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# How can I obtain representatives of a quotient ring?

I want to compute quotient of integer ring of $\mathbb{Q}(\omega) (\omega^3=1)$ by a prime ideal $(-4-3\omega)$. Especially I want to compute representatives.

N=3
x=polygen(ZZ,'x')
K.<a>=CyclotomicField(N)
O = K.ring_of_integers()
p=-4-3*a
R.<b,c>=QuotientRing(O, K.ideal(p))


What should I do next?

More, I want to compute its cardinality (=13), But

R.cardinality()


made error.

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## 1 Answer

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Go this way instead:

sage: P = K.ideal(p)
sage: R = P.residue_field()
sage: R.cardinality()
13
sage: [z.lift() for z in R]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]


Why representatives are integers:

sage: R(a)
3
sage: (-a)*p
a - 3
sage: (-a)*P in P
True


So in the quotient $a = 3$.

In general $a$ is mapped to one of K.defining_polynomial().roots(R, multiplicities=False).

more

## Comments

Thank you. What does R(a) mean? By the way this way seems very ad hoc. Isn’t there any ways to compute representatives of quotient of general (polynomial) ring?

( 2023-10-09 06:48:13 +0200 )edit

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Asked: 2023-10-08 10:52:01 +0200

Seen: 120 times

Last updated: Oct 08 '23