$\ker \phi = (x_1-a_1, ..., x_n-a_n)$ for a ring homomorphism $\phi: R[x_1, ..., x_n] \to R$ Let $R$ be a commutative ring, $a_1, ..., a_n$ its elements and
$\phi: R[x_1, ..., x_n] \to R$
defined by
$ \phi(f(x_1, ..., x_n)) = f(a_1, ... ,a_n)$
a ring homomorphism.
Prove: $\ker \phi = (x_1-a_1, ..., x_n-a_n)$
It is obvious that $(x_1 - a_1, ..., x_n -a_n) \subseteq \ker \phi$. I'm not sure how to prove the converse.
At this point I don't know any division algorithms for multivariable polynomials, only for the ones in $R[x]$(and the book from where I taken the exercise doesn't assume the reader to know something beyond basics of rings and ideals, and the division algorithm for $R[x]$). Though I know this could be solved for $i = 1$ by dividing by $x-a$: 
Let $f(x) \in \ker \phi$, divide by $x-a: f(x) = q(x)(x-a) + r, f(a) = r = 0$, so $f(x) = q(x)(x-a) \in (x-a)$.
 A: This proof is pretty notation-heavy for which I apologize, but the idea is not complicated: a polynomial $f$ is usually written as a Taylor series centered at $(0,\ldots, 0)$.  We use the Binomial Theorem to re-center it at $(\alpha_1, \ldots, \alpha_n)$ and show that the constant term of this series is $f(\alpha_1, \ldots, \alpha_n)$.
Lemma:
Let $R$ be a unital commutative ring and $R[x_1, \ldots, x_n]$ be the $n$-variable polynomial ring over $R$.  For $\alpha_1, \ldots, \alpha_n \in R$, let
\begin{align*}
\varphi = \text{eval}_{\alpha_1, \ldots, \alpha_n}: R[x_1, \ldots, x_n] &\to R\\
x_1, \ldots, x_n &\mapsto \alpha_1, \ldots, \alpha_n
\end{align*}
be the evaluation homomorphism.  Then $\ker(\varphi) = (x_1-\alpha_1, \ldots, x_n-\alpha_n)$ and the induced map $\overline{\varphi}: R[x_1, \ldots, x_n]/(x_1-\alpha_1, \ldots, x_n-\alpha_n) \to R$ is an isomorphism.
Proof:
Certainly $(x_1-\alpha_1, \ldots, x_n-\alpha_n) \subseteq \ker(\varphi)$ so it remains to show the reverse inclusion.  Given $f \in \ker(\varphi)$, we may write $f(x_1, \ldots, x_n) = \sum_{\gamma} a_\gamma {x_1}^{\gamma_1} \cdots {x_n}^{\gamma_n}$ where $a_\gamma = 0$ for all but finitely many multi-indices $\gamma$.  Writing $x_i = (x_i - \alpha_i) + \alpha_i$ and expanding by the binomial theorem, we have
\begin{align*}
f(x_1, \ldots, x_n) &= \sum_{\gamma} a_\gamma {x_1}^{\gamma_1} \cdots {x_n}^{\gamma_n} = \sum_{\gamma} a_\gamma ((x_1 -\alpha_1) +\alpha_1)^{\gamma_1} \cdots ((x_n -\alpha_n) +\alpha_n)^{\gamma_n}\\
 &= \sum_{\gamma} a_\gamma \left(\sum_{\kappa_1=0}^{\gamma_1} \binom{\gamma_1}{\kappa_1} (x_1 -\alpha_1)^{\kappa_1}\alpha_1^{\gamma_1 - \kappa_1}\right) \cdots \left(\sum_{\kappa_n=0}^{\gamma_n} \binom{\gamma_n}{\kappa_n} (x_n -\alpha_n)^{\kappa_n}\alpha_n^{\gamma_n - \kappa_n}\right)\\
