Find a factorisation in a cubic field. Take $\alpha$ a root of $x^3-7$ and consider $\mathbb{Q}(\alpha)$. This is a pure cubic field. I am interested in the ideals which ramify. Since the discriminant here is $27\cdot7^2$, I have to look at the ideals generated by 3 and 7 in $\mathbb{Z}[\alpha]$, which is the ring of integers. Surely $(7)=(\alpha)^3$. 
Now we consider $(3)$. I think I have to find $a,b,c\in \mathbb{Z}$ such that $$(3)=(3,\alpha+a)(3,\alpha+b)(3,\alpha+c).$$ Computing, I get a system of modular equation which solution is $a,b,c=2$. Thus $$(3)\supset(3,\alpha+2)(3,\alpha+2)(3,\alpha+2).$$ For the other inclusion, is there a better way or have I to find 3 in this product by hand? I think that not always is simple. And more, I am always sure that the ideals in the factorisation are of the form $(3,\alpha+a)$? Why not, for example, $(3,\alpha^2+a)$?
 A: Are you aware of Kummer's factorization theorem? This gives you the answer instantly. Since $x^2-7$ is the minimal polynomial of $\alpha$ over $\Bbb Q$, we can take its reduction mod $3$ and reduce it into a product of irreducibles to get the prime factorization of $(3)$ in $\Bbb Q(\alpha)$: as polynomials in $\Bbb Z/3\Bbb Z$ we have 
$$x^3-7=x^3+2=(x+2)^3$$
and therefore $(3)=\mathfrak p^3$ where $\mathfrak p=(3,\alpha+2)$.
Edit: You could also look at the "norm" function, it sends an ideal $I\subset\Bbb Z[\alpha]$ to the cardinality of $\Bbb Z[\alpha]/I$ (this is always finite if $I$ is nonzero since $\Bbb Z[\alpha]$ is the ring of integers of a number field). This norm is multilpicative with respect to products of ideals, and as a result, in general we have that if
$$I\subseteq J\text{ and } N(I)=N(J)\implies I=J.$$
Applied to our case, if you could show that
$$N((3))=N((3,\alpha+2)^3)=(N(3,\alpha+2))^3$$
(which is the same as showing that $N(3,\alpha+2)=3$), then it follows that the two ideals are equal.
