When is $\mathbb{Z} [x]/f(x) $ a Dedekind domain? Given a monic separable irreducible polynomial $f$ with integer coefficients, when  $\mathbb{Z} [x]/f(x)$ is a Dedekind domain? And when it happens to be a Dedekind domain, how to know its class number?
The question is too broad. But do we have some useful criterion if we impose more conditions on $f$?
 A: Edit (2016-05-05): I feel I should return to this question to give a better answer.  What I had initially written (below the line) is, as far as I know, correct, but it's far from the best way of saying things.  An excellent place for clear and detailed explanations on this question is Peter Stevenhagen's paper "The Arithmetic of Number Rings" in Algorithmic Number Theory: Lattices, Number Fields, Curves and Cryptography (Buhler & Stevenhagen eds., MSRI publications 44, Cambridge 2008): Stevenhagen discusses both the question of computing integral closure and that of computing the class number.  But for the completeness of MSE, let just restate two things I might have said.
I explain below why $\mathbb{Z}[x]/(f)$ is normal (i.e., integrally closed, i.e., a Dedekind domain) iff $\mathbb{Z}_p[x]/(f)$ is normal (i.e., a discrete valuation ring — or perhaps the zero ring) for every $p$.  I point out that this is automatically true for primes which don't divide the discriminant (I wrote "or the leading coefficient", that's redundant), so we need only worry about primes which do.  I should have added: in fact, we need only worry about primes whose square divides the discriminant, because those that do not are automatically regular.  So in particular, if the discriminant of $f$ is not a square, then $\mathbb{Z}[x]/(f)$ is normal.
The second thing I missed is part of the "Kummer-Dedekind" theorem.  To state this, assume $f$ is monic (I didn't think too carefully about what happens if it is not, but essentially some primes disappear "at infinity").  Let $\bar f = \prod_{i=1}^r \varphi_i^{e_i} \in \mathbb{F}_p[x]$ be the factorization of $f$ mod $p$, where $\varphi_i$ are the distinct monic irreductible factors of $\bar f$, and let $g_i \in \mathbb{Z}[x]$ be an arbitrary monic lift of $\varphi_i \in \mathbb{F}_p[x]$; finally, let $f = q_i g_i + r_i$ be the division of $f$ by $g_i$ in $\mathbb{Q}[x]$ (which is, in fact, in $\mathbb{Z}[x]$ because $g_i$ is monic).  Then: $\mathbb{Z}_p[x]/(f)$ is normal iff $e_i = 1$ or $r_i$ is not divisible by $p^2$ (in $\mathbb{Z}[x]$).
(Furthermore, if such is the case, then $(p)$ splits as the product of the prime ideals $(p,\;g_i\mathbin{\mathrm{mod}}f)$ with exponents $e_i$; and if such is not the case, then $\frac{1}{p}\cdot(q_i\mathbin{\mathrm{mod}}f)$ is an element of the field of fractions of $\mathbb{Z}[x]/(f)$ which is integral over the latter but does not belong to it.)
Again, at some deeper level this is equivalent to what I was saying below, but it's much clearer that way, and simple enough to test by hand for not-too-complicated polynomials $f$.

The class number is a completely different question, so let me concentrate just on how to decide whether $\mathbb{Z}[x]/(f)$ is integrally closed (i.e., a Dedekind domain).
First, normality is a local condition: $\mathbb{Z}[x]/(f)$ is normal (i.e., integrally closed) iff $\mathbb{Z}_{(p)}[x]/(f)$ is normal for every prime $p$, where $\mathbb{Z}_{(p)}$ refers to the localization of $\mathbb{Z}$ at $p$ (i.e., the ring of rationals such that $p$ does not divide the denominator).  Second, normality passes to the completion (the words "excellent ring" are a magic phrase here): we can, if we wish, replace $\mathbb{Z}_{(p)}[x]/(f)$ by $\mathbb{Z}_p[x]/(f)$ in the above.
