I am having some trouble understanding the solution to this problem.
Problem: Let $\varphi$ denote the homomorphism $\mathbb{Z}[x] \to \mathbb{R}$ defined by $\varphi(f(x)) = f(1 + \sqrt{2})$. Is the kernel of $\varphi$ a principal ideal? If so, find a generator.
Solution:
Line 1: We want to find the polynomials $f(x)$ such that $f(1+\sqrt{2}) = 0$.
Line 2: Using the canonical embedding of $\mathbb{Z}[x]$ into $\mathbb{R}[x]$, we see that $(x – (1+\sqrt{2}))$ would generate the kernel of $\varphi$ in $\mathbb{R}[x]$.
Line 3: But then, since $(x – (1+\sqrt{2}))(x – (1-\sqrt{2})) = x^2-2x-1$, and $(x^2-2x-1) \subseteq \mathbb{Z}[x]$, we obtain $(x^2-2x-1) \subseteq \ker\varphi$, as an ideal in $\mathbb{Z}[x]$.
Line 4: But since $x^2-2x-1$ is primitive and irreducible in $\mathbb{Q}[x]$, since its roots are $(1\pm\sqrt{2})$, $x^2-2x-1$ is irreducible in $\mathbb{Z}[x]$.
Line 5: Now suppose that $f(x) \in \ker\varphi \setminus (x^2-2x-1)$.
Line 6: By the Euclidean algorithm dividing $f(x)$ by $x^2-2x-1$, we can assume without loss of generality that $f(x) = ax+b$.
Line 7: But then, $\varphi(f) = a + a\sqrt{2} + b \ne 0$ unless $a=b=0$, and so we see that $(x^2-2x-1) = \ker\varphi$.
Here are my questions in order:
Line 2: What's the simplest way to rephrase it, without the fancy terminology?
Line 3: We obtain $(x^2-2x-1) \subseteq \ker\varphi$, as an ideal in $\mathbb{Z}[x]$. Why?
Line 6: Why use "without loss of generality"? It is not clear enough for me – how can I prove it without using that?
Thanks for any help.
Best Answer
Line 3: This is because $\phi(x^2-2x-1)=0.$
Line 6: Using division algorithm, write $f(x)=g(x)(x^2-2x-1) + r(x)$ for some $g(x), r(x) \in \mathbb Z[x]$ with deg $r(x) \leq 1.$ Now apply $\phi$ in the above equation. We get $\phi(f(x))=\phi(r(x))$ and $r(x)$ is of the form $ax+b, a, b \in \mathbb Z.$