# irreductible polinomials

• Oct 21st 2012, 06:02 AM
pollipato
irreductible 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 AM
Bingk
Re: irreductible polinomials
You're looking for a polynomial with rational coefficients that has $-1 + \sqrt{2}$ as a root.
That means, in some extension field, you can factor out $x - (-1 + \sqrt{2})$ from the polynomial.
Can you think of another root, say $u$ (which may be in the extension), so that when you expand $(x - (-1 + \sqrt{2}))(x-u)$, the polynomial will have rational coefficients?
• Oct 22nd 2012, 07:52 AM
Deveno
Re: irreductible polinomials
note that if x = √2 - 1

x2 = (√2 - 1)2 = 2 - 2√2 + 1 = 3 - 2√2

so x2 + 2x = 3 - 2√2 + 2√2 - 2 = 3 - 2 = 1

thus: x2 + 2x - 1 = 0

that is, √2 - 1 is a root of x2 + 2x - 1 in Q[x].

note that if x2 + 2x - 1 is reducible over Q, it must factor as x2 + 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.