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Edit: This seems to be wrong, as pointed out by Will Sawin in the comments.

The prime ideals of $\mathbb{Z}$ and $\mathbb{Z}[x]$ are well-known. It is also not too hard to compute the underlying set of $\operatorname{Spec}(\mathbb{Z}[x_1, \dots, x_n])$ for arbitrary $n$; the description is as follows:

Proposition Each prime ideal of $\operatorname{Spec}(\mathbb{Z}[x_1, \dots, x_n])$ has the form $(f_0, f_1, \dots, f_n)$ where

  • $f_0$ is $0$ or a prime number,
  • $f_1 \in \mathbb{Z}[x_1]$ is $0$ or a polynomial which is irreducible over the residue field of $(f_0) \trianglelefteq \mathbb{Z}$,
  • $f_2 \in \mathbb{Z}[x_1][x_2]$ is $0$ or a polynomial which is irreducible over the residue field of $(f_0, f_1) \trianglelefteq \mathbb{Z}[x_1]$,
  • $f_3 \in \mathbb{Z}[x_1, x_2][x_3]$ is $0$ or a polynomial which is irreducible over the residue field of $(f_0, f_1, f_2) \trianglelefteq \mathbb{Z}[x_1,x_2]$,
  • etc.

Proof. We induct on $n$. The base case $n = 0$ is easy. Now suppose this classification is correct for some $n \geq 0$. There is an inclusion $i : \mathbb{Z}[x_1, \dots, x_n] \to \mathbb{Z}[x_1, \dots, x_{n+1}]$, which gives a morphism $$i^{-1} : \operatorname{Spec}(\mathbb{Z}[x_1, \dots, x_{n+1}]) \to \operatorname{Spec}(\mathbb{Z}[x_1, \dots, x_n]).$$ Take an arbitrary element $\mathfrak{p}$ of $\operatorname{Spec}(\mathbb{Z}[x_1, \dots, x_n])$, which by inductive hypothesis can be written as $\mathfrak{p} = (f_0, \dots, f_n)$ with $f_0, \dots, f_n$ as described above. The fiber of $i^{-1}$ over $\mathfrak{p}$ is homeomorphic to $$\operatorname{Spec}(k(\mathfrak{p}) \otimes_{\mathbb{Z}[x_1, \dots, x_n]} \mathbb{Z}[x_1, \dots, x_{n+1}]) \cong \operatorname{Spec}(k(\mathfrak{p})[x_{n+1}])$$ where $k(\mathfrak{p})$ denotes the residue field of $\mathfrak{p} \trianglelefteq \mathbb{Z}[x_1, \dots, x_n]$. Since $k(\mathfrak{p})$ is a field, an element of $\operatorname{Spec}(k(\mathfrak{p})[x_{n+1}])$ has the form $(\bar{f}_{n+1})$ where $\bar{f}_{n+1}$ is $0$ or an irreducible polynomial in $k(\mathfrak{p})[x_{n+1}]$.

Thus, an arbitrary element of the fiber of $i^{-1}$ over $\mathfrak{p}$ has the form $(f_0, \dots, f_n, f_{n+1})$, where $f_{n+1}$ is $0$ or a polynomial in $\mathbb{Z}[x_1, \dots, x_n][x_{n+1}]$ which is irreducible over $k(\mathfrak{p})$. $\square$

I am seeking a reference for this result. Does it appear in some classic tome on commutative algebra, for example?

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    $\begingroup$ Your statement is not quite right. For example for $n=1$ the ideal generated by $2 x_1-1$ should be generated by a monic polynomial in $x_1$, but of course $2 x_1-1$ is not monic, and no monic alternative exists. The issue is that while every ideal of $k(\mathfrak p)[x_{n+1}]$ is generated by a monic polynomial, that polynomial need not lie in $\mathbb Z[x_1,\dots, x_{n+1}]$ and putting it into $\mathbb Z[x_1,\dots, x_{n+1}]$ may make it nonmonic. $\endgroup$
    – Will Sawin
    Commented Jun 26 at 21:03
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    $\begingroup$ I don't think so - this leads to a subtler issue where if $\mathbb Z[x_1,\dots, x_n]/(f_0,\dots, f_n)$ is not a UFD, then there is no reason to think there is a unique optimal way to take a polynomial in $k(\mathfrak p)[x_{n+1}]$ and clear denominators to obtain a polynomial in $\mathbb Z[x_1,\dots,x_{n+1}]$. So the ideal could have multiple generators involving $x_{n+1}$. On the level of fields it's correct that each generator $x_i$ is algebraic or transcendental over the field generated by the previous ones but for rings more can happen. $\endgroup$
    – Will Sawin
    Commented Jun 26 at 21:08
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    $\begingroup$ Prime ideals in $\mathbb C[x_1,x_2,x_3]$ can require arbitrarily many generators math.stackexchange.com/a/2030114/84942 and the same should extend to $\mathbb Z[x_1,x_2,x_3]$, and possibly even $\mathbb Z[x_1,x_2]$. $\endgroup$
    – Will Sawin
    Commented Jun 26 at 21:10
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    $\begingroup$ In $R[x,y]$, $R$ any nonzero commutative ring, the ideal $J_n$ generated by $\{x^ay^b:a+b=n\}$ is not generated by less than $n+1$ elements (because $J_n/J_{n+1}$ is isomorphic to $R^{n+1}$ as $R[x,y]$-module, where $R$ is identified to $R[x,y]/(x,y)$. $\endgroup$
    – YCor
    Commented Jun 26 at 23:27
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    $\begingroup$ @YCor But this is in general not a prime ideal (I guess that it is prime only when $n=1$). $\endgroup$
    – Z. M
    Commented Jun 27 at 11:18

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You might want to look at Ferrero, "Prime ideals in polynomial rings in several indeterminates", Proc. Amer. Math. Soc. 125 (1997), 67-74, https://doi.org/10.1090/S0002-9939-97-03663-0. I think what he's doing there may be what you're trying to do.

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