides n, so that P contains I.

John Cremona

>
> On Sun, Mar 13, 2011 at 9:30 PM, John Cremona <john.crem...@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 <avishek....@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.
>
> > > Waiting for your reply.
> > > Best regards,
> > > Avishek
>
> > --
> > To post to this group, send email to sage-s...@googlegroups.com
> > To unsubscribe from this group, send email to
> > sage-support...@googlegroups.com
> > For more options, visit this group at
> >http://groups.google.com/group/sage-support
> > URL:http://www.sagemath.org
>
> --
> Dr. Avishek Adhikari
> website :http://imbic.org/avishek.html

> Secretary, IMBIChttp://www.imbic.org/forthcoming.html