Can someone give an example of an ideal $I \subset R= \Bbb{Z}[x_1,...x_n]$ with $R /I \cong \Bbb{Q}$?

Question

Question in the title. I have never seen a quotient of $R$ by a maximal ideal $I$ that is an infinite field, so I would also be interested in the case that $R/I$ is any infinite field. If no such examples exist, I would also like to hear about the case where $R$ is a polynomial ring over $\Bbb{Z}$ in infinitely many variables.

Answer 1

Proposition
If $I\subset R=\mathbb Z[X_1,\cdots X_n]$ is a maximal ideal, then the quotient $R/I$ is a finite field.
Proof
The ring $R$ is Jacobson : this means that each of its prime ideals is the intersection of the maximal ideals which contain it.
The canonical ring morphism $f:\mathbb Z\to R$ is a morphism between Jacobson rings and thus has the wonderful property that the inverse image of a maximal ideal is maximal.
This implies in our case that $f^{-1}(I)\subset \mathbb Z$ is a maximal ideal, necessarily of the form $p\mathbb Z$ for some prime integer $p$, so that $\mathbb Z/f^{-1}(I)=\mathbb Z/p\mathbb Z=:\mathbb F_p$.
But then the extension field $\mathbb F_p \subset R/I$ is finitely generated as an algebra over $\mathbb F_p$, and is thus a finite-dimensional vector space over $\mathbb F_p$ by Zariski's version of the Nullstellensatz.
Thus we conclude that $R/I$ is a finite field $\mathbb F_{p^n}$

Bibliography
The best reference on Jacobson rings is the very last section of Chapter 5 of Bourbaki's Commutative Algebra.
Zariski's Theorem is Proposition 7.9 in Atiyah-Macdonald.
Wikipedia has the cheek to call it Zariski's lemma :-)

Answer 2

You can't find such an example: a maximal ideal $\mathfrak m$ in $\mathbf Z[x_1,\dots,x_n]$ has a non-zero intersection with $\mathbf Z$, which is a prime ideal $p\mathbf Z$, hence the quotient $\mathbf Z[x_1,\dots,x_n]/\mathfrak m$ is a finitely generated algebra over the finite field $\mathbf F_p$, which has characteristic $p$.

The same argument is valid for polynomial rings with an infinite number of indeterminates.