Surjective homomorphism on Commutative Ring

Question

Let $A$ be a commutative ring, $R= A[x_{1},...,x_{n}]$ and $(a_{1},...a_{n}) \in A^{n}$ . Let $\phi : R \to A$ be defined by $\phi (f(x_{1},...,x_{n}) = f(a_{1},...,a_{n})$. Then show that $\phi$ is a surjective homomorphism and $Ker\phi = \sum _{i=1} ^{n} A(x_{i} -a_{i}) = (x_{1}-a_{1},...,x_{n}-a_{n})$. In particular, if $A$ is a field then $Ker\phi$ is a maximal ideal

Answer 1

It is easy to show that $ϕ$ is a surjective homomorphism.
since $\phi (x_{1}-a_{1},...,x_{n}-a_{n})=(a_{1}-a_{1},...,a_{n}-a_{n})=0$, $(x_{1}-a_{1},...,x_{n}-a_{n}) \subseteq Ker\phi $. for the convers, assume $g \in Ker\phi$ and use division algorithm (Theorem 2.2.1 of Herzog-Hibi book "monomial ideals") to see that $g=h(x_{1}-a_{1},...,x_{n}-a_{n})$ for some $h\in R$.
note that if $A$ is a field then $R/Kerϕ$ is a field and so $Kerϕ$ is a maximal ideal