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How can I find the elements in a factor ring?

asked 2020-08-04 04:10:29 -0500

moondog 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 13:04:23 -0500

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 $\mathcal O_K$ 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 04:10:29 -0500

Seen: 101 times

Last updated: Aug 05