Factorizing ideal in pure cubic field 
If $K=\mathbb{Q}[\sqrt[3]{19}]$, what are the prime ideals in the factorization of $3\mathcal{O}_K = \mathfrak{p}_1^2\mathfrak{p}_2$. 

I'm not able to determine $\mathfrak{p}_1$ and $\mathfrak{p}_2$ using the methods suggested here: $\mathbb{Q}(\sqrt[3]{17})$ has class number $1$
I can factorize $2\mathcal{O}_K, 5\mathcal{O}_K$ and $7\mathcal{O}_K$ using standard theorem (factorizing $x^3-19 \mod p$).
I can factorize it using SageMath but don't know the algorithm it is using to calculate factors (since it can compute factors for any number field).
 A: EDIT: The following is a complete rewrite of my original posting here. Not that I told any (or very many) lies, it’s just that what I wrote was woefully incomplete. I originally said that doing this computation shouldn’t be hard once you knew the integral basis, and I find myself rather humbled to discover that the task was not at all easy. 
The field is $\Bbb Q(z)$, where $z^3=19$, and a good integral basis is $\{1,z,\frac{1+z+z^2}3\}$. For convenience, I’ll call $(1+z+z^2)/3=b$.
To shortcut the whole argument below, let me say that $\mathfrak p_1=(3,b)$ and $\mathfrak p_2=(3,b-1)$. If you want to see why, read on.
The first thing to notice is that $(z-1)b=6$, which allows you to compute $(3,z-1)(3,b)=(9,3b,3(z-1),6)=(3,3b,3(z-1))=(3)$. If we call $I=(3,z-1)$ and $J=(3,b)$, then $IJ=(3)$, which is as far as I got in my original answer to this question. Now, the norm of $(3)$ is $27$, and so one of $I,J$ has norm $3$, the other has norm $9$. But which is which? The one of norm $3$ is maximal, so prime.
But notice that $b^2=\frac{13 + 7z+z^2}3=4+2z+b$, so that $4+2z\in(3,b)=J$, so $-2+2z\in J$, and consequently $z-1\in J$. It follows that $I\subset J$, properly, so that $J$ is the maximal ideal, and since $IJ=(3)=\mathfrak p_1^2\mathfrak p_2$, it follows that $J=\mathfrak p_1$ and $I=\mathfrak p_1\mathfrak p_2$.
How to find $\mathfrak p_2$? I looked at the minimal polynomial for $b$, which is $X^3-X^2-6X-12$, and modulo $3$, this factors as $X^2(X-1)$. The $X^2$-factor has something to do with the ideal that $b$ is in, namely $J=\mathfrak p_1$, and the factor $X-1$ should have to do with the other prime. Anyway, I tried the ideal $(3,b-1)=(3,\frac{-2+z+z^2}3)=\mathfrak a$, and by fairly laborious hand computation, I saw, first, that $\mathfrak a\mathfrak p_1=(3,z-1)=I$, so that $\mathfrak a=\mathfrak p_2$, and to be sure, I calculated $\mathfrak p_1^2\mathfrak p_2$ and sure enough got $(3)$. You do see at a glance that $\mathfrak p_1+\mathfrak p_2=(1)$, the full ring of integers. I haven’t included here any of the ideal-multiplication computations beyond the one I showed at the beginning for $IJ=(3)$, but they all go like that, they’re just more complicated.
Notice that I didn’t use any systematic method, I just thrashed around till I got the right combination of ingredients.
