Prime decomposition of pR in $\mathbb{A}\cap \mathbb{Q}[\alpha]$ for $\alpha={^3\sqrt{hk^2}}$ if p is a prime such that $p^2|m$

Question

I'm going through Marcus number Field chapter 3 an I'm finding very hard to understand the part about the decomposition of pR (theorem 27) that tells us that if $p\not||R/Z[\alpha ]|$ then we can decompose $pR$ by looking at a factorization of it's minimal polynomial (Kummer's theorem?)

In partcular I'm stuck on exercise 26

Let $\alpha={ ^{3}\sqrt{m}}$ where m is a cubefree integer, $K = \mathbb{Q}[\alpha]$, $R = \mathbb{A} \cap \mathbb{Q}[\alpha]$

1. Show that if p is a prime $\neq 3$ and $p^2 \not|m$ , then the prime decomposition of pR can be determined by factoring $x^3 − m\; mod\; p.$ (See Theorem 27 and exercise 41, chapter 2 (this tells us the discriminat and the integral bases I write below).)

2. Suppose $p^2 | m$. Writing $m = hk^2$ as in exercise 41, chapter 2, set $ \gamma= \frac{\alpha^2}{k}.$ Show that p does not divide $|R/Z[\gamma ]|$; use this to determine the prime decomposition of pR.

3. Determine the prime decomposition of 3R when $m\not\equiv \pm 1$ (mod 9).

4. Determine the prime decomposition of 3R when m = 10. (Hint: Set $\beta = (\alpha − 1)^2/3$ and use exercise 18 to show that disc(β) = 4 disc(R). Also note exercise 41(d), chapter 2 (this tells us that $\beta^3-\beta^2+\left(\frac{ 1+2m}{3}\right)\beta-\frac{(m-1)^2}{27}=0))$ Show that this always works for $m\equiv \pm 1\; (mod\; 9)$ except possibly when $m\equiv \pm 8\; (mod\; 27)$.
5. Show that 9 $\not|$ disc(R) when $m\equiv \pm 1\; (mod\; 9)$; use this to show that 3R is not the cube of a prime ideal. Assuming the converse of Theorem 24, show that 3R = $P^2Q$ where P and Q are distinct primes of R.

I think I've done point 1) using the fact that $p^2\not| disc(\alpha)$ implies we can use theorem 27 that tells us exactly that we can decompose pR simply by factorizing the minimal polynomial of $\alpha$, but the problem is now point 2) (and the ones after since they rely on 2).

I was able to prove that $\gamma=\sqrt{h^2k}$ and that $p^2\not| h^2k$ so either we can use the fact above or $p^2|disc(\alpha)=-27^2*k^2h\Rightarrow p^2|27^2$ so p=3, but now I don't know how to prove that 3 doesn't divide $|R/\mathbb{Z}[\alpha]|$ since for me the latter is always divisible by 3.

An integral base of the above is either $$\left(1,\alpha,\frac{\alpha^2+k^2\alpha+k^2}{3k}\right),\quad \left(1,\alpha,\frac{\alpha^2-k^2\alpha+k^2}{3k}\right),\quad \left(1,\alpha,\frac{\alpha^2}{k}\right) $$ if respectivly $m\equiv 1\; (mod\; 9)$, $m\equiv -1\; (mod\; 9)$, $m\not\equiv \pm1\; (mod\; 9)$

Any help would be welcomed, even more if quite specific on the calculations since I think there is something I miss on a theoretical level.

Exercise 18 Let K be a number field of degree n over $\mathbb{Q}$ , and let $\alpha_1, \dots , \alpha_n \in K.$

1. Show that $disc(r\alpha_1, \alpha_2, \dots , \alpha_n) = r^2 disc(\alpha_1, \dots , \alpha_n)$ for all r $\in \mathbb{Q}$.

2. Let $\beta$ be a linear combination of $\alpha_2, \dots , \alpha_n$ with coefficients in $\mathbb{Q}$. Show that $disc(\alpha_1 + \beta, \alpha_2, \dots , \alpha_n) = disc(\alpha_1, \dots , \alpha_n).$

Theorem 24 Let p be a prime in $\mathbb{Z}$, and suppose p is ramified in a number ring R. Then p | disc(R).

UPDATE: The question is still without answer so for now I'll post my solution to the first two points, then if a better one comes I'll be happy to set it as solving the question.

Answer 1

1. Uniforming the notation between this exercise and therem 27 of Marcus we have $$ L=\mathbb{Q}[\alpha]\quad S=\mathbb{A}\cap \mathbb{Q}[\alpha]\quad K=\mathbb{Q}\quad R=\mathbb{Z}$$ so to use theorem 27 we have to check $$ p\not|\left|\frac{\mathbb{A}\cap \mathbb{Q}[\alpha]}{\mathbb{Z}[\alpha]}\right|$$ but actually first we can use the corollary telling us that the hypotesis are satisfied if $p^2\not|disc(\alpha)$, exercise 41 in chapter 2 tells us that in our case $disc(\alpha)=-27^2m$ and so if $p\neq 3\wedge p^2|m$ we are in the hypotesis of the corollary and thus of the theorem and so we can decompose pR by factoring $x^3-m$;

2. In this case the hypotesis of the corollary are not satisfied.\ We have also that $p^2|m\iff p^2|h\vee p^2|k^2$ but since h is squarefree we have that it has to be $p^2|k^2\iff p|k\iff p\not|h$ since they are coprime.\ Now we can write $$ \alpha=\sqrt{hk^2}\iff \alpha^2=\sqrt{h^2k^4}\iff \gamma=\frac{\alpha^2}{k}=\sqrt{h^2k}$$ and we have that $p^2|h^2k\iff p^2|h^2\iff p|h$ which is not true so $p^2\not| h^2k$. But now $p|h^2k=n$ but $p^2\not|h^2k$ so $x^3-n$ is a p-Eisentstein polynomial and we can use the following theorem to deduce $$p\not| |\mathbb{A}\cap \mathbb{Q}[\gamma]/\mathbb{Z}[\gamma]|$$

Let K = $\mathbb{Q}(\alpha)$ where $\alpha\in \mathbb{A}\cap \mathbb{Q}[\alpha]$ is the root of an Eisenstein polynomial at p, with degree n. Then $p \not| |\mathbb{A}\cap \mathbb{Q}[\alpha] / \mathbb{Z}[\alpha]|$.