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 $f(x) = q(x)(x-a) \in (x-a)$.
Best Answer
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.