Find a polynomial $g(x) \in \Bbb Q [x]$ such that $I = g(x) $

Question

Find a polynomial $g(x) \in \Bbb Q [x]$ such that ideal $I = (g(x)) $, where

1. $I = \{f(x) \in \mathbb Q[x] : f(\sqrt2) = 0\}$
2. $ I = \{f(x) \in \mathbb Q[x] : f(1-i) = f(1+i) = 0 \}$

For 1, I think $g(x)$ will be $x^2-2$. For 2, I have no idea where to start.

Answer 1

The idea here is that the minimal polynomial of an element $ \alpha $ over a field $ K $ divides any polynomial in $ K[X] $ which has $ \alpha $ as a root. (1) is the set of all polynomials in $ \mathbb{Q}[X] $ which have $ \sqrt{2} $ as a root, but the minimal polynomial of $ \sqrt{2} $ is $ X^2 - 2$ which means that $ I = (X^2 - 2) $.

For (2), simply observe that $ (X - (1+i))(X - (1-i)) = X^2 - 2X + 2 $ is the minimal polynomial of both of these elements, so that the ideal is $ I = (X^2 - 2X + 2) $.