Can $\operatorname{Spec}(R[X])$ ever be finite? I have a quick question. 

Suppose $R$ is a nonzero commutative ring. Is it possible that $|\operatorname{Spec}(R[X])|<\infty$?

 A: Suppose that $R$ is non-zero. Then, we can find a surjection $R\to k$ for some field $k$, which induces a surjection $R[x]\to k[x]$. From this we obtain a closed embedding $\text{Spec }k[x]\to\text{Spec }R[x]$ and thus it suffices to prove that $\text{Spec }k[x]$ is infinite for $k$ a field. 
To do this we can proceed as follows. If $k$ is infinite then $(x-a)\in\text{spec }k[x]$ for all $a\in k$. If $k$ is finite, say $k=\mathbb{F}_q$, then we have infinite elements of $\text{MaxSpec }k[x]$ corresponding to the field extensions $\mathbb{F}_{q^n}$ for every $n\in\mathbb{N}$.
Note that the above actually shows that even $\text{MaxSpec }R[x]$ is infinite if $R\ne 0$.
A: Let $J=\operatorname{Rad}(R[X])$ be the Jacobson radical. I claim $R[X]/J$ has infinitely many ideals. 
Consider the principal ideal $I=(X+J)$ in $R[X]/J$. I claim that $I^n\supsetneq I^{n+1}$ for all $n\geq 1$. The containment follows immediately from definition. Suppose that $I^n=I^{n+1}$ for some $n$. Then $X^n+J\in I^n=I^{n+1}$, so $X^n+J=pX^{n+1}+J$ for some $p\in R[X]$. This implies $X^n-pX^{n+1}\in J$. Recalling $x\in J$ if $1-xy$ is a unit for all $y$ in the ring (cf. Atiyah & MacDonald Chapter 1), we have that $1-(X^n-pX^{n+1})$ is a unit in $R[X]$. However, $1-(X^n-pX^{n+1})$ is a unit iff the constant coefficient is invertible, and all other coefficients are nilpotent (cf. A&M Chapter 1, Exercises 1,2). But $X^n$ has coefficient $-1$, which is not nilpotent. So necessarily $I^n\supsetneq I^{n+1}$, and we get an infinite descending chain of ideals $I\supsetneq I^2\supsetneq I^3\supsetneq\cdots$. 
Now suppose that $R[X]$ has only finitely many prime ideals, and thus only finitely many maximal ideals, since maximal ideals are prime. Enumerate the maximal ideals as $\mathfrak{m}_1,\dots,\mathfrak{m}_n$. Since the maximal ideals are clearly pairwise coprime, the Chinese Remainder Theorem implies 
$$
R[X]/J=R[X]/\bigcap_{i=1}^n\mathfrak{m}_i\simeq R[X]/\mathfrak{m}_1\times\cdots\times R[X]/\mathfrak{m}_n=:S.
$$
It is a standard result that the ideals of $S$ have form $I_1\times\cdots\times I_n$ where $I_i$ is an ideal of $R[X]/\mathfrak{m}_i$. But each $R[X]/\mathfrak{m}_i$ is a field, hence has only trivial ideals. So $S$ has only $2^n$ ideals, a contradiction since $R[X]/J$ has infinitely many ideals. So $R[X]$ must have infinitely many prime ideals, in fact, it must even have infinitely many maximal ideals.
