Ask Your Question
0

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

asked 2019-03-11 21:07:03 -0500

Llew 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?

edit retag flag offensive close merge delete

1 answer

Sort by ยป oldest newest most voted
2

answered 2019-03-12 04:11:26 -0500

rburing gravatar image

updated 2019-03-12 04:38:29 -0500

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 $$\sum_{a,b} 2ab = \sum_{a,b} ab + \sum_{a,b} ab = \sum_{a,b} ab + \sum_{a,b} (-a)b = \sum_{a,b} ab - ab = 0.$$

edit flag offensive delete link more

Comments

1

Thank you, that is very helpful.

Llew gravatar imageLlew ( 2019-03-12 06:23:14 -0500 )edit

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Question Tools

Stats

Asked: 2019-03-11 21:07:03 -0500

Seen: 35 times

Last updated: Mar 12