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we know that the maximal ideals of ${\mathbb Z}[x]$ are of the form $(p, f(x))$ where $p$ is a prime number and $f(x)$ is a polynomial in ${\mathbb Z}[x]$ which is irreducible modulo $p$.

Is it true that:

the maximal ideals of ${\mathbb Z}[x,y]$ are of the form $(p, f(x,y),g(x,y))$ where $p$ is a prime number and $f(x,y), g(x,y)$ are polynomials in ${\mathbb Z}[x,y]$ which are irreducible modulo $p$.

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Better yet, you can replace $f(x,y)$ with $f(x)$. See the answer to this question.

Edited to add: At Martin Brandenburg's request, I'm expanding this to add the details I thought were too obvious to mention:

1) A maximal ideal $M$ of ${\mathbb Z}[X,Y]$ is the kernel of a map to a field $k$.

2) Any field of characteristic zero contains ${\mathbb Q}$ and hence is not finitely generated as a ${\mathbb Z}$-algebra.

3) Therefore the field $k$ has finite characteristic $p$; therefore $M$ contains $p$.

4) Now $M/(p)$ is a maximal ideal in $({\mathbb Z}/p{\mathbb Z})[X,Y]$ and therefore (by the answer to the question linked above) has the form $(\overline{f}(X),\overline{g}(X,Y))$.

5) We can lift $\overline{f}$ and $\overline{g}$ to polynomials $f,g\in M$.

6) It is easy to check that $p,f,g$ generate $M$.

7) Because ${\overline f}(X,Y)={\overline f}(X,0)$, it follows that $f(X,Y)-f(X,0)$ maps to zero mod $p$.

8) By 7) and 6), $(p,f(X,0),g(X,Y))=(p,f(X,Y),g(X,Y))=M$, so that $M$ has generators of the advertised form.

Edited to add further: As Yves Cornulier points out in comments, step 2) above is less trivial than both I and Will Sawin made it out to be. The key additional point is that a field $k$ finitely generated over ${\mathbb Z}$ must have finite characteristic because --- by the generalized Nullstellensatz --- the unique closed point in $Spec(k)$ must map to a closed point in $Spec({\mathbb Z})$.

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    $\begingroup$ $Z$ is not field. $\endgroup$
    – Stella
    May 19, 2012 at 20:54
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    $\begingroup$ Anna: "$Z$ is not a field". But $Z/pZ$ is. $\endgroup$
    – Steven Landsburg
    May 19, 2012 at 20:55
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    $\begingroup$ One can use the lemma: Let $R$ be a ring finitely generated over $\mathbb Z$. Then a maximal ideal of $R[x]$ is a maximal ideal of $R$ plus a polynomial in $R[x]$ that is irreducible modulo that maximal ideal. Apply to $R=\mathbb Z$, then to $R=\mathbb Z[x]$, etc. Proof: The quotient is some finitely generated field, thus some finite field, so the image of $R$ is a subfield, thus the quotient by a maximal ideal $I$. The field extension is generated by $x$, so it is the quotient of $F[x]$ by the minimal polynomial $m_x$, so it is $R[x]/(I,m_x)$. $\endgroup$
    – Will Sawin
    May 19, 2012 at 21:17
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    $\begingroup$ -1 since this answer doesn't explain what the maximal ideals of $Z[x,y]$ are; the linked question only talks about maximal ideals in polynomial rings over fields. Perhaps Will Sawin can post his comment as a real answer. $\endgroup$
    – Martin Brandenburg
    May 20, 2012 at 7:30
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    $\begingroup$ It was not obvious to me, neither to anna who asked this question. Each maximal ideal in Z[x,y] intersects Z in a prime ideal, and one has to argue that this prime ideal is not 0. This has been done by Will Sawin above. Perhaps you can just add these "obvious" arguments to your answer so that the reader doesn't have to struggle through our comments. Sorry for yapping.. $\endgroup$
    – Martin Brandenburg
    May 21, 2012 at 6:30

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