We let $R$ be a commutative ring with unity, and $R[X_1,\dots, X_n]$, $a=(a_1,\dots, a_n)\in R^n$ and $\phi_a:R[X_1,\dots, X_n]\rightarrow R$, $\phi_a(f)=f(a)$. I want to show that $\ker(\phi_a)=(X_1-a_1,\dots, X_n -a_n)$.
I know that to be true for $n=1$. I want to assume that WLOG, $a=(0,\dots,0)$, but I can't see why I'm allowed to do so (in this case it'll be easy to solve) - I didn't learn about maximal ideals yet.
Is my approach correct? How can I justify assuming $a=(0,\dots,0)$?
The universal property of $R[X_1,\ldots,X_n]$ says that for every commutative ring $S$, every ring homomorphism $\varphi\colon R\to S$, and every choice of elements $s_1,\ldots,s_n$, there exists a unique ring homomorphism $\Phi\colon R[X_1,\ldots,X_n]\to S$ such that $\Phi(X_i) = s_i$ for $i=1,\ldots, n$ and $\Phi(r)=\varphi(r)$ for each $r\in R$.
So take $S=R[X_1,\ldots,X_n]$, $\varphi\colon R\to S$ the standard embedding of $R$ into the polynomial ring as the constant polynomials, and take $s_i=X_i-a_i$. Let $\Phi\colon R[X_1,\ldots,X_n\to R[X_1,\ldots,X_n]$ be the corresponding morphism.
Verify that $\Phi$ is an isomorphism of $R[X_1,\ldots,X_n]$ to itself (for example, verify that the map $\Psi$ sending $X_i$ to $X_i+a_i$ is its inverse).
Now verify that $\phi_{\mathbf{0}}\circ \Phi = \phi_{a}$ and use $\Phi$ and $\Psi$ to transfer ideal information from one to the other.