How to prove $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$ is an integral basis of $\mathbb{Q}(\sqrt[3]{2})$

Solution 1:

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