To answer Question 1: Yes, there do exist integrally closed abstract number rings with infinite class group.
By factorization of ideals, for $R$ to be an abstract number ring it is enough that it is a Dedekind domain with finite residue field $R/\mathfrak{p}$ at each prime $\mathfrak{p}$. Theorem B of the paper mentioned by Hagen Knaf in his answer actually gives what you ask for (R. C. HEITMANN, PID’S WITH SPECIFIED RESIDUE FIELDS, Duke Math. J. Volume 41, Number 3 (1974), 565-582).
Theorem B: Let G be a countable abelian torsion group. Then there is a countable Dedekind domain of characteristic 0 whose class group is G, and whose residue fields are those of the integers (i.e. one copy of $\mathbb{Z}/p\mathbb{Z}$ for each prime $p$).
As such rings have finite residue fields, this gives an integrally closed abstract number ring with class group any countable torsion group you like.
We can do much better than this though. After thinking about your question for a bit, I see how we can construct the following, so that all countable abelian groups occur as the class group of such rings.
Let G be a countable abelian group. Then, there is a Dedekind domain $R$ with finite residue fields such that $\mathbb{Z}[X]\subseteq R\subseteq\mathbb{Q}(X)$ and ${\rm Cl}(R)\cong G$.
I see some surprise mentioned in the comments below that it is enough to look at over-rings of $\mathbb{Z}[X]$ to find Dedekind domains with any countable class group. In fact, over-rings of $\mathbb{Z}[X]$ are very general in terms of prime ideal factorization, and can show the following. I'll use ${\rm Id}(R)$ for the group of fractional ideals of $R$ and $R_{\mathfrak{p}}$ for the localization at a prime $\mathfrak{p}$, with $\bar R_{\mathfrak{p}}$ representing its completion (which is a compact discrete valuation ring (DVR) in this case).
Let $R$ be a characteristic zero Dedekind domain with finite residue fields. Then, there is a Dedekind domain $R^\prime$ with $\mathbb{Z}[X]\subseteq R^\prime\subseteq\mathbb{Q}(X)$ and a bijection $\pi\colon {\rm Id}(R)\to{\rm Id}(R^\prime)$ satisfying
- $\pi(\mathfrak{ab})=\pi(\mathfrak{a})\pi(\mathfrak{b})$.
- $\pi(\mathfrak{a})$ is prime if and only if $\mathfrak{a}$ is.
- $\pi(\mathfrak{a})$ is principal if and only if $\mathfrak{a}$ is.
- If $\mathfrak{p}\subseteq R$ is a nonzero prime then $\bar R_{\mathfrak{p}}\cong\bar R^\prime_{\pi(\mathfrak{p})}$.
In particular, the class groups are isomorphic, ${\rm Cl}(R)\cong{\rm Cl}(R^\prime)$.
The idea is that we can construct Dedekind domains in a field $k$ by first choosing a set $\{v_i\colon i\in I\}$ of discrete valuations on $k$ and, letting $k_v=\{x\in k\colon v(x)\ge0\}$ denote the valuation rings, we can take $R=\bigcap_ik_{v_i}$. Under some reasonably mild conditions, this will be a Dedekind domain with the valuations $v_i$ corresponding precisely to the $\mathfrak{p}$-adic valuations, for prime ideals $\mathfrak{p}$ of $R$. In this way, we can be quite flexible about constructing Dedekind domains with specified prime ideals (and, with a bit of work, specified principal ideals and class group). Constructing discrete valuations $v$ on $k=\mathbb{Q}(X)$ is particularly easy. Given a compact DVR $R$ of characteristic 0 and field of fractions $E$, every extension $\theta\colon k\to E$ gives us a valuation $v(f)=u(f(X))$ where $u$ is the valuation in $E$. To construct such an embedding only requires choosing $x\in E$ which is not algebraic over $\mathbb{Q}$ and, if we want the localization $k_v$ to have completion isomorphic to $R$, then we just need $\mathbb{Q}(x)$ to be dense in $E$. There's plenty of freedom to choose $x\in R$ like this. In fact, there's uncountably many $x$, as they form a co-meagre subset of $R$. So, we have many many valuations on $\mathbb{Q}(X)$ corresponding to any given compact DVR. In this way, we have a lot of flexibility in constructing Dedekind domains in over-rings of $\mathbb{Z}[X]$.
