Determining the Ring of Integers Let $a,b>1$ be two relatively prime integers with $3$ dividing the product $ab$. Set $f(X) = X^3 - ab^2$. Pick a root $\alpha \in \mathbb{C}$ and consider the extension $K = \mathbb{Q}(\alpha)$. How does one show that,
$$ \mathbb{Z}_K = \mathbb{Z}[\alpha,\tfrac{\alpha^2}{b} ]$$
At first I tried to compute the discriminant of the basis $(1,\alpha,\tfrac{\alpha^2}{n})$ and checking if it is square-free. But it is not.
The second approach I tried was to somehow show that $$\mathbb{Z}[\alpha,\tfrac{\alpha^2}{b} ]$$ is a Dedekind domain by localizing at maximal ideals and showing they are all PIDs. The problem is it is not clear why $\mathbb{Z}[\alpha,\tfrac{\alpha^2}{b} ]$ is a ring of dimension $1$, and if we somehow accept that, then it is not clear what that maximal ideals of this ring are.
 A: As stated, this is not quite true. For example, if $a=p^3$ for a prime $p$ then your ring is $\mathbb Z[p\sqrt[3]{b^2},p^2\sqrt[3]{b}]$ which is certainly not integrally closed. Because of this counterexample, the only interesting case of this problem is when $a$ and $b$ can be chosen to be squarefree, so we will assume this going forward.
Let $\alpha = \sqrt[3]{ab^2}$, $\beta = \sqrt[3]{a^2b} = \alpha^2/b$, so that $1,\alpha,\beta$ is a $\mathbb Q$-basis for $K$. First we obviously have the inclusion $\mathbb Z[\alpha,\beta] \subseteq \mathcal O_K$. Now for $x,y\in K$, let $T(x,y) = \text{tr}_{K/\mathbb Q}(xy)$. Let $A$ be the largest subset of $K$ for which $T(\mathcal O_K,A)\subseteq \mathbb Z$. Then $\mathcal O_K\subseteq A$. If $\gamma = r + s\alpha+t\beta \in \mathcal O_K$, then the conditions $T(1,\gamma), T(\alpha,\gamma), T(\beta,\gamma)\in \mathbb Z$ give $r\in \frac{1}{3}\mathbb Z, s\in \frac{1}{3ab} \mathbb Z, t\in \frac{1}{3ab}\mathbb Z$, so $$A = \frac{1}{3}\mathbb Z + \frac{1}{3ab}\mathbb Z\alpha + \frac{1}{3ab}\mathbb Z\beta.$$ In this way, one can prove the more general fact that if $\alpha$ is a primitive element, then $A= \frac{1}{f'(\alpha)}\mathbb Z[\alpha]$.
So we need to determine "which denominators are allowed." We can see that the possible denominators that can appear come from factors of $a$, and factors of $b$ (since $3$ is known to be a factor of $ab$). Let $p$ be a prime dividing $a$. Then in $\mathbb Z_p$ we have $\alpha^3=ab^2=up$ for some $u\in \mathbb Z_p^\times$, so $L:=\mathbb Q_p(\alpha)$ is a totally ramified extension of $\mathbb Q_p$ and $\alpha$ is a uniformizer. Via general theory, then, $\mathcal O_L = \mathbb Z_p[\alpha]$. In this way, one sees that denominators involving $p$ (in the expression for $A$) cannot appear in an element of $\mathcal O_K$. The same argument works for a prime $q$ dividing $b$. So $\mathcal O_K = \mathbb Z[\alpha,\beta]$.

It's actually relatively doable to compute the ring of integers without the assumption that $3$ divides $ab$. For this, we will WLOG assume $3$ doesn't divide $ab$ and $a,b\equiv 1\bmod 3$ since we can achieve this by negating $a,b$ if necessary.
Let's look at our local argument with the prime $3$ specifically. In this case, $L$ is still totally ramified, but it is less clear what the uniformizer is since $3$ no longer divides $ab^2$. If $ab^2\not \equiv 1\bmod 9$, then $$f(x-1) = x^3-3x^2+3x-1+ab^2$$ is Eisenstein at $3$, so $\mathcal O_L = \mathbb Z_3[\sqrt[3]{ab^2}+1]$ and we see as before that there can be no denominators containing a $3$. If $a^2b\not \equiv 1\bmod 9$, the same argument works.
So in fact we have shown that $\mathcal O_K = \mathbb Z[\alpha,\beta]$ except possibly in the case where $ab^2\equiv a^2b\equiv 1\bmod 9$. This is equivalent to $a\equiv b\bmod 9$. Even in this case, we know $\mathcal O_K\subseteq \frac{1}{3}\mathbb Z[\alpha,\beta]$. Another thing we can use to our advantage in this case is that $\mathcal O_K$ is in fact a ring, so we must have $\mathcal O_K\cdot \mathcal O_K\subseteq \mathcal O_K \subseteq \frac{1}{3}\mathbb Z[\alpha,\beta]$, and in particular if $J=3\mathcal O_K$ is an ideal of $\mathbb Z[\alpha,\beta]$, then $J^2 \subseteq 3\mathbb Z[\alpha,\beta]$. So if $\gamma  = r + s\alpha + t\beta \in J$, we must have $$3\mathbb Z[\alpha,\beta] \ni \gamma^2 \equiv r^2-st + (s^2-rt)\beta + (t^2-rs)\alpha \bmod 3.$$
It is easy to show that all coefficients are $0\bmod 3$ iff $r\equiv s\equiv t\bmod 3$. Thus $\mathcal O_K \subseteq \mathbb Z[\alpha,\beta,\frac{1+\alpha+\beta}{3}]$. In fact, $\mathcal O_K = \mathbb Z[\alpha,\beta,\frac{1+\alpha+\beta}{3}]$. To see this, let $n=\frac{ab^2-1}{9}$. Let $$\delta=\frac{1+\alpha+\alpha^2}{3} = \frac{1+\alpha+\beta}{3} + \frac{b-1}{3}\beta.$$ Then $\delta(\alpha-1) = 3n$, so $\delta$ is a root of $f(\frac{3n}{x}+1) = \frac{27n^3}{x^3}+\frac{27n^2}{x^2}+\frac{9n}{x}-9n$, so $\delta$ is a root of $g(x) = x^3-x^2-3nx-3n^2$, and in particular $\mathcal O_K = \mathbb Z[\alpha,\beta,\delta]$.
So when $a,b\equiv 1\bmod 3$, we have shown that $\mathcal O_K = \mathbb Z[\alpha,\beta]$ except in the exceptional circumstance where $a \equiv b\bmod 9$, and have handled that case appropriately to show that $\mathcal O_K=\mathbb Z[\alpha,\beta,\delta]$.
