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Let $\mathfrak{m}$ be a maximal ideal of $\mathbb{Z}[X,Y]$. Then does $\mathfrak{m}$ contain a prime number?

I think that for a prime ideal $\mathfrak{p} = (f_1, \dots, f_m)$ and a prime number $p$ which divides no coefficients of $f_i$. But I can't show it.

k.j.
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2 Answers2

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Hint: $\mathbb Z[X,Y]/\mathfrak m$ is a field. If $\mathfrak m$ didn't contain a prime, then this field would have characteristic zero, and hence would contain a subfield isomorphic to $\mathbb Q$. Is that possible?

Wojowu
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  • Thank you very much. Essentially I have thought it, but I can't show.... So would you tell me in a bit more detail? – k.j. Aug 28 '19 at 13:09
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    This hint is very far from a solution. Maybe you give the OP a hint for how to show that $\mathbb Q\subset \mathbb Z[X,Y]/\mathfrak m$ is not possible. (E.g. Use Zariski Lemma, Artin-Tate Lemma, etc.) – user26857 Aug 28 '19 at 21:21
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    @user26857 I only came to realize that myself a few hours ago. What I had in mind is that $\mathbb Q$ is not finitely generated as a ring over $\mathbb Z$, but of course that doesn't exclude the possibility of it lying in a larger such field. Thank you for the comment though, today I've learned! – Wojowu Aug 29 '19 at 08:15
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Let $\frak m$ be a maximal ideal of $\Bbb Z[X,Y]$.

The intersection $I=\frak m\cap\Bbb Z$ is a prime ideal of $\Bbb Z$ and there's a canonical injective map $\Bbb Z/I\rightarrow\Bbb Z[X,Y]/\frak m$.

If $I=(0)$ then the map becomes an inclusion $\Bbb Z\subset\Bbb Z[X,Y]/\frak m$ and this contradicts the fact that $\Bbb Z[X,Y]/\frak m$ should be a field because the non-zero elements from $\Bbb Z$ have no inverse.

Therefore we must have $I=(p)$ for some prime number $p$, so that $p\in\frak m$.

Andrea Mori
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  • Why $\mathbb{Z}[X,Y] / \mathfrak{m}$ don't have inverse of non-zero $n$? – k.j. Aug 28 '19 at 13:25
  • It is possible for some integers $\notin { 0, \pm 1 }$ to have inverses in the quotient, for example if $2x-1 \in \mathfrak{m}$. So, you would need some argument to show it can't happen for all integers - possibly some argument based on the fact that $\mathbb{Z}[x,y]$ is Noetherian so $\mathfrak{m}$ is finitely generated. – Daniel Schepler Aug 28 '19 at 21:23