TITLE: Why if $\alpha$ is a root of $x^3-5x^2+2$, then $\mathcal O_{\Bbb Q[\alpha]}=\mathbb Z[\alpha]$?
QUESTION [9 upvotes]: I'm trying to show that if $\alpha$ is a root of the polynomial $x^3-5x^2+2$ and $K=\Bbb Q[\alpha]$, then $\mathcal O_K=\Bbb Z[\alpha]$. This is homework, and one of the previous exercises asks to show that if we define $disc(\Bbb Z[\alpha])=disc(1,\alpha,\ldots,\alpha^{n-1})$, then $$disc(\Bbb Z[\alpha])=[\mathcal O_K:\Bbb Z[\alpha]]^2disc(\mathcal O_K).$$
So I guess what has to be done is to show that $disc(\Bbb Z[\alpha])=disc(\mathcal O_K)$. However I can't see why this is true.
What I want is a HINT on how to solve this problem. Please DO NOT post a full answer. I have only read Neukirch's Algebraic Number theory until section 3 of chapter 1, so I would like to know a hint only using what I know so far.
Thanks

REPLY [2 votes]: Here’s another argument, based partly on the ideas of @guest’s response. You find that the discriminant of $\Bbb Z[\alpha]$ is $4$ times a prime, and you only want to know that the $4$ belongs there in the discriminant of $\mathscr O$. But look at your polynomial locally at $2$, where the Newton Polygon has its vertices at $(0,1)$, $(2,0)$, and $(3,0)$. In particular, the first slope is $-1/2$, requiring ramification at $2$ to get all the roots. So the discriminant must be divisible by $2$, and hence the $4$ needs to be there.