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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.

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

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

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

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

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

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

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  • $\begingroup$ 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 $\endgroup$ – rmdmc89 Mar 12 '19 at 0:29

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