I've written out proofs of these statements. As it is much too long to fit here, I'll link to my write-up: Constructing Dedekind domains with prescribed prime factorizations and class groups. Hopefully there's no major errors. I'll also mention that this is an updated and hopefully rather clearer write-up than my initial link (which were very rough notes skipping over many steps).
I think also that my linked proof can be modified to show that you can simultaneously choose any prescribed unit group of the form $\{\pm1\}\times U$ where $U$ is a countable free abelian group.
Yes.
Consider the adjacency matrices
$$ A = \left[\begin{array}{rrrrrrrrrrr}
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\
1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\\\
0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\\\
0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0
\end{array}\right] $$
and
$$ B = \left[ \begin{array}{rrrrrrrrrrr}
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\
1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\
0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\\\
0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \\\\
0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0
\end{array}\right]. $$
These are both the adjacency matrices of trees, and both have characteristic polynomial
$$\lambda^{11}-10\lambda^9+34\lambda^7-47\lambda^5+25\lambda^3-4\lambda.$$
Each tree
has exactly two vertices of degree 3, separated by a path of length 1 in the case of $A$ but length 2 in the case of $B$. In particular, the trees are not isomorphic.
Now consider the [EDIT: improved, much nicer] matrix
$$ C = \left[\begin{array}{rrrrrrrrrrr}
1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & -1 \\\\
0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\\\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\\\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\\\
0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\\\
\end{array}\right] $$
with determinant $-1$.
Since $C^{-1}AC = B$, the two trees (on 11 vertices) are non-isomorphic but
have adjacency matrices that are conjugate over $\mathbb Z$.
Now to explain the where the example comes from. The pair of graphs
was constructed by a method, attributed to Schwenk, that I found in Doob's
chapter of Topics in algebraic graph theory (edited by Beineke and Wilson).
The first 9 rows and columns of $A$, in common with $B$, come from a particular tree on 9 vertices that has a pair of attachment points such that extending the tree in the same way from either point gives isomorphic spectra.
Adding a single pendant vertex cannot work for this problem, as I found using Brouwer and van Eijl's trick, mentioned
by Chris Godsil, of comparing the Smith normal forms of (very) small polynomials in $A$ and $B$, in this case $A+2I$ and $B+2I$. When a path of length two is added at either of the two special vertices, however, there doesn't seem to be any obstruction of this type.
I then set about trying to conjugate both $A$ and $B$, separately, to the companion matrix of their mutual characteristic polynomial, by looking for a random small integer vector $x$ for which the matrix $X_A = [ x\ Ax\ A^2x\ \ldots\ A^{10}x]$ has determinant $\pm 1$, and similarly $y$ giving $Y_B$. (The fact that I succeeded fairly easily may have something to do with the fact that $A+I$ is invertible over $\mathbb Z$.) The matrix
$X_AY_B^{-1}$ then acts like the $C$ above.
[EDIT: The actual matrix $C$ I found at random and first posted was not nearly so pretty, with a Frobenius norm nearly ten times the current example. But taking powers 0 to 10 of $A$ times $C$ gave a $\mathbb Q$-basis for the full space of conjugators, whose Smith normal form (as 11 vectors in $\mathbb R^{121}$) was all 1's down the diagonal, so in fact it was a $\mathbb Z$-basis. Performing an LLL reduction on this lattice basis then gave a list of smaller-norm matrices, the third of which is the more illuminating $C$ given above, of determinant $-1$. The other determinants from the reduced basis were all $0$ and $\pm 8$.]