Also, we don't need to test every prime $p$, only the finitely many which divide the discriminant of $f$ or its leading coefficient: the other primes are unramified and necessarily normal (see, e.g., Serre, Local Fields, I, §6, (i), prop. 15 and cor. 1 and 2, which is also relevant in what follows) (these unramified primes appear as $e=1$ in what follows).
So now let's concentrate on a single prime $p$.
Suppose $f$ for the time being that $f$ is monic (now that we're at a single prime $p$, it's not such a strong condition: we just want the leading coefficient not to be a multiple of $p$).  If $\bar f = \prod_{i=1}^r \varphi_i^{e_i} \in \mathbb{F}_p[x]$ where $\varphi_i$ are the distinct irreductible factors of $\bar f$.  By Hensel's lemma (since $\mathbb{Z}_p$ is complete) we can lift this factorization to one of $f = \prod_{i=1}^r f_i \in \mathbb{Z}_p[x]$ where $f_i$ reduces to $\varphi_i^{e_i}$ mod $p$, and the Chinese remainder theorem tells us that $\mathbb{Z}_p[x]/(f) = \prod_{i=1}^r \mathbb{Z}_p[x]/(f_i)$, which is normal iff every factor is, so we can assume that $f = f_i$ (i.e, $\bar f = \varphi^e$ where $\varphi\in\mathbb{F}_p[x]$ is irreducible and $e$ is the ramification).  If $f$ isn't monic in the start of this paragraph, it should at least be primitive (i.e., gcd of coefficients $=1$), otherwise $\mathbb{Z}_p[x]/(f)$ certainly can't be integrally closed; even if it is primitive, some degree can be lost "at infinity": think of $f$ as a homogeneous polynomial in two variables instead, Henselize as such, and ignore the part at infinity.
So now we have $\mathcal{O} := \mathbb{Z}_p[x]/(f)$ which is a local noetherian domain of dimension $1$, whose maximal ideal $\mathfrak{P}=(p)+(\varphi)$ is the set of elements which are divisible by $\varphi$ after reduction mod $p$.  So it is normal iff it is regular iff $\mathfrak{P}/\mathfrak{P}^2$ is $1$-dimensional as a vector space over $\mathcal{O}/\mathfrak{P} = \mathbb{F}_p[x]/(\varphi)$.  At this stage, everything is quite explicit and it should be at least clear that the question is algorithmic (by doing linear algebra on $\mathbb{Z}_p$-modules, which is algorithmic); since every step thus far (factorization over finite fields, Hensel's lemma) was also algorithmic, and since I explained why we need to check only finitely many primes, the overall question is algorithmic.
A particularly important case is that when $\varphi = x$, in other words $f = x^e + a_1 x^{e-1} + \cdots + a_e$ with $a_i$ multiple of $p$ for $1\leq i \leq e$.  Then $\mathfrak{P}$ consists of those $c_0 + c_1 \xi + \cdots + c_{e-1} \xi^{e-1}$ (where $\xi$ is the class of $x$ mod $f$) such that $c_0$ is a multiple of $p$, and $\mathfrak{P}^2$ is generated by $p^2$, $p\xi$ and $\xi^i$ for $2\leq i \leq e$.  So the quotient $\mathfrak{P}/\mathfrak{P}^2$ is the $\mathbb{F}_p$-vector space of $(c_0,c_1)$ with $c_0 \in p\mathbb{Z}_p/p^2\mathbb{Z}_p$ and $c_1 \in \mathbb{Z}_p/p\mathbb{Z}_p$ quotiented out by the image of $\xi^e$ thither, which is $(-a_e,0)$.  So it is $1$-dimensional iff $a_e$ is not a multiple of $p^2$, which is to say, $f$ is an Eisenstein polynomial.  The general case will be similar, but more complicated to write down (if, instead of $\mathbb{Z}_p[x]/(f)$ we had a complete discrete valuation ring with an algebraically closed residue field, we could always reduce to the case just described).
