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How can I obtain representatives of a quotient ring?

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

Ys1123 gravatar image

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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answered 2023-10-08 12:08:29 +0200

rburing gravatar image

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).

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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?

Ys1123 gravatar imageYs1123 ( 2023-10-09 06:48:13 +0200 )edit

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

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Last updated: Oct 08 '23