Taking rational $x$ and not restricting the determinant of $X_A$ gives a space of possible rational matrices $C$ of dimension 11, which are generically invertible; varying $y$ gives the same space [EDIT: as does multiplying on the left by powers (or in the more general case commutants) of $A$]. Since the spectrum of $A$ has no repeated roots, this is also the dimension of the commutant of $A$, and every matrix conjugating $A$ to $B$ lies in this space. Starting with a rational basis, it is not hard to find an exact basis for the integer sublattice, and taking the determinant of a general point in the integer lattice gives an integer polynomial in 11 variables which takes the value $1$ or $-1$ if and only if the matrices $A$ and $B$ are conjugate over $Z$. If there are repeated roots, you have to work a little harder; in general the full space has dimension the sum of the squares of the multiplicities, and is generated by multiplying on the left by a basis for the commutator space of $A$. A basis for the commutant can be produced (for a diagonalizable matrix) by first conjugating $A$ to a direct sum of companion matrices for the irreducible factors of the characteristic polynomial, and then one at a time, for each $k$-by-$k$ block corresponding to a $k$-times repeated factor of degree $m$, replacing each of the $k^2$ blocks with powers $0$ to $m-1$ of the companion matrix for that factor, with $0$ everywhere elsewhere.
Best Answer
A very belated answer to 1), but: I just saw that this is treated very nicely in Curtis and Reiner's Representation Theory of Finite Groups and Associative Algebras. Theorem 20.6 therein not only works with non-maximal orders $\mathbb{Z}[\theta]$ but even does the non-commutative case as well. And the argument is very clean and simple.
2) is a nice question by the way, and one that I have wondered about over the years. As Franz says, you need to be careful about what you mean by the "ideal class group" in this case. For any domain $R$ the quotient of the monoid of the nonzero ideals of $R$ by the submonoid of nonzero principal ideals gives you a monoid, say $H(R)$: let's call it the ideal class monoid. This is the monoid that your question 1) asks for a proof of the finiteness of when $R$ is an order in a number field $K$. (Well, your question asks about the monogenic case, but the proof of course doesn't need that.) It is known that $H(R)$ is a group iff $R$ is a Dedekind domain, so in the arithmetic case iff $R$ is integrally closed in its fraction field. In every other case there are non-invertible ideal classes, so to get a group you should pass to the group of units of the monoid $H(R)$: by definition this group is the Picard group of $R$ (it is also the group of isomorphism classes of rank one projective $R$-modules under tensor product, or if you like of isomorphism classes of line bundles on the scheme $\operatorname{Spec} R$). If e.g. $R$ is Noetherian, integrally closed, there is also a divisor class group $\operatorname{Cl}(R)$ whose nonvanishing is the obstruction to $R$ being a UFD. There is a canonical injection $\operatorname{Pic}(R) \hookrightarrow \operatorname{Cl}(R)$ which need not be an isomorphism: the most famous example is probably $\mathbb{C}[x,y,z]/(xy-z^2)$ which has two-element divisor class group but trivial Picard group. The map is an isomorphism if $R$ is "locally factorial", so e.g. if $R$ is nonsingular, hence always in dimension one.
The one example I know of $H(R)$ for a nonmaximal order is $R = \mathbb{Z}[\sqrt{-3}]$, where the monoid $H(R)$ has order $2$. There is up to isomorphism exactly one order $2$ commutative monoid which is not a group: the nonidentity element $\bullet$ must be "absorbing": $x \bullet = \bullet$ for all $x$. In this case a representative for the absorbing element is the prime ideal $\mathfrak{p} = \langle 2, 1+ \sqrt{-3} \rangle$. This is the unique non-principal prime ideal in $R$ (something which cannot happen in a Dedekind domain!). I would be interested to know if anything whatsoever is known or conjectured about which finite non-cancellative monoids arise as $H(R)$ as $R$ ranges over orders in algebraic number fields.
If $R$ is a nonmaximal order in a number field $K$ with maximal order $\tilde{R}$, there is a nonzero ideal $\mathfrak{f}$, the conductor, which is the set of all $x \in \tilde{R}$ such that $\tilde{R} x \in R$. This is an ideal in both $\tilde{R}$ and $R$, and in fact it is the largest ideal of $\tilde{R}$ which is also an ideal of $R$. The primes $\mathfrak{p}$ of $R$ containing $\mathfrak{f}$ are precisely the singular points on $\operatorname{Spec} R$. Every ideal prime to $\mathfrak{f}$ is invertible and factors uniquely into primes, and every invertible ideal can be adjusted in its equivalence class to be prime to $\mathfrak{f}$. Thus the noninvertible ideals are the ones with support which cannot be "moved off of $\mathfrak{f}$". This material can be found in Neukirch's Algebraic Number Theory.