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.

Thank you in advance!
 A: 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. $\square$
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 $\mathbb{Z}[1/N]$.
