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.

  • $\begingroup$ Polynomials that evaluate to $0$ at $1$; i.e., multiples of $x-1$ $\endgroup$ – J. W. Tanner Nov 14 '19 at 5:11
  • $\begingroup$ A polynomial $f$ is in the kernel of $(\phi)_1$ if f(1) = 0. What can you say about $f$ in this case? $\endgroup$ – paul blart math cop Nov 14 '19 at 5:11
  • $\begingroup$ @paulblartmathcop f is equal to the algebra generated by x-1. Would 0 or 1, however, be contained in ker($\phi$)? Just a thought. $\endgroup$ – asuhdude Nov 14 '19 at 6:14
  • $\begingroup$ Well what is f(x) = 1 evaluated at 1? How about g(x) = 0 evaluated at 1? $\endgroup$ – paul blart math cop Nov 14 '19 at 6:38
  • $\begingroup$ @paulblartmathcop so would 0 be in the ker($\phi$) as g(x) = 0 evaluated at 1 = 0 and 1 not in the ker($\phi) as f(x) = 1 evaluated at 1 = 1? $\endgroup$ – asuhdude Nov 14 '19 at 6:44

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

  • $\begingroup$ Would 0 or 1 be contained in ker($\phi$)? Or is ker($\phi$) just polynomials? $\endgroup$ – asuhdude Nov 14 '19 at 6:23
  • $\begingroup$ $0$ must be contained in $\text{ker}(\phi)$ as $0=0\cdot (x-1)$ but $1\not\in \text{ker}(\phi)$ as the polynomial $1\in\Bbb Z[x]$ does not vanishes at point $1\in \Bbb Z$. $\endgroup$ – Sumanta Nov 14 '19 at 6:54

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