# Are there Infinite Quotients of Algebraic Extensions of $\mathbb{Z}$?

It is well known that $\mathbb{Z}[a_1, \dots, a_n]/(a)$ is a finite ring if each $a_i$ is an algebraic integer and $a \neq 0.$

I suppose this statement becomes wrong if we just require those $a_i$ to be arbitrary algebraic numbers, but unfortunately I failed in finding any counterexample until now. I would appreciate it if somebody could name one. I am particularly interested in the case where $a$ is irreducible.

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Undoubtedly you know that any algebraic number can be written in the form "an algebraic integer divide by a rational integer". Somehow this suggests to me that you could learn something by first looking at the toy case $n=1$, $a_1=1/2$. May be you already did study it? –  Jyrki Lahtonen Aug 3 '13 at 17:49
@JyrkiLahtonen: Thank you for this hint! I already had similar thoughts and considered some quotients of say $\mathbb{Z}[\frac{1}{2}], \mathbb{Z}[\frac{\sqrt{2}}{3}]$ and so on - but without success. –  Dune Aug 3 '13 at 17:53
@Dune Let me point something out, that may be helpful. The fact that $\mathbb{Z}[a_1,\ldots,a_n]/(a)$ is always a finite ring relies heavily upon the fact that $\mathbb{Z}[a_1,\ldots,a_n]$ is a finitely generated group. Because then, $(a)$, being just $a\mathbb{Z}[a_1,\ldots,a_n]$ has the same finite rank as $\mathbb{Z}[a_1,\ldots,a_n]$. Thus, their quotient is a finitely generate, rank 0 abelian group, which forces it to be finite. Thus, when you start moving to non-integers, so tha the adjunction ring is no longer finite rank/free that's where you're problem lies. –  Alex Youcis Aug 3 '13 at 18:22
If $R$ is a ring, $a\in R$, and $S\subset R$ is multiplicatively closed, then $S^{-1}R/aS^{-1}R\cong S^{-1}\left(R/aR\right)$ (where the $S$ on the right hand side really means the image of $S$ in $R/aR$). In particular, if $R/(a)$ is finite, the any localization of it is also finite, which means $S^{-1}R/aS^{-1}R$ is finite. This implies any localization of a ring of the form $\mathbb{Z}[a_1,\ldots,a_n]$ (with the $a_i$ algebraic integers) has the property that the quotient by a non-zero ideal is finite. –  Julian Rosen Aug 3 '13 at 18:41
@Dune If a ring $R=\mathbb{Z}[a_1,\ldots,a_n]$ with the $a_i$ algebraic numbers is a countexample (i.e., contains a non-zero element such that the quotient by that element is infinite), then $R$ cannot be a localization of an algebra generated over $\mathbb{Z}$ by algebraic integers. In particular, the rings $\mathbb{Z}[1/2]$ and $\mathbb{Z}[\sqrt{3}/2]=\mathbb{Z}[\sqrt{3}][1/2]$ have the property that the quotient by a non-zero ideal is finite. –  Julian Rosen Aug 3 '13 at 19:59

Every ring of the form $\mathbb{Z}[a_1,\ldots,a_n]$ with the $a_i$ algebraic numbers has the property that the quotient by a non-zero ideal is finite.

Fix an algebraic closure $\overline{\mathbb{Q}}$ of $\mathbb{Q}$, and let $a_1,\ldots,a_n\in \overline{\mathbb{Q}}$. Set $R=\mathbb{Z}[a_1,\ldots,a_n]$. One can check that $R$ is a $1$-dimensional Noetherian domain. This means that for $0\neq a\in R$, $R/(a)$ is $0$-dimensional, and therefore Artinian.

Artinian rings are finite products of Artin local rings, and each Artin local factor of $R/(a)$ has the property that its residue field is finitely-generated as a $\mathbb{Z}$-algebra, and hence finite. Artin local rings with finite residue field are finite, so this completes the proof.

EDIT: I've added a proof that $\mathbb{Z}[\alpha_1,\ldots,\alpha_n]$ is $1$-dimensional.

Lemma: Suppose $S$ is a Noetherian domain of Krull dimension at most $1$. Let $K$ be an algebraic closure of the fraction field of $S$, and take $s\in K$. Then $S[s]$ is a Noetherian domain of dimension at most $1$.

Proof: Let $x$ be an indeterminate, and consider the surjection of $S$-algebras $\varphi:S[x]\to S[s]$, $x\mapsto s$, inducing an isomorphism $S[x]/\ker(\varphi)\cong S[s]$.

Now, $S[x]$ has dimension at most 2. Because $s$ is algebraic over the fraction field of $S$, $\varphi$ is not injective. This means that $\ker(\varphi)\neq (0)$, so $S[x]/\ker(\varphi)$ has dimension strictly smaller than the dimension of $S[x]$, i.e., $S[x]/\ker(\varphi)\cong S[s]$ has dimension at most $1$.

Finally, $S[s]$ is a domain because it is a subring of a field, and is Noetherian, as it is finitely generated over a Noetherian ring.

By induction, this lemma implies that for $S$ a $1$-dimensional Noetherian domain and $s_1,\ldots,s_k$ in an algebraic closure of the fraction field of $S$, $S[s_1,\ldots,s_k]$ is a Noetherian domain of dimension at most $1$.

Note that $R=\mathbb{Z}[a_1,\ldots,a_n]$ cannot be $0$-dimensional (i.e., a field) because $a_1,\ldots,a_n$ have some common denominator $N\in\mathbb{Z}$, and $R[1/N]$ is integral over the $1$-dimensional ring $\mathbb{Z}[1/N]$.

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That is surprising! I would be happy if you add a hint how to show $\dim(R)=1$. :) –  Dune Aug 3 '13 at 20:15
@Dune If $S$ is a 1-dimensional Noetherian domain, and $f(x)\neq 0\in S[x]$ is prime, then $S[x]$ is 2-dimensional, and $S[x]/f(x)$ is a 1-dimensional Noetherian domain (because $(f(x))$ is a height 1 prime). Using this and induction on the number of generators of $R$ shows that $\dim(R)=1$. I bet there's a simpler argument, though. –  Julian Rosen Aug 3 '13 at 20:24
Great, thank you very much! –  Dune Aug 3 '13 at 20:30
@PinkElephants: Thank you for this detailed explanation! But can't we simply deduce that $R$ is one dimensional since $R[\frac{1}{N}]$ is integral over $\mathbb{Z}[\frac{1}{N}]$ and therefore one dimensional by the going up theorem? –  Dune Aug 3 '13 at 23:48