Spectrum of $R[x]$ The spectrum of $\Bbb Z[x]$ is well known : a prime ideal of $\Bbb Z[x]$ is or $(Q, p)$, with $Q \in \Bbb Z[x]$ zero or irreducible modulo $p$, and $p$ prime or zero.
If I'm not mistaken, we have a similar result for $R[x]$, when $R$ is a domain with finite Krull dimension : under this hypothesis, a prime ideal of $R[x]$ is  $(Q) + \mathfrak p$, with $\mathfrak p$ a prime ideal of $R$ and $Q$ zero of irreducible modulo $\mathfrak p$.
Some years ago I saw a paper dealing with this question of the spectrum of $R[x]$ in great details, but I'm unable to find it again. Could you help me ?
 A: By factoring out primes then localizing, it reduces to a simple case, e.g. see the exposition below from Kaplansky's Commutative Rings

A: The conjecture that every prime ideal $\mathfrak P\subset R[z]$ is of the form $\mathfrak P  =\mathfrak p+(Q)$ is not true.  
Let $k$ be a field and $R=k[x,y]$, the polynomial ring over $k$ with two indeterminates.
Consider the curve in $C\subset \mathbb A^3_k$ given parametrically by $(t^3,t^4,t^5) \quad (t\in A^1_k)$.
   Its ideal is the prime ideal $I(C)=\mathfrak P=(y^2-xz,x^3-yz,z^2-x^2y)\subset k[x,y,z]=R[z]$.  
The important point is that this ideal cannot be generated by two polynomials, i.e.  $C$ is not an ideal-theoretic  complete intersection: see a proof here.
So it is certainly not true that $\mathfrak P$ is of the required form $ \mathfrak P=\mathfrak p+(Q)$ (with $\mathfrak p$ prime in $R=k[x,y]$), because we would have $\mathfrak p=(f(x,y))$  and thus $\mathfrak P=(f,Q)$, which would falsely imply  that $C$ is a complete intersection.
A: Alternatively, you could look at pages 22-23 of Miles Reid's "Undergraduate Commutative Algebra" (LMS Student Texts 29).
A: I still haven't found again the paper I was looking for, but the answers of Bill and Georges show me that I was mistaken and also show me how to fix it.
Finally, it has nothing to with the Krull dimension.
So let $R$ be a ring (commutative, unital).
The prime ideals of $R[x]$ are precisely :
  $$\mathfrak p R[x] + R[x]\cap(Q R_{\mathfrak p}[x])$$ with $\mathfrak p$ a prime ideal of $R$ and $Q$ a polynomial of $R[x]$ which is zero irreducible in $k(\mathfrak p)[x]$, where $k(\mathfrak p)$ is the residual field of $\mathfrak p$.
More over, we have $$ \mathfrak p_1 R[x] + R[x]\cap(Q_1 R_{\mathfrak p_1}[x]) \varsubsetneq \mathfrak p_2 R[x] + R[x]\cap(Q_2 R_{\mathfrak p_2}[x])$$ if and only if one of the following holds :


*

*$\mathfrak p_1 = \mathfrak p_2$ and $Q_1 = 0$

*$\mathfrak p_1 \varsubsetneq \mathfrak p_2$ and $Q_2$ divides $Q_1$ in $k(\mathfrak p_2)$.


In the example of Georges, we have
$$ I(C) = (x^4-y^3) + R[z]\cap\left( \left(z-\frac{x^2}{y}\right) R_{(x^4-y^3)}[z] \right). $$
