Let $R$ be a commutative ring and $a_1, \dots, a_n$ elements of $R$. Prove that $$\frac{R[x_1, \dots, x_n]}{(x_1-a_1, \dots, x_n-a_n)} \cong R.$$
We can look at this as \begin{align*} \frac{R[x_1, \dots, x_n]}{(x_1-a_1, \dots, x_n-a_n)} &\cong \frac{R[x_1]\cdots[x_n]}{(x-a_1, \dots, x-a_n)}\\ &\cong \frac{R[x_1]\cdots[x_n]/(x_n-a_n)}{(x-a_1, \dots, x-a_{n-1})}\\ &\cong \frac{R[x_1]\cdots[x_{n-1}]}{(x-a_1, \dots, x-a_{n-1})}\\ &\cong\\ &\ \ \vdots\\ &\cong R. \end{align*}
However, I want to use an explicit map $\varphi: R[x_1, \dots, x_n] \twoheadrightarrow R$ where $f(x_1, \dots, x_n) \mapsto f(a_1, \dots, a_n)$ and show $\ker \varphi = (x_1-a_1, \dots, x_n-a_n)$.
I can see $(x_i-a_i) \mapsto 0$ for $1 \le i \le n$ and so $(x_1-a_1, \dots, x_n-a_n) \subset \ker \varphi$.
How can I show that if $f(a_1, \dots, a_n) =0$, then $f(x_1, \dots, x_n) \in (x_1-a_1, \dots, x_n-a_n)$?
