Integral extensions Let $p\neq1$ be an integer and let $\beta$ be a root of $x^6-p$. 

What is the difference, in terms of $\mathbb{Z}$-modules, between $\mathbb{Z}[\beta]$ and $\mathbb{Z}[\beta^2,\beta^3]$?

I can see that the latter does not contain $\beta$, so I conclude that it is strictly contained in the former. However, they both have 6 generators (as $\mathbb{Z}$-modules). Why?
 A: In the following I am assuming $p$ is prime, so that by Eisenstein $x^6 - p$ is irreducible. (@HagenvonEitzen has a comprehensive answer for all cases.)
$A = \Bbb{Z}[\beta]$ has a $\Bbb{Z}$-basis given by $$1, \beta, \beta^{2}, \beta^{3}, \beta^{4}, \beta^{5},$$ while one for $B = \Bbb{Z}[\beta^{2}, \beta^{3}]$ is $$1, p \beta, \beta^{2}, \beta^{3}, \beta^{4}, \beta^{5}.$$
This is because in $B$ you can find $\beta^{4} = (\beta^{2})^{2}$, $\beta^{5} = \beta^{2} \beta^{3}$, but then $\beta^{7} = \beta^{3} \beta^{4} = p \beta$, and you have noticed that $\beta \notin B$.
So $B$ is a subgroup (sub-$\Bbb{Z}$-module) of $A$ of index $p$, but still free on $6$ generators. Note the comment of @Hurkyl, to the effect that over $\Bbb{Q}$ the difference vanishes, as $p$ becomes invertible.
A: First assume that the polynomial $X^6-p$ is irreducible.
Then as $\mathbb Z$-module (aka. abelian group), 
$$\tag1\mathbb Z[\beta]=\mathbb Z\oplus\beta\mathbb Z\oplus\beta^2\mathbb Z\oplus\beta^3\mathbb Z\oplus\beta^4\mathbb Z\oplus\beta^5\mathbb Z$$
and
$$\tag2\begin{align}\mathbb Z[\beta^2,\beta^3]&=\mathbb Z\oplus\beta^{2+2+3}\mathbb Z\oplus\beta^2\mathbb Z\oplus\beta^3\mathbb Z\oplus\beta^{2+2}\mathbb Z\oplus\beta^{2+3}\mathbb Z\\&=\mathbb Z\oplus p\beta\mathbb Z\oplus\beta^2\mathbb Z\oplus\beta^3\mathbb Z\oplus\beta^4\mathbb Z\oplus\beta^5\mathbb Z.\end{align}$$
The directness of the sum follows from the irreducibility of $X^6-p$.
Hence $\mathbb Z[\beta^2,\beta^3]$ is a subgroup of index $p$.
On the other hand, if $X^6-p$ is reducible, several cases are possible:
1) $X^6-p$ has a root $b\in\mathbb Z$ (for example if $p=64$). Then $p=b^6$ and $-b$ is another root. If $\beta=\pm b$, then clearly $$\mathbb Z[\beta]=\mathbb Z[\beta^2,\beta^3]=\mathbb Z,$$
i.e. the groups are the same (or the subgroup index is $1$).
Otherwise $\beta=b\zeta$ with $b^6=p$ and $\zeta$ a primitive third root of unity, $\zeta^2+\zeta+1=0$. Then $$\mathbb Z[\beta]=\mathbb Z[b\zeta]=\mathbb Z+ \mathbb Z[b\zeta]+\mathbb Z[b^2\zeta^2]=\mathbb Z\oplus b\zeta\mathbb Z$$
and $$\mathbb Z[\beta^2,\beta^3]=\mathbb Z[\beta^2] =\mathbb Z[b^2\zeta^2]=\mathbb Z[b^2\zeta]=\mathbb Z\oplus b^2\zeta \mathbb Z,$$
hence $\mathbb Z[\beta^2,\beta^3]$ is a subgroup of index $|b|=\sqrt[6]p$ in $\mathbb Z[\beta]$.
2) $X^6-p$ has no linear, but a quadratic factor (example: $p=8$). Then $p=b^3$ for some nonsquare $b\in\mathbb Z$ and $X^6-p=X^6-b^3=(X^2-b)(X^4+X^2b+b^2)$. Then $\beta =\pm\sqrt b$ if it is in fact a root of the first factor and multiplied by a third root of unity if it is a root of the second factor. This can be treated similarly to case 1. We obtain a subgroup index $|b|=\sqrt[3]p$.
3) $X^6-p$ is the product of two irreducible cubic polynomials (example: $p=4$). Then $p=b^2$ for some $b\in\mathbb Z$ and $X^6-p=(X^3-b)(X^3+b)$. Again we need to work with third roots of unity and find $|b|=\sqrt[2]p$ as subgroup index.
