Spectrum of $R[x]$

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.

By factoring out primes then localizing, it reduces to a simple case, e.g. see the exposition below from Kaplansky's Commutative Rings enter image description here

Alternatively, you could look at pages 22-23 of Miles Reid's "Undergraduate Commutative Algebra" (LMS Student Texts 29).

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 :