I have some questions on the following problem regarding polynomial factorizations.
Problem:
Find the kernal of the map $\phi$ $\mathbb{Z}$[x] -> $\mathbb{R}$ where $f(x)$ -> $f(1 + \sqrt{2})$.
Solution + questions:
So we are looking for polynomials with coefficients over $\mathbb{Z}$ that have $1 + \sqrt{2}$ as a root. $\mathbb{Z}$ is not a field so we cannot assume the kernel is a principle ideal, though of course the kernel will be some ideal, perhaps it is principle.
The function f = ($x - (1 + \sqrt{2}$))($x - (1- \sqrt{2}$)) is an example of one function that has our desired root with coefficients over the integers, once we compute product: f = $x^2 -2x -1$.
Question 1: Is the reasoning below correct?
Now I believe this polynomial is minimal over $\mathbb{Z}$ since the minimal polynomials over the integers are the linear and quadratic polynomials. Hence since polynomial factorizations are unique, the previous form I have written $f$ in would be irreduicble over $\mathbb{R}$ as it is the product of two degree $1$ polynomials. Since the form f = ($x - (1 + \sqrt{2}$))($x - (1- \sqrt{2}$)) is irreducible over R, it is unique, and so there is no further reduction possible over $\mathbb{Z}$ from the listed quadratic.
Next since the kernal, $K$, of a homomorphism is an ideal, we know all products of $f$ are contained within $K$.
Question 2: How to I find the rest of the kernel, $K$?
Thanks!
First, prove (by brute force) that $\psi: \Bbb Q[ \sqrt 2] \to \Bbb Q [\sqrt 2]$ defined by $\psi(a+b \sqrt 2)= a-b \sqrt 2 ~(\text{for } a, b \in \Bbb Q)$ is a field automorphism and note that $\forall a \in \Bbb Q(\psi(a)=a)$. Thus, $\forall g \in \Bbb Q[x]~\forall \alpha \in \Bbb Q[\sqrt 2], \psi(g(\alpha))=g(\psi(\alpha))$.
Thus, if $f \in \Bbb Z[x] \subseteq \Bbb Q[x]$ satisfies $f(1 + \sqrt 2)=0$, then:
$$0=\psi(0)=\psi(f(1+ \sqrt 2)) = f(\psi(1+ \sqrt 2))=f(1- \sqrt 2),$$ so $1 - \sqrt 2$ must also be a root of $f$, which means that $(x^2-2x-1) \vert f$, so $\ker(\phi) \subseteq \langle x^2-2x-1 \rangle$. You've already shown that $\langle x^2-2x-1 \rangle \subseteq \ker(\phi)$, so equality follows.