Does a maximal ideal of $\mathbb{Z}[X,Y]$ contain a prime number?

Question

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.

Answer 1

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?

Answer 2

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 $p\in\frak m$.