<s>[INST] Why is any prime factor of $(3)$ also a prime factor of $(2 + \sqrt[3]{19})$ in the ring of integers of $\mathbb{Q}(\sqrt[3]{19})$ I am reading this blurb from K. Conrad's expository notes on the conductor ideal: http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/conductor.pdf
In page 3 (example 2.3) he shows that $(2 + \root 3 \of{19})$ has norm $3^3 = 27$ and that $(-1 + \root 3 \of{19})$ has norm $2 \times 3^2 = 18$, and then states that any prime factor of $(3)$ is a factor of both $(2 + \root 3 \of{19})$ and $(-1 + \root 3 \of{19})$.
Of course, if a prime factor of $(3)$ divides any one of these ideals, it will divide the other too since their generators have a difference of 3. But I am not sure why a prime divisor of $(3)$ necessarily divides any of the two ideals. Clearly $3$ itself does not divide them because $\frac{2 + \root 3 \of{19}}{3}$ is not an algebraic integer.
I suspect it must have something to do with the norm of $(2 + \root 3 \of{19})$ being a power of $3$, but perhaps I am missing something obvious and can't figure out what's happening here.
 [/INST] After a bit of experimentation, I noticed that
$$ (2 + \sqrt[3]{19})(-1 + \sqrt[3]{19})^2 = 3 \times (7 - \sqrt[3]{19}).$$
So $(3)$ divides $(2 + \sqrt[3]{19})(-1 + \sqrt[3]{19})(-1+\sqrt[3]{19})$.
Therefore, any prime ideal $\mathfrak p$ that divides $(3)$ must divide at least one of $(2 + \sqrt[3]{19})$ or $(-1 + \sqrt[3]{19})$.
But as you pointed out, the difference between $2 + \sqrt[3]{19}$ and $-1 + \sqrt[3]{19}$ is $3$, so every such $\mathfrak p$ will actually divide both $(2 + \sqrt[3]{19})$ and $(-1 + \sqrt[3]{19})$.
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