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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- Oct 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?

- Oct 22nd 2012, 04:33 AMBingkRe: irreductible polinomials
You're looking for a polynomial with rational coefficients that has $\displaystyle -1 + \sqrt{2}$ as a root.

That means, in some extension field, you can factor out $\displaystyle x - (-1 + \sqrt{2})$ from the polynomial.

Can you think of another root, say $\displaystyle u$ (which may be in the extension), so that when you expand $\displaystyle (x - (-1 + \sqrt{2}))(x-u)$, the polynomial will have rational coefficients? - Oct 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.