Is it true that $\mathbb Z[x_1, \ldots, x_n] / \mathcal P \simeq (\mathbb Z/ p) [ x_1, \ldots, x_n]$?

Question

Here $\mathcal P$ is a prime ideal of $\mathbb Z[x_1, \ldots, x_n]$ and $p = \mathcal P \cap \mathbb Z$ (and assume $p \neq (0)$). I think I can prove this using the 3rd part of the second isomorphism theorem for rings taking $S = \mathbb Z$ and $I = \mathcal P$, but it was a bit messy and I see this statement (or maybe it was one like it) so often without proof/explanation that I wonder if it's obvious and I'm just missing something..

If it helps, instead of indeterminants we can replace the $x_1, \ldots, x_n$ with algebraic integers $\alpha_1, \ldots, \alpha_n$ if the statement needs this in order to hold. But I didn't think it was necessary?

Answer 1

If $P$ is the ideal $(p,x)$ of $R = \mathbb Z[x]$, then $R/P \cong \mathbb Z/p\mathbb Z$.

Answer 2

For algebraic integers it is not more true, try $\Bbb{Z}[2^{1/3}]/(3,2^{1/3}-1)\cong \Bbb{F}_3$ which is neither $\Bbb{Z}[2^{1/3}]/(2)$ nor the splitting field of $x^3-2\in \Bbb{F}_3[x]$.

What is true is that for $P$ a non-zero prime ideal of $\Bbb{Z}[\alpha_1,\ldots,\alpha_n]$ (some algebraic integers) then $\Bbb{Z}[\alpha_1,\ldots,\alpha_n]/P\cong \Bbb{F}_p[\beta_1,\ldots,\beta_n]$ where $p$ is the prime number $(p) = P\cap \Bbb{Z}$ and the $\beta_j$ are some roots of the reduction $\bmod (P\cap R_{j-1})$ of the $R_{j-1}=\Bbb{Z}[\alpha_1,\ldots,\alpha_{j-1}]$ minimal polynomials of the $\alpha_j$, moreover every prime ideal containing $p$ is obtained this way.