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How do I compute the index of $\mathbf{Z}[\alpha]$ in $\mathbf{Z}[\alpha,\beta,\gamma]$?

For example, $\alpha = \sqrt[3]{-19}$ and $\beta = (\alpha^2 - \alpha + 1)/3$ satisfy $(\alpha + 1)\beta = -6$, so $\alpha + 1 = -6/\beta = (-\beta^2 + \beta + 6)/2$.

In these notes (example 3.7, p. 32) it is claimed that $\mathbf{Z}[\alpha]$ has index $3$ in $\mathbf{Z}[\alpha,\beta]$ and $\mathbf{Z}[\beta]$ has index 2 in $\mathbf{Z}[\alpha,\beta]$, but I don't follow the reasoning.

I do see that $3\beta \in \mathbf{Z}[\alpha]$ and $2\alpha \in \mathbf{Z}[\beta]$, but where do I go from there?

In another example, I have $2\beta \in \mathbf{Z}[\alpha]$ and $3\gamma \in \mathbf{Z}[\alpha]$ (and again some algebraic relations), and I want to show that the index of $\mathbf{Z}[\alpha]$ in $\mathbf{Z}[\alpha,\beta,\gamma]$ is $6$.

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This might not be the best we to look at this, but these results follow at least by writing down explicit $\mathbb{Z}$-bases for these rings.

Notice that $\mathbb{Z}[\alpha]$ has rank $3$ as $\mathbb{Z}$-module; a basis is given by $\{1,\alpha,\alpha^2\}$ but also by $\{1,\alpha,\alpha^2-\alpha+1\}$.

Let us find a basis for $\mathbb{Z}[\alpha, \beta]$. Since $\alpha$ and $\beta$ have degree $3$, $\mathbb{Z}[\alpha,\beta]$ is surely generated by $1$, $\alpha$, $\alpha^2$, $\beta$, $\beta^2$, $\alpha \beta$, $\alpha \beta^2$, $\alpha^2 \beta$, $\alpha^2 \beta^2$. Since $\alpha \beta = - \beta - 6$, it is also generated by $1$, $\alpha$, $\alpha^2$, $\beta$, $\beta^2$. Since $\alpha^2 = 3\beta + \alpha - 1$, it is also generated by $1$, $\alpha$, $\beta$ and $\beta^2$. Because $\beta^2 = \frac{13-7\alpha +\alpha^2}{3} = \beta - 2\alpha + 4$ we can even generate $\mathbb{Z}[\alpha, \beta]$ by $1$, $\alpha$, $\beta$. Since these elements are independent over $\mathbb{Z}$, the elements $\{1, \alpha, \frac{\alpha^2-\alpha+1}{3} \}$ form a basis of $\mathbb{Z}[\alpha,\beta]$ over $\mathbb{Z}$.

One now easily sees that $\mathbb{Z}[\alpha]$ has index $3$ in $\mathbb{Z}[\alpha,\beta]$.

Similarly, $\mathbb{Z}[\beta]$ has $\{1, 2\alpha, \beta\}$ as a $\mathbb{Z}$-basis, so it has index $2$ in $\mathbb{Z}[\alpha,\beta]$ which has basis $\{1,\alpha,\beta\}$.

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