The class group of the number field $\mathbb{Q}[X]/(f(X))$ Let $\alpha$ be a root of $f(X)=X^3-3X^2+X-14$ and let $F=\mathbb{Q}(\alpha)$. 
I have to find the class gloup of the number field $F$.
My attempt 
A standard calculation shows that the discriminant of $F$ is given by $\triangle_F=-6043.$ Since $\triangle_F$ is sqauare free we get $O_F=\mathbb{Z}[\alpha]\cong \mathbb{Z}[X]/(f(X))$. Now the spectrum of $O_F$ is given by $V(f(X))\subset \mathsf{Spec} \: \mathbb {Z}[X]$, therefore the prime ideals of $O_F$ are the those of the type $(p)$, such that $p\in\mathbb{Z}$ is irreducible and $f(X)$ is irreducible $\mathsf{mod} \: p$.
Now, the class group $\mathsf{Cl}(O_F)$ is generated by the prime ideals of $O_F$ such that $N((p))=p^3<M_F$, where $M_F$ is the Minkowski constant associated to $F$, that in this case is given by $$M_F=\dfrac{3!}{3^3}\dfrac{4}{\pi}\sqrt{6043}\approx 21.994996\ldots$$
Since $(2)\subset O_F$ is not prime, because $f(X)=X(X^2-3X+1)$ in $\mathbb{F}_2[X]$ and $N((3))=27>M_F$, we obtain $$\mathsf{Cl}(O_F)=0.$$
Is my attempt correct?
 A: The two main results that you need are:
1) The Kummer-Dedekind Theorem which, as carmichael561 mentions in the comments, allows you to determine the factorization of a rational prime $p$ in $\mathcal{O}_F$ from the factorization of $f(X)$ mod $p.$
2) The application of Minkowski's theorem which says that "the ideal class group is generated by the classes of the prime ideals of norm less than the Minkowski bound $M_F$".

STEP 1: Find a small set of generators for $Cl_F.$
In order to find the prime ideals of norm less than $M_F\approx 21.99,$ we can factorize the rational primes 2, 3, 5, 7, 11, 13, 17, 19 using Kummer-Dedekind. This gives the following prime ideal factorizations:
$2O_F = P_2Q_2$ where $P_2$ has norm 2 and $Q_2$ has norm $2^2$
$3O_F = P_3Q_3$ where $P_3$ has norm 3 and $Q_3$ has norm $3^2$
$5O_F = P_5Q_5$ where $P_5$ has norm 5 and $Q_5$ has norm $5^2$
$7O_F = P_7Q_7$ where $P_7$ has norm 7 and $Q_7$ has norm $7^2$
$11O_F = P_{11}Q_{11}$ where $P_{11}$ has norm 11 and $Q_{11}$ has norm $11^2$
$13O_F = P_{13}Q_{13}$ where $P_{13}$ has norm 13 and $Q_{13}$ has norm $13^2$
$17O_F = P_{17}$ where $P_{17}$ has norm $17^3$
$19O_F = P_{19}Q_{19}$ where $P_{19}$ has norm 19 and $Q_{19}$ has norm $19^2$
It thus follows that the class group $Cl_F$ is generated by the classes of the prime ideals:
$$\{P_2,P_3,P_5,P_7,P_{11},P_{13},P_{19}\}.$$

STEP 2: Find relations amongst the generators found in STEP 1.
Let $\alpha$ be a root of $f(X).$ We observe that for any $n \in \mathbb{Z}$ we have
$$N((\alpha+n)O_F)=|f(-n)|.$$
In particular, we have
$N(\alpha O_F)=14 \Rightarrow \alpha O_F=P_2P_7 \Rightarrow$ we can discard $P_7$ from our generating set.
$N((\alpha+1)O_F)=19 \Rightarrow (\alpha+1) O_F = P_{19} \Rightarrow$ we can discard $P_{19}$ from our generating set.
$N((\alpha-1)O_F)=15 \Rightarrow (\alpha-1) O_F = P_3P_5 \Rightarrow$ we can discard $P_5$ from our generating set.
$N((\alpha-3)O_F)=11 \Rightarrow (\alpha-3) O_F = P_{11} \Rightarrow$ we can discard $P_{11}$ from our generating set.
$N((\alpha+4)O_F)=130 \Rightarrow (\alpha+4) O_F = P_2P_5P_{13} \Rightarrow$  we can discard $P_{13}$ from our generating set.
$N((\alpha-4)O_F)=6 \Rightarrow (\alpha-1) O_F = P_2P_3 \Rightarrow$ we can discard $P_3$ from our generating set.
It therefore follows that $Cl_F$ is cyclic generated and that the class of $P_2$ is a generator.
It remains to compute the order of the class of $P_2$ in $Cl_F.$
Observe that $N((\alpha-2)O_F)=16.$
The Kummer-Dedekind theorem tells us that 
$$Q_2=(2,\alpha^2+\alpha+1)O_F$$
and so we see that $Q_2$ doesn't divide $(\alpha-2)O_F.$ 
[Proof: $\alpha-2 \in Q_2 \Rightarrow \alpha=(\alpha-2)+2 \in Q_2 \Rightarrow1=(\alpha^2+\alpha+1)-(\alpha+1)\alpha \in Q_2,$ contradiction.]
Hence we must have $(\alpha-2)O_F=P_2^4$ and so the order of the class of $P_2$ is either 1, 2 or 4.
We now show that neither $P_2$ nor $P_2^2$ is principal...
