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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.


marked as duplicate by user26857 abstract-algebra Jan 18 '17 at 20:25

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    $\begingroup$ Since $R$ is finitely generated and $\mathbb{Q}$ is not, this is impossible. The quotient of any finitely generated ring modulo any ideal is finitely generated. $\endgroup$ – Lee Mosher Jan 17 '17 at 21:06
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    $\begingroup$ Ah, that's a pretty easy argument. Can $R/I$ ever be an infinite field other than the rationals? $\endgroup$ – Vik78 Jan 17 '17 at 21:08

If $I\subset R=\mathbb Z[X_1,\cdots X_n]$ is a maximal ideal, then the quotient $R/I$ is a finite field.
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 $\mathbb F_{p^n}$

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 :-)

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    $\begingroup$ Dear Bernard, thank you and above all bravo for your sense of fairplay, which I value higher than any post on mathematics ! $\endgroup$ – Georges Elencwajg Jan 17 '17 at 22:53

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.

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    $\begingroup$ I think you meant to write characteristic $p$ instead of zero. So you have shown that when the number of intermediates is finite, the quotient must be a finite field, and when the number of intermediates is infinite the quotient has positive characteristic. Is it possible that when there are infinitely many indeterminates, the quotient is an infinite field? If so, can you give an example? $\endgroup$ – Vik78 Jan 17 '17 at 21:30
  • $\begingroup$ Sorry, meant "indeterminates" instead of "intermediates." $\endgroup$ – Vik78 Jan 17 '17 at 21:35
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    $\begingroup$ I've added a sketch of why the intersection is non-zero. As to the quotient being an infinite field, I think the construction of the algebraic closure of $\mathbf F_p$ would work ( the algebraic closure of any field begins with considering the ring pf polynomials with indeterminates indexed by irreducible univariate polynomials with coefficients in the field). $\endgroup$ – Bernard Jan 17 '17 at 21:50
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    $\begingroup$ Dear Bernard, your "Added" section is not convincing: you might have $\mathfrak p\cap \mathbb Z=0$ and $\mathfrak m+p\mathbf Z[x_1,\dots,x_n]=\mathbf Z[x_1,\dots,x_n]$ for all prime $p$. There is no elementary argument for proving that $\mathfrak m\cap \mathbb Z=0$ $\endgroup$ – Georges Elencwajg Jan 17 '17 at 21:58
  • $\begingroup$ Dear Bernard, I didn't downvote your answer: I prefer having a courteous discussion with users rather than downvoting them. $\endgroup$ – Georges Elencwajg Jan 17 '17 at 22:32

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