I am trying to prove that the a maximal ideal $m \subset \mathbb{Z}[x, y]$ is of the form $m = (p, f(x), g(x, y))$, where $p \in \mathbb{Z}$ is a prime number, $f(x) \in \mathbb{Z}[x]$ is a monic irreducible polynomial which is also irreducible as $f(x) \in \mathbb{F}_p[x]$, and $g(x, y)$ satisfies some conditions.
I read an outline of a proof and I realize that it is a critical point to show that $p \in m$. This follows from the fact that $\mathbb{Q}$ cannot be a subfield of a field $\mathbb{Z}[x, y]/m$, a finitely generated $\mathbb{Z}$-algebra. In fact, $\mathop{\mathrm{ch}} \mathbb{Z}[x, y]/m$ is positive and $\mathbb{Z}[x, y]/m$ is a finite field. cf. A question on MSE, A question on MO.
Then, I wonder whether we can obtain the form of $m$ by using the inclusion $\mathbb{Z}[x] \subset \mathbb{Z}[x, y]$. Let $i \colon \mathbb{Z}[x] \to \mathbb{Z}[x, y]$ be its inclusion map. If we succeed in showing that $i^{-1}(m) \subset \mathbb{Z}[x]$ is a maximal ideal, we obtain that $m \cap \mathbb{Z}[x] = (p, f(x))$. Then, we can consider $m / i^{-1}(m) \in \mathbb{Z}[x, y] / i^{-1}(m) = F[y]$ where $F = \mathbb{Z}[x]/i^{-1}(m)$ is a field.
However, it is not necessarily true that a preimage of a maximal ideal is also a maximal ideal. I know that there are sufficient conditions that guarantee this statement as follows.
- $\varphi \colon R \to S$ is a ring homomorphism where $S$ is integral over $R$
- $\varphi \colon R \to S$ is a $k$-algebra homomorphism where $R$ and $S$ are affine $k$-algebra over an algebraically closed field $k$.
However, we cannot apply them for our argument, at least directly. I would appreciate if you could tell me a proof based on our strategy.
P.S. After I wrote this question, I found an answer for a question that a preimage of a maximal ideal is also a maximal ideal on a $\mathbb{Z}$-algebra homomorphism between finitely generated $\mathbb{Z}$-algebras. If this is true, our strategy works. I will check details when I try this problem next time. Your additional comments are welcome.
