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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?

bateman
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  • The obvious things aren't seeming to work. What is the motivation of this? It's not hard to show that $\mathcal{O}_K=\mathbb{Z}[\sqrt[3]{7}]$ independent of discriminant – Alex Youcis Sep 25 '13 at 08:07
  • @AlexYoucis Yes, I thought it was possible with "the obvious things", but it is not...so I decided to edit the question. – bateman Sep 25 '13 at 08:14

2 Answers2

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One can also see how $3$ is ramified. The polynomial $(y+1)^3-7=y^3-3y^2+3y-6$ is $3$-Eisenstein, so $3$ is fully ramified (i.e. $(3)=\mathfrak P^3$ for some prime ideal $\mathfrak P$), so $3^2$ has to divide the discriminant. So indeed, $\mathscr O_K=\mathbb Z[\sqrt[3]{7}]$.

user8268
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  • Hi :) I'm having a little difficulty following. So, I'm not aware of what form of Dedekind-Kummer theorem you're using. I know that if $3$ were to factor into primes in $\mathbb{Z}[\sqrt[3]{7}]$ it would have to be totally ramified, and then $3$ would obviously have to totally ramify in $K$. But I don't see why a priori you know it factors into primes there. Also, is there some theorem that says that if you are a totally ramified prime, then you must actually divide the discriminant more than once? Thanks! – Alex Youcis Sep 25 '13 at 08:51
  • @AlexYoucis Hi Alex. I'm using: if $\mathfrak P^e|p$ then $\mathfrak P^{e-1}|\delta$ ($\delta$ is the different) and that if $f$ is $p$-Eisenstein then $\mathbb Q_p[x]/(f)$ is totally ramified over $\mathbb Q_p$. I am in no imaginable way a number theorist; in particular, I don't know what Dedekind-Kummer theorem is :) – user8268 Sep 25 '13 at 09:08
  • Ah, I see! I must have misread. You used the fact that it totally ramified (since the field was generated by a $3$-eisenstein polynomial). But, then since the extension is degree $3$, and $3\equiv 0\mod 3$, you have that $\mathfrak{p}^e\mid\delta$. Since $\text{disc}(K)=|\delta|$ you get that $|\mathfrak{p}|^3\mid \text{disc}(K)$, since $|\mathfrak{p}|$ must be $3$ (by consideration of our $efg$s) we get that $3^3\mid \text{disc}(K)$. Well, this isn't exactly what you said, but the same idea. It's nice to note that in your statement about $\mathfrak{P}^e\mid p$, you actually get – Alex Youcis Sep 29 '13 at 00:39
  • $\mathfrak{P}^e\mid\delta$ if $e\equiv 0\mod p$ (I used this earlier in my comment). You do know the Dedekind-Kummer theorem, just not by that name. It's just the various theorems relating to simple integral extensions of Dedekind domains, and how to use the min poly of the generator of the extension to deduce things (e.g. is the extension normal, if so, how do primes split, etc.). – Alex Youcis Sep 29 '13 at 00:41
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    @AlexYoucis Thanks a lot for reminding me that $\mathfrak P^e|\delta$ if $p|e$; normally I run away screaming when I hear "wild ramification", but maybe it's time for me to use some courage and read about it. – user8268 Sep 29 '13 at 13:23
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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.

Alex Youcis
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