What is the norm of a number in a cubic integer ring? 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?
 A: 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 $$\left(\frac{5}{3}\right)^3 + \left(\frac{11}{3}\right)^3 10 + \left(\frac{8}{3}\right)^3 100 - 3 \left(\frac{5}{3}\right)\left(\frac{11}{3}\right)\left(\frac{8}{3}\right)100$$ $$=\frac{21445}{9} - \left(\frac{440}{9}\right) 10 = 1905.$$
I found this answer in a question from 2013: Ring of integers of a cubic number field
A: 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.
A: The number
$\dfrac{a+\sqrt[3]{10}(b+\sqrt[3]{10}c)}{3}: a,b,c\in\mathbb Z$
will be an algebraic integer if and only if $a+b+c$ is a multiple of $3$.
Clearly adding $3$ to any of $a,b,c$ 0}will not alter the integral ir nonintegral nature of the fraction above, so we only need to test the cases given by $a,b,c\in\{-1,0,1\}$.
It is not difficult to see that additive inversion and cyclic permutation of the parameters will not affect the integral nature (the cyclic permutation $(a,b,c)\to(c,a,b)$ is a multiplication by $\sqrt[3]{10}$ and then subtraction of $3c$), so the only required test cases for $(a,b,c)$ are
$(1,1,1)$
$(1,1,0)$
$(1,1,-1)$
$(1,0,0)$
$(1,0,-1)$
The first and last of these work, and they cover the cases where the sum is a multiple of $3$.
