Prime ideals in polynomial rings over integers

Im trying to find a characterization of the prime ideals in the polynomial ring $R = \mathbb Z[X,Y]$ in two variables over the integers.

Actually I need to find the maximal ideals in quotient rings $R / I$ where $I$ has codimension 2 such that $R/I$ is free and finitely generated as a $\mathbb Z$-module, i.e. I need to find the maximal ideals $I \subseteq \mathfrak m \subseteq R$. An example of such an $I$ is the ideal generated by $X^2 - XY - Y - 1$ and $X^3 - X^2 - 2XY - X - Y$.

In Mumford's Red Book of Varieties and Schemes, II, §1, Example H there is a characterization of prime ideals in $\mathbb Z[X]$. Do anyone know how to generalize that to two indeterminates?

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Let $P$ be a prime ideal of $R := \mathbb{Z}[X,Y]$. Then $P' := P \cap \mathbb{Z}$ is a prime ideal of $\mathbb{Z}.$ If $P' = (p)$ for a rational prime $p,$ then $P/p$ is a prime of $\mathbb{F}_p[X,Y]$. Otherwise, let $S = \mathbb{Z} \setminus {0} \subset \R.$ It follows that $S^{-1}P$ is a prime of $S^{-1}R = \mathbb{Q}[X,Y].$ This reduces the problem to polynomial rings over fields. –  Tom Bachmann Apr 11 '13 at 9:40
Let $k$ be a field and $P$ a prime of $k[X,Y].$ Then $p$ contains an irreducible element $f$, hence $(0) \subset (f) \subset P \subset M,$ where $M$ is a maximal ideal containing $P.$ The first inclusion is strict. Since $k[X,Y]$ has dimension two, $P=M$ or $P=(f).$ This reduces the problem to classifying irreducible polynomials and maximal ideals. Maximal ideals of $k[X_1, \dots, X_n]$ are well-known, by the Nullstellensatz. –  Tom Bachmann Apr 11 '13 at 9:43
So you are looking at the fibers of $\mathbb{A}^2_{\mathbb{Z}} \to \mathrm{Spec}(\mathbb{Z})$, which shows that $\mathbb{A}^2_{\mathbb{Z}} = \coprod_p \mathbb{A}^2_{\mathbb{F}_p}$ (as sets) with $\mathbb{F}_0 := \mathbb{Q}$. But this is trivial and works over every ring and in any dimension. It is a completely different matter to write down the prime ideals in terms of generators. For example, in dimension $1$ one uses Gauss' Lemma in the generic fiber. –  Martin Brandenburg Apr 11 '13 at 15:35