I found this question, which is my question but for a different homomorphism. Let me briefly restate the question with my homomorphism:
Let $\phi$ be the homomorphism $\mathbb{Z}[x]\rightarrow \mathbb{R}$ defined by $\phi(f(x))=f(\frac{1}{2}+\sqrt{2})$. Is the kernel of $\phi$ a principal ideal? If so, find the generator.
I take the same approach as in the example question, namely I consider that $(x-(\frac{1}{2}+\sqrt{2}))$ would generate the kernel of $\phi$ in $\mathbb{R}[x]$. Then using the same trick as in line 3 in the example question, I have that $(x-(\frac{1}{2}+\sqrt{2}))(x-(\frac{1}{2}+-\sqrt{2}))=x^2-x-2+\frac{1}{4}=g(x)$, but unlike the example question, here $g(x)\not\in\mathbb{Z}[x]$. However, $4g(x)\in\mathbb{Z}[x]$. So, is that something I can do and just proceed as they do in the example question. Or for this homomorphism a completely different approach is necessary?
I'd appreciate any help. This topic still confuses me.
Best Answer
The same approach works. If $f := 4(x-(\frac{1}{2}+\sqrt{2}))(x-(\frac{1}{2}-\sqrt{2})) \in \mathbb Z[x]$, we claim that $\ker \phi = (f)$. By construction, $f \in \ker \phi$, so $(f) \subseteq \ker \phi$. On the other hand, if $p \in \ker \phi$, by the division algorithm in $\mathbb Q[x]$ there are $q,r \in \mathbb Q[x]$ such that $p = fq+r$ and $r = r_1+r_2x$ for some $r_1,r_2 \in \mathbb Q$. Now note that $$0 = p(\tfrac12+\sqrt2) = r_1 + r_2 \cdot (\tfrac12+\sqrt2)$$ so $$r_2 \neq 0 \implies\sqrt2 = -\tfrac12 -\tfrac{r_1}{r_2} \in \mathbb Q$$ meaning that $r_2=0$, and then $r_1 = 0$. Hence $p=fq \in (f)$.