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

asked 2019-03-12 03:07:03 +0200

Llew gravatar image

updated 2020-06-01 10:57:04 +0200

FrédéricC gravatar image


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 2019-03-12 10:11:26 +0200

rburing gravatar image

updated 2019-03-12 10:38:29 +0200

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)

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 ( 2019-03-12 12:23:14 +0200 )edit

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Asked: 2019-03-12 03:07:03 +0200

Seen: 291 times

Last updated: Mar 12 '19