 &= \sum_{\gamma} a_\gamma \sum_{\kappa_1=0}^{\gamma_1} \cdots \sum_{\kappa_n=0}^{\gamma_n} \binom{\gamma_1}{\kappa_1} \cdots \binom{\gamma_n}{\kappa_n}\alpha_1^{\gamma_1 - \kappa_1} \cdots\alpha_n^{\gamma_n - \kappa_n}  (x_1 -\alpha_1)^{\kappa_1} \cdots (x_n -\alpha_n)^{\kappa_n}\\
 &= \sum_{\gamma} a_\gamma \sum_{\kappa} \binom{\gamma_1}{\kappa_1} \cdots \binom{\gamma_n}{\kappa_n}\alpha_1^{\gamma_1 - \kappa_1} \cdots\alpha_n^{\gamma_n - \kappa_n}  (x_1 -\alpha_1)^{\kappa_1} \cdots (x_n -\alpha_n)^{\kappa_n}\\
  &=\sum_{\kappa} \underbrace{\left(\sum_{\gamma}a_\gamma\binom{\gamma_1}{\kappa_1} \cdots \binom{\gamma_n}{\kappa_n}\alpha_1^{\gamma_1 - \kappa_1} \cdots\alpha_n^{\gamma_n - \kappa_n}\right)}_{b_\kappa}  (x_1 -\alpha_1)^{\kappa_1} \cdots (x_n -\alpha_n)^{\kappa_n}\\
 &=\sum_{\kappa} b_\kappa (x_1 -\alpha_1)^{\kappa_1} \cdots (x_n -\alpha_n)^{\kappa_n}.
\end{align*}
Note that the constant term $b_{(0,\ldots, 0)}$ is
\begin{align*}
b_{(0,\ldots, 0)} = \sum_{\gamma} a_\gamma\binom{\gamma_1}{0} \cdots \binom{\gamma_n}{0}\alpha_1^{\gamma_1} \cdots\alpha_n^{\gamma_n} = f(\alpha_1, \ldots,\alpha_n) = \varphi(f) = 0
\end{align*}
since $f \in \ker(\varphi)$.  Now all other terms in the sum $\sum_{\kappa} b_\kappa (x_1 -\alpha_1)^{\kappa_1} \cdots (x_n -\alpha_n)^{\kappa_n}$ have a factor of $(x_i - \alpha_i)^{\kappa_i}$ for some $i$ with $\kappa_i \geq 1$, hence belong to $(x_1 - \alpha_1, \ldots, x_n - \alpha_n)$.  Thus $f(x_1, \ldots, x_n) = \sum_{\kappa} b_\kappa (x_1 -\alpha_1)^{\kappa_1} \cdots (x_n -\alpha_n)^{\kappa_n} \in (x_1 - \alpha_1, \ldots, x_n - \alpha_n)$.
A: In problems like this, instead of explicitly showing that every element of the kernel is in $I=(x_1-a_1,\dots,x_n-a_n)$, it is easier to think about the quotient ring $S=R[x_1,\dots,x_n]/I$.  Since $I\subseteq\ker(\phi)$, $\phi$ induces a surjective homomorphism $\tilde{\phi}:S\to R$ such that $\phi$ is equal to the composition of $\tilde{\phi}$ with the quotient map $\psi:R[x_1,\dots,x_n]\to S$.  Let us now show that $\tilde{\phi}$ is injective, so it is actually an isomorphism and $\ker(\phi)=\ker(\psi)=I$.
To prove this, note that in $S$, $\psi(x_i)=\psi(a_i)$ for each $i$.  Since $\psi$ is a homomorphism, it follows that given any polynomial $f(x_1,\dots,x_n)$, $\psi(f(x_1,\dots,x_n))=\psi(f(a_1,\dots,a_n))$.  This means that every element of $S$ is actually $\psi(r)$ for some $r\in R$.  But for $r\in R$, $\tilde{\phi}(\psi(r))=\phi(r)=r$.  In particular, $\tilde{\phi}(\psi(r))=0$ iff $r=0$.  It follows that the kernel of $\tilde{\phi}$ consists of only $\psi(0)=0$, so $\tilde{\phi}$ is injective.
