Thread: irreductible polinomials

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

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?

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.