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 | 1 | initial version |  |
You can do it like this (also simplifying the notation a bit):
sage: R.<x> = PolynomialRing(GF(3))
sage: S.<t> = R.quotient(x^2 + 1)
sage: sum(p^2 for p in S)
0| | 2 | No.2 Revision |
You can do it like this (also simplifying the notation a bit):
sage: R.<x> = PolynomialRing(GF(3))
sage: S.<t> S.<i> = R.quotient(x^2 + 1)
sage: sum(p^2 for p in S)
0We can explain the result (for this choice of f) as follows: p(i)=a+bi implies p(i)2=a2−b2+2abi and the sum runs over all a and b, so the "real part" of the sum is ∑a,ba2−b2=∑aa2−∑bb2=0,
and the "imaginary part" is ∑a,b2ab=∑a,bab+∑a,bab=∑a,bab+∑a,b(−a)b=∑a,bab−ab=0.
