hi! i have to find the polinomial ireductible over the rationals asosiated with -1+2^[1/2] can any body heplme?

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- October 21st 2012, 06:02 AMpollipatoirreductible polinomials
hi! i have to find the polinomial ireductible over the rationals asosiated with -1+2^[1/2] can any body heplme?

- October 22nd 2012, 04:33 AMBingkRe: irreductible polinomials
You're looking for a polynomial with rational coefficients that has as a root.

That means, in some extension field, you can factor out from the polynomial.

Can you think of another root, say (which may be in the extension), so that when you expand , the polynomial will have rational coefficients? - October 22nd 2012, 07:52 AMDevenoRe: irreductible polinomials
note that if x = √2 - 1

x^{2}= (√2 - 1)^{2}= 2 - 2√2 + 1 = 3 - 2√2

so x^{2}+ 2x = 3 - 2√2 + 2√2 - 2 = 3 - 2 = 1

thus: x^{2}+ 2x - 1 = 0

that is, √2 - 1 is a root of x^{2}+ 2x - 1 in Q[x].

note that if x^{2}+ 2x - 1 is reducible over Q, it must factor as x^{2}+ 2x - 1 = (x - a)(x - b) for rational a, b.

by a theorem of Gauss, a and b must actually be integers, so either:

a = 1, b= -1, or a = -1, b = 1, both of which lead to a contradiction.