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How to compute the sums of squares of elements of a quotient ring?

asked 5 years ago

Llew gravatar image

updated 4 years ago

FrédéricC gravatar image

Hi,

I'm new to Sage, and I would like to be able to test, given some q, whether \sum_{p(t) \in \mathbb{F}_{q}[t]/(f)}^{}{p^2(t)}=k \bmod (f) for some fixed f \in \mathbb{F}_q [t] and k \in \mathbb{F}_q.

I can get as far as (for q=3 and f=x^2+1):

sage:    R = PolynomialRing(GF(3),'x'); x = R.gen()
sage:    S = R.quotient(x^2 + 1)

But I'm not sure how to sum over all the elements of the quotient ring, let alone their squares.

Any hints?

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answered 5 years ago

rburing gravatar image

updated 5 years ago

You can do it like this (also simplifying the notation a bit):

sage: R.<x> = PolynomialRing(GF(3))
sage: S.<i> = R.quotient(x^2 + 1)
sage: sum(p^2 for p in S)
0

We can explain the result (for this choice of f) as follows: p(i) = a+bi implies p(i)^2 = a^2 - b^2 + 2abi and the sum runs over all a and b, so the "real part" of the sum is \sum_{a,b} a^2 - b^2 = \sum_a a^2 - \sum_b b^2 = 0, and the "imaginary part" is \sum_{a,b} 2ab = \sum_{a,b} ab + \sum_{a,b} ab = \sum_{a,b} ab + \sum_{a,b} (-a)b = \sum_{a,b} ab - ab = 0.

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Thank you, that is very helpful.

Llew gravatar imageLlew ( 5 years ago )

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Asked: 5 years ago

Seen: 381 times

Last updated: Mar 12 '19