Problem:
Solution:
This is basically exercising the algorithm we developed in the last problem, the flow goes like this
$ I(V(f_1, f_2)) = I(V(g)) $
$ I(V(g)) = \langle g_{\text{red}} \rangle $
$ g_{\text{red}} = \frac{g}{d} $
$ d = gcd(g, g') $
$ g = x^4 - 2x^3 + 2x - 1 $
$ g' = 4x^3 - 6x^2 + 2 $
$ d = x^2 - 2x + 1 $
$ g_{\text{red}} = x^2 - 1 $
Therefore $ I(V(x^5 - 2x^4 + 2x^2 - x, x^5 - x^4 - 2x^3 + 2x^2 + x - 1)) = \langle x^2 - 1 \rangle $
All these are computed using this program
print "Problem 17"
f = polynomial.from_string("x^5 - 2x^4 + 2x^2 - x")
g = polynomial.from_string("x^5 - x^4 - 2x^3 + 2x^2 + x - 1")
gcd = polynomial.polynomial_gcd(f, g)
gcdd = gcd.derivative()
d = polynomial.polynomial_gcd(gcd, gcdd)
(gred, r) = polynomial.polynomial_divide(gcd, d)
print gcd
print gcdd
print d
print gred
Solution:
This is basically exercising the algorithm we developed in the last problem, the flow goes like this
$ I(V(f_1, f_2)) = I(V(g)) $
$ I(V(g)) = \langle g_{\text{red}} \rangle $
$ g_{\text{red}} = \frac{g}{d} $
$ d = gcd(g, g') $
$ g = x^4 - 2x^3 + 2x - 1 $
$ g' = 4x^3 - 6x^2 + 2 $
$ d = x^2 - 2x + 1 $
$ g_{\text{red}} = x^2 - 1 $
Therefore $ I(V(x^5 - 2x^4 + 2x^2 - x, x^5 - x^4 - 2x^3 + 2x^2 + x - 1)) = \langle x^2 - 1 \rangle $
All these are computed using this program
print "Problem 17"
f = polynomial.from_string("x^5 - 2x^4 + 2x^2 - x")
g = polynomial.from_string("x^5 - x^4 - 2x^3 + 2x^2 + x - 1")
gcd = polynomial.polynomial_gcd(f, g)
gcdd = gcd.derivative()
d = polynomial.polynomial_gcd(gcd, gcdd)
(gred, r) = polynomial.polynomial_divide(gcd, d)
print gcd
print gcdd
print d
print gred
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