Show that $\mathbb{Q}(6^{1/3})$ and $\mathbb{Q}(12^{1/3})$ have the same degree and discriminant but are not isomorphic.

Solution 1:

To find the first two rings of integers, one may use the following result, which is quite helpful at times: if $ K $ is a number field with $ [K : \mathbf Q] = n $ and $ a $ is an element of $ \mathcal O_K $ with degree $ n $ over $ \mathbf Q $ such that its minimal polynomial is Eisenstein at a prime $ p $, then the index of $ \mathbf Z[a] $ in $ \mathcal O_K $ is not divisible by $ p $.

This settles the first question immediately: $ X^3 - 6 $ has discriminant which is only divisible by $ 2 $ and $ 3 $, thus the index of $ \mathbf Z[6^{1/3}] $ in $ \mathcal O_{\mathbf Q(6^{1/3})} $ can only be divisible by $ 2 $ or $ 3 $, as it divides the discriminant. However, it cannot be divisible by either of those primes since $ X^3 - 6 $ is Eisenstein at both of them, so the ring of integers is indeed $ \mathbf Z[6^{1/3}] $, with discriminant $ -2^2 \cdot 3^5 $.

For the second question, drawing a Newton polygon shows that the extension still remains totally ramified at $ 2 $, and since $ 2 $ is coprime to the degree of the extension, it is tamely ramified. It follows that the exact power of $ 2 $ dividing the discriminant is $ 2^2 $. Moreover, the