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How can I find the elements in a factor ring?

asked 2020-08-04 11:10:29 +0200

moondog gravatar image

updated 2022-10-15 14:11:53 +0200

FrédéricC gravatar image

Let $K=\mathbb{Q}(\sqrt{2})$ and $\mathfrak{a}=(3)=3\mathcal{O}_K$ an ideal in $K$. How can I find the elements in $\mathcal{O}_K/\mathfrak{a}$ using sage?

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answered 2020-08-05 20:04:23 +0200

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()
sage: Q.is_prime_field()
sage: Q.inject_variables()
Defining abar0, abar1
sage: abar0                                                                                                                                 
sage: abar1                                                                                                                                 

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 $\mathcal O_K$ is generated by $1$ and $a=(x\text{ mod }(3))$.) $$ \mathcal O_K/(3) =(\Bbb Z[x]/(x^2-2))/3 =\Bbb Z[x]/(x^2-2,\ 3) =(\Bbb Z[x]/3)/(x^2-2) =\Bbb F_3[x]/(x^2-2) \ . $$ 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.

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Asked: 2020-08-04 11:10:29 +0200

Seen: 327 times

Last updated: Aug 05 '20