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Let's say $K = \mathbb{Q}(\root 3 \of {10})$, and then $\mathcal{O}_K$ is a ring of algebraic integers. My calculations suggest numbers of the form $$\frac{a}{3} + \frac{a \root 3 \of {10}}{3} + \frac{a (\root 3 \of {10})^2}{3}$$ with $a \in \mathbb{Z}$ are algebraic integers. I imagine a few other forms are also algebraic integers, but I haven't been able to discover the pattern. Maybe congruence modulo $3$? Such as in $$\frac{5}{3} + \frac{11 \root 3 \of {10}}{3} + \frac{8 (\root 3 \of {10})^2}{3},$$ which has a minimal polynomial of $x^3 - 5x -285x + 1905$, unless I made a mistake somewhere along the way.

How do I compute the norms of numbers in $\mathcal{O}_K$? Is there a way to grab it off the minimal polynomial in a manner similar to quadratic integer rings?

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    $\begingroup$ The norm is the constant term of the characteristic polynomial (it's not the minimal polynomial even in the quadratic case); in other words, it's the determinant of an element acting on a number field by multiplication. Usually it's easier to compute traces though. $\endgroup$ – Qiaochu Yuan Feb 24 '15 at 22:47
  • $\begingroup$ I think there's a little mistake in that minimal polynomial. $\endgroup$ – Robert Soupe Feb 25 '15 at 1:18
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All of this is known in detail for a general extension $\Bbb Q(\sqrt[3] n\,)$, but not to me. I can only make remarks germane to this particular extension, with $n=10$.

To save typing, I’m going to write $\sqrt[3]{10}=\rho$. The minimal polynomial of $a+b\rho+c\rho^2$ is $X^3-3aX^2+(3a^2 - 30bc)X+(-a^3 + 30abc - 10b^3 - 100c^3)$, where I’ve put the constant term in parentheses to isolate it, as the norm of the general irrational quantity.

From this, it turns out that the characteristic polynomial of $(1+\rho+\rho^2)/3$ is $X^3-X^2-3X-3$. Let’s call that irrationality $\tau$, and I’ll call its characteristic polynomial there $f(X)$. Because $f$ has integer coefficients, $\tau$ is an algebraic integer.

You can find the discriminant of the ring $\Bbb Z[\tau]$ as the absolute value of the norm down to $\Bbb Q$ of $f'(\tau)$. This turns out to be $300$, a very nice number, because I can now argue that $\Bbb Z[\tau]$ is the ring of integers $\mathcal O$ of $\Bbb Q(\rho)$. It’s contained in $\mathcal O$, because it’s generated by an algebraic integer, and any larger ring will have discriminant that varies from $300$ by a square in $\Bbb Z$. But if you divide $300$ by any square, you’ll either get a noninteger, or something not divisible by all of the three primes $2$, $3$, and $5$, which must appear in the discriminant because they’re all ramified in $\Bbb Q(\rho)$.

So this tells you which things in $\Bbb Q(\sqrt[3]{10}\,)$ are algebraic integers, and I’ll leave it to you to get conditions on $a$, $b$, and $c$ that are equivalent to the integrality of $a+b\sqrt[3]{10} + c\sqrt[3]{100}$.

One more remark: although I used a computation package to get the characteristic polynomials quickly, all the computations I used can be done by hand. I know this, ’cause I have done them that way in the past. And as a firm believer in the benefit to be derived from hand computation, I recommend this to all.

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If $\mathcal{O}_K$ with $K = \mathbb{Q}(\root 3 \of d)$ where $d > 1$ is a squarefree integer, then the norm of a number $a + b \root 3 \of d + c(\root 3 \of d)^2$ is $a^3 + b^3 d + c^3 d^2 - 3abcd$.

So then for example, the norm of $$\frac{5}{3} + \frac{11 \root 3 \of {10}}{3} + \frac{8 (\root 3 \of {10})^2}{3}$$ (which has a minimal polynomial of $x^3 - 5x^2 - 285x + 1905$, by the way) is $$\left(\frac{5}{3}\right)^3 + \left(\frac{11}{3}\right)^3 10 + \left(\frac{8}{3}\right)^3 100 - 3 \left(\frac{5}{3}\right)\left(\frac{11}{3}\right)\left(\frac{8}{3}\right)100$$ $$=\frac{21445}{9} - \left(\frac{440}{9}\right) 10 = 1909.$$

I found this answer in a question from 2013: Ring of integers of a cubic number field

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    $\begingroup$ I searched for this because I got the same answer from a different direction. I was looking at the three-dimensional number system with axes 1, x and x^2 with x^3 = 1, and the norm in this system comes out as N[a,b,c] = a^3 + b^3 + c^3 - 3abc. If you measure b and c in cube roots, it works out the same as this answer, and of course, there will be no zero divisors (since zero divisors in this system are numbers where a + b + c = 0 or where a = b = c, which is impossible if irreducible cube roots are used). $\endgroup$ – Marek14 Aug 11 '17 at 10:16

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