Ker($\phi$) of a ring homomorphism

Question

Let $\phi_1 : \mathbb{Z}[x]\rightarrow \mathbb{Z}$ be the evaluation homomorphism at $1$. What's $\ker(\phi_1)$ ?

I know the kernel of a ring homomorphism $\phi : R\rightarrow S$ is the set $\{a \in R \mid \phi(a) = 0_S\}$. But I'm having a hard time finding what elements are contained in $\ker(\phi_1)$. Thanks in advance.

Answer 1

Let $f(x)\in \text{ker}(\phi)\implies \phi(f)=0\implies f(1)=0$. Now by division algorithm we have, $$f(x)=q(x)(x-1)+r(x)$$ for some $q(x),r(x)\in \Bbb Z[x]$ with $r=0$ or $\text{deg}(r)<\text{deg}(x-1)=1.$ In other words, $r$ is a constant polynomial in either cases. But, $0=f(1)=q(1)(1-1)+r(1)\implies r=0$. So that, $f(x)=(x-1)q(x)$. That is, $\text{ker}(\phi)\subseteq \{g(x)(x-1):g(x)\in \Bbb Z[x]\}$.

Conversely for any $g(x)\in \Bbb Z[x]$ we have, $\phi\big(g(x)(x-1)\big)=(1-1)g(1)=0\implies \{g(x)(x-1):g(x)\in \Bbb Z[x]\}\subseteq\text{ker}(\phi).$

Combining all these $\{g(x)(x-1):g(x)\in \Bbb Z[x]\}=\text{ker}(\phi)$.