Processing math: 100%

First time here? Check out the FAQ!

Ask Your Question
1

How can I find the elements in a factor ring?

asked 4 years ago

moondog gravatar image

updated 2 years ago

FrédéricC gravatar image

Let K=Q(2) and a=(3)=3OK an ideal in K. How can I find the elements in OK/a using sage?

Preview: (hide)

1 Answer

Sort by » oldest newest most voted
2

answered 4 years ago

dan_fulea gravatar image

Here is a possibility...

sage: K.<a> = QuadraticField(2)
sage: OK = K.OK()

sage: OK.gens()                                                                                                                             
(1, a)    sage: J = OK.ideal(3)

sage: Q = OK.quotient(J, names='abar')
sage: Q
Quotient of Maximal Order in Number Field in a with defining polynomial x^2 - 2 with a = 1.414213562373095? by the ideal (3)
sage: Q.is_field()
True
sage: Q.is_prime_field()
False
sage: Q.inject_variables()
Defining abar0, abar1
sage: abar0                                                                                                                                 
1
sage: abar1                                                                                                                                 
a

But do not expect too much functionality from this construction. Instead, why not construct "with bare hands" the ring last listed in the following chain of isomorphisms. (We know that OK is generated by 1 and a=(x mod (3)).) OK/(3)=(Z[x]/(x22))/3=Z[x]/(x22, 3)=(Z[x]/3)/(x22)=F3[x]/(x22) . Which is:

R.<x> = PolynomialRing(GF(3))                                                                                                         
F.<a> = GF(9, modulus=x^2-2)

Of course, knowing the purpose for the needed code snippet may change the situation.

Preview: (hide)
link

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Question Tools

Stats

Asked: 4 years ago

Seen: 409 times

Last updated: Aug 05 '20