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This an exercise from Jürgen Neukirch.

(1.2.5) Show that $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$ is an integral basis of $\mathbb{Q}(\sqrt[3]{2})$

And what's more, i'm still stuck on this problem.

(1.2.6) Show that $\{1,\theta,\frac{1}{2}(\theta+\theta^2)\}$ is an integral basis of $\mathbb{Q}(\theta)$, $\theta^3-\theta-4$.

My try:

The classical method. Assume that $a+b\sqrt[3]{2}+c\sqrt[3]{4}$ is integral over $\mathbb{Z}$, but the coefficients of its minimal polynomial forms a such complex equation to solve that i cannot conclude $a,b,c\in \mathbb{Z}$.

The discriminant method. A remark result about this is that if the discriminant of $f(x)$ is square-free then $\{1,x,\ldots,x^n\}$ is an integral basis of $\mathbb{Q}[X]/(f(x))$. But the discriminant of $X^3-2$ is $-108$, which fails to be square-free.

Cubic Bear
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  • You should see the answers to this question – Mathmo123 Oct 24 '17 at 22:19
  • Possible duplicate of [Easy way to show that $\mathbb{Z}[\sqrt[3]{2}]$ is the ring of integers of $\mathbb{Q}[\sqrt[3]{2}]$](https://math.stackexchange.com/questions/99913/easy-way-to-show-that-mathbbz-sqrt32-is-the-ring-of-integers-of-mat) – Mathmo123 Oct 24 '17 at 22:19
  • Prove that three given elements are linearly independent over $\mathbb Q$ (just like in a vectorial space). – Piquito May 26 '18 at 11:39

1 Answers1

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As pointed out already, this is equivalent to showing that $\mathbb{Z}[\sqrt[3]{2}]$ is the ring of integers of $\mathbb{Q}[\sqrt[3]{2}]$

If not, then the discriminant of the actual ring of integers will divide the discriminant of $\mathbb{Z}[\sqrt[3]{2}]$, which is -108, by a square factor. The only squares dividing -108 are 4 and 9.

It is sufficient to check that $\frac{1 + \sqrt[3]{2} + \sqrt[3]{4}}{x}$ is not an algebraic integer for $x = 2,3$. The norm will give you this answer:

$$N(\frac{1 + \sqrt[3]{2} + \sqrt[3]{4}}{x}) = \frac{1 +2 + 4 - 6}{x^3} = \frac{1}{x^3}$$

Clearly not an integer for $x = 2,3$, so $\frac{1 + \sqrt[3]{2} + \sqrt[3]{4}}{x}$ is not an algebraic integer in these cases and the ring of integers must be $\mathbb{Z}[\sqrt[3]{2}]$.

This is the general idea. You can find out more here (page 34 onwards): http://www.jmilne.org/math/CourseNotes/ANT210.pdf

hbghlyj
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TCiur
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    How did you conclude so immediately that $N(1+\sqrt[3]{2}+\sqrt[3]{4})=1+2+4-6$? I wanna know how to do it too – rmdmc89 Mar 12 '19 at 00:29
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    You need to check that $\frac{a+b \sqrt[3]{2} + c \sqrt[3]{4}}{x}, \forall 0 \leq a,b,c \leq x-1$(not all $a,b,c$ zero) is not an algebraic integer, not just for $a=b=c=1$. – user600016 Feb 19 '23 at 06:19
  • Using the identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$, where $a=1,b=\sqrt[3]{2},\sqrt[3]{4}$ along with the fact that norm is multiplicative and that $N(\sqrt[3]{2}-1)=1$. – user600016 Apr 22 '23 at 09:12