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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.

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  • $\begingroup$ Just to be clear, $$\frac{2 + \root 3 \of {19}}{3}$$ has a minimal polynomial of $3x^3 - 6x^2 + 4x - 3$. However, $$\frac{1 + \root 3 \of {19} + (\root 3 \of {19})^2}{3}$$ is an algebraic integer, check $x^3 - x^2 - 6x - 12$. $\endgroup$ – Mr. Brooks May 20 '17 at 21:14
  • $\begingroup$ I'd note that $(3) = \mathfrak{p}^3$ in $\mathbb{Z}[\sqrt[3]{19}]$ because $x^3-19 = (x-1)^3$ in $\mathbb{F}_3[x]$. Is there some way to extend this to $\mathcal{O}_K = \mathbb{Z}[\sqrt[3]{19}, \frac{1 + \root 3 \of {19} + (\root 3 \of {19})^2}{3}]$ ? $\endgroup$ – reuns May 20 '17 at 21:47
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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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  • $\begingroup$ Thanks, this answers my question! Though I am still not completely sure whether or not this computation was coincidental. $\endgroup$ – Tob Ernack May 19 '17 at 19:00
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It is precisely due to the fact that $27$ is the norm of $\alpha=2+\sqrt[3]{19}$.

From this, it follows that if $\mathfrak p$ divides $\alpha$, the norm of $\mathfrak p$ must be $3$, $9$, or $27$ (the last is excluded ’cause you’ve verified that $\alpha/3$ is not integral). But the only primes with norm a power of $3$ are divisors of $3$.

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