<s>[INST] The discriminant of a cubic extension Let $K=\mathbb{Q}[\sqrt[3]{7}]$. I want to find the discriminant $d_K$ of the number field $K$.
I have computed $\operatorname{disc}(1,\sqrt[3]{7},\sqrt[3]{7^2})=-3^3\cdot 7^2$. I know that $d_K$ must divide this quantity, and may differ form it by a square, hence we have 3 possibilities: $d_K=-3\cdot 7^2$ or $-3^3$ or $-3^3\cdot 7^2$. Since $7$ ramifies in $K$, I can exclude the case $-3^3$. Now, how can I exclude also $-3\cdot 7^2$?
I need this because I want to prove that $\mathscr{O}_K=\mathbb{Z}[\sqrt[3]{7}]$. Is it possible to prove this without the discriminant?
 [/INST] Yes, it is possible to prove it independent of computing the discriminant.
It suffices to check that $\mathbb{Z}[\sqrt[3]{7}]$ is normal (why?). To see this, we a priori need to check that $M\mathbb{Z}[\sqrt[3]{7}]_M$  is principal for all $M$ maximal in $\mathbb{Z}[\sqrt[3]{7}]$ (note $\mathbb{Z}[\sqrt[3]{7}]$ is already Noetherian and dimension $1$, so this is checking for Dedkind domain, which is equvalent). But, by standard theory related to Dedekind-Kummer it suffices to check that $M\mathbb{Z}[\sqrt[3]{7}]_M$ is principal only when $f'(\sqrt[3]{7})\in M$, where $f(x)$ is the min poly of $\sqrt[3]{7}$ (i.e. $x^3-7$).
But, evidently $f'(\sqrt[3]{7})=3\sqrt[3]{7}^2$. So, evidently if $M$ contains $f'(\sqrt[3]{7})$ it contains $\sqrt[3]{7}f'(\sqrt[3]{7})=3\cdot 7$. So, $M$ either lies above $3$ or $7$.
Using more standard Dedekind-Kummer theory we know that the maximals lying above $3$ are $(3,\sqrt[3]{7}+2)$ (since $x^3-7=(x+2)^3$ in $\mathbb{F}_3[x]$) and the maximals lying above $7$ are $(7,\sqrt[3]{7})$ (since $x^3-7=x^3$ in $\mathbb{F}_7[x]$). 
The latter is already principal (why?), and so we need only check that the first is principal in its localization. Namely, we need to check that $(3,\sqrt[3]{7}+2)\mathbb{Z}[\sqrt[3]{7}]_{(3,\sqrt[3]{7}+2)}$ is principal. But, it's easy to check that 
$$(\sqrt[3]{7}+2)^2=3\cdot(5+4\sqrt[3]{7}+2\sqrt[3]{7^2})\quad(\ast)$$
But, note that $5+4\sqrt[3]{7}+2\sqrt[3]{7^2}\notin (3,\sqrt[5]{7}+2)$ else evidently $5\in(3,\sqrt[3]{7}+2)$ which is false since $(3,\sqrt[3]{7}+2)\cap\mathbb{Z}=(3)$. Thus, we see that $5+4\sqrt[3]{7}+2\sqrt[3]{7^2}$ is invertible in $\mathbb{Z}[\sqrt[3]{7}]_{(3,\sqrt[3]{7}+2)}$ and thus $(\ast)$ obviously implies that 
$$(3,\sqrt[3]{7}+2)\mathbb{Z}[\sqrt[3]{7}]_{(3,\sqrt[3]{7}+2)}=\sqrt[3]{7}\mathbb{Z}[\sqrt[3]{7}]_{(3,\sqrt[3]{7}+2)}$$
Thus, by prior comment we see that $\mathbb{Z}[\sqrt[3]{7}]$ is integrally closed, and thus $\mathcal{O}_K=\mathbb{Z}[\sqrt[3]{7}]$ as desired.
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