# Ker($\phi$) of a ring homomorphism

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.

• Polynomials that evaluate to $0$ at $1$; i.e., multiples of $x-1$ – J. W. Tanner Nov 14 '19 at 5:11
• A polynomial $f$ is in the kernel of $(\phi)_1$ if f(1) = 0. What can you say about $f$ in this case? – paul blart math cop Nov 14 '19 at 5:11
• @paulblartmathcop f is equal to the algebra generated by x-1. Would 0 or 1, however, be contained in ker($\phi$)? Just a thought. – asuhdude Nov 14 '19 at 6:14
• Well what is f(x) = 1 evaluated at 1? How about g(x) = 0 evaluated at 1? – paul blart math cop Nov 14 '19 at 6:38