Problem on polynomial in several variable

Question

Let $A$ be a commutative ring and let $R= A[x_{1},...,x_{n}]$. Let $a \in A$ and $f \in R$ satisfy $f(x_{1},..x_{n-1},a) = 0$ (as a polynomial in the variables $x_1,\ldots,x_{n-1}$). How do we show that $f= (x_{n} - a) g$ for some $g \in R$?

Answer 1

First, prove this with $n=1$: Suppose $p(x)\in B[x]$, where $B$ is a commutative ring, and $a\in B$ such that $p(a)=0$. Then $p(x)=(x-a)h(x)$ for some $h\in B[x]$. This is proven by induction in the degree of the polynomial: If $degree(p)=1$, the result is easily obtained. If $p(x)=\sum_{i=0}^{n+1}\alpha_i x^i$ and $p(a)=0$, then $q(x)=p(x)-\alpha_{n+1}x^{n+1} +\alpha_{n+1}a^{n+1}$ is of degree $n$ and $q(a)=0$, so you can use the induction hipothesis.

Now, for $n>1$: We have an isomorphism $A[x_1,\ldots,x_n]\simeq(A[x_1,\ldots,x_{n-1}])[x_n]$: Given $p(x_1,\ldots,x_n)\in A[x_1,\ldots,x_n]$, we can think of $p$ as a polynomial in the variable $x_n$ with coefficients in $A[x_1,\ldots,x_{n-1}]$. So, let $B=A[x_1,\ldots,x_{n-1}]$. If $p\in A[x_1,\ldots,x_n]=B[x_n]$ satisfies $p(a)=0$ (remeber that $A\subseteq B$), then, by the case above, $p$ can be written as $p=(x_n-a)h$ for some $h\in B[x_n]$.