Maximal ideals of $R[x_1,\ldots,x_n]$ that is $R$ is a commutative rings with identity

Question

Let $R$ be a commutative ring with identity and $R[x_1,\ldots,x_n]$ a polynomial ring over $R$. What are maximal ideals in $R[x_1,\ldots,x_n]$? How are, if $R$ is a Hilbert ring (Jacobson ring)?

Answer 1

0) I don't think there is a reasonable general, non tautological answer to this question.
Here, however, are some generalities about your problem.

1) The basic tool is certainly to use the morphism of schemes $f:\mathbb A^n_R\to \operatorname {Spec} R$ associated to the inclusion $R\subset R[x_i,\dots, x_n]$.
In some sense one can reduce the problem to the case where $R=\kappa$ is a field:

2) For each maximal ideal $\mathfrak m\subset R[x_1,\dots, x_n]$ the image $f([\mathfrak m])\in \operatorname {Spec} R$ of the closed point $[\mathfrak m]\in \mathbb A^n_R$ is the point $[\mathfrak p]\in \operatorname {Spec} R$ corresponding to the prime ideal $\mathfrak p=\mathfrak m \cap R$.
The problem is that $\mathfrak p$ has no reason to be maximal!
[That's exactly why (affine) schemes were created: to take care of such prime but not maximal ideals.]
So we have to look at all the fibers $f^{-1}[\mathfrak p]\; (\mathfrak [p]\in \operatorname {Spec} R)$ of $f$ and these fibers are the affine spaces $\mathbb A^n_{\kappa(p)}$ (where $\kappa(p)=\operatorname {Frac}(R/\mathfrak p)$ )

3) All the closed points $[\mathfrak m]\in \mathbb A^n_R$ you are interested in will be among the closed points of their fiber $f^{-1}(f(\mathfrak m))=f^{-1}(\mathfrak p)=\mathbb A^n_{\kappa(p)}$ but, alas, you will also get parasitic points closed in the fiber but not closed in $\mathbb A^n_R$:

4) This can happen even for a Jacobson ring as simple as $R=\mathbb Z$ and when $n=1$.
For example the prime ideal $\mathfrak p=(x) \subset \mathbb Z[x]$ is not closed in $\operatorname {Spec} \mathbb Z[x]$ but it is closed in its fiber $\operatorname {Spec}\mathbb Q[x]$, which is the generic fiber of $f$, i.e. the fiber over $\eta=[(0)]\in \operatorname {Spec} \mathbb Z$.