TITLE: How to prove that $\frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$ is an algebraic integer
QUESTION [7 upvotes]: As the title says, I'm trying to show that $\frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$ is an algebraic integer.  
I suppose there's probably some heavy duty classification theorems that give one line proofs to this but I don't have any of that at my disposable so basically I'm trying to construct a polynomial over $\mathbb{Z}$ which has this complex number as a root.
My general strategy is to raise both sides of the equation 
$$x = \frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$$
to the $n^{th}$ power and then break up the resulting sum in such a way as to resubstitute back in smaller powers of $x$.  Also since this root is complex I know it must come in a conjugate pair for the coefficients of my polynomial to be real, thus I know that
$$x = -\frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$$
must also be a root of my polynomial.  Hence from this I obtain:
$$3x^2 = (10^{\frac{2}{3}} -1 )^2$$
However since my root is pure imaginary I don't really get any more information from this, so I'm a bit stumped, I tried raising both sides of $3x^2 -1 = 10^{\frac{4}{3}} - (2)10^{\frac{2}{3}}$ to the third power but it doesn't look like it's going to break up correctly, can anyone help me with this?  Thanks.

REPLY [5 votes]: A somewhat less computational approach can be based on the idea that an algebraic number is an algebraic integer if and only if it is everywhere locally integral. Certainly $\sqrt{-3}$ is a $p$-adic unit (possibly in a quadratic extension) everywhere except $3$, so the only issue is at $3$.
Second, observe/speculate that there is a cube root $\alpha$ of $10$ in $\mathbb Q_3$, using Hensel's Lemma: $10$ is pretty close to $1$ modulo powers of $3$, that is, $3$-adically. A useful form of Hensel's lemma here is that a monic polynomial $f$ with coefficients in $\mathbb Z_3$ and $x_o\in \mathbb Z_3$ such that $|f(x_o)/f'(x_o)^2|_3<1$ produces a root of $f(x)=0$ in $\mathbb Z_3$.
From this, we see that if we can find a cube root of $10$ mod $3^3=27$, we have a $3$-adic cube root. Indeed, $4^3=64=10+2\cdot 3^3$.
So there is $\alpha\in \mathbb Z_3$ with $\alpha-4=0 \mod 3$, so ${\alpha-1\over 3}$ is a $3$-adic integer, and certainly ${\alpha-1\over \sqrt{-3}}$ is integral over $\mathbb Z_3$, although in a quadratic extension.
A similar pattern occurs for all odd primes: $(1+p)^p=1+p^2 \mod p^3$, so there is a $p$-th root of $1+p^2$ in $\mathbb Z_p$.