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Ideal moduli and residue symbols

asked 2015-03-05 19:37:25 -0600

jwiebe gravatar image

updated 2015-03-08 14:01:00 -0600

Hi, everyone;

I'm fairly new to sage, but I feel like I have some heavy lifting to do. I'm attempting to do a couple of things:

1) I need to see if two complex numbers are equivalent mod an ideal, eg pi == 1 mod 2+2i. It might be a dumb question but my searching thus far has come up short

2) I need to compute residue symbols, and I'm using the Number Field residue symbol method, but I'm having trouble. I have the following: C = ComplexField() I = C.0 r = C.ideal(b).residue_symbol(D,4) with a and b complex numbers. Help!

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Could you please define what you mean by "ideal". The complex plane is a field, so every ideal in the usual sense is trivial here.

tmonteil gravatar imagetmonteil ( 2015-03-06 11:36:00 -0600 )edit

Okay, after reading a bit more I can clarify. The ambient space is the ring of integers of Q(zeta_m) - mth root of unity. So these ideals are in the ring.

jwiebe gravatar imagejwiebe ( 2015-03-06 15:00:57 -0600 )edit

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answered 2015-03-08 14:00:23 -0600

It is fairly straightforward once you know Sage standards

sage: K = CyclotomicField(9)
sage: O = K.ring_of_integers()
sage: zeta9 = O.gen(1)

At this point you have three objects defined in the console/notebook: the cyclotomic field K, its ring of integers O and the generator zeta9. Now you can defined ideals and quotients as follows.

sage: I = O.ideal(3*zeta9^2 + 2*zeta9^3 + 5)
sage: R = O.quotient(I, 'a')

note: I am not sure why, but the arguemnt 'a' is mandatory in the method quotient.

To check equality modulo I just do it in the quotient

sage: R(3*zeta9^2 + 7) == R(2 - 2*zeta9^3)
True
sage: R(zeta9) == R(2)
False

Vincent

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Asked: 2015-03-05 19:37:25 -0600

Seen: 46 times

Last updated: Mar 08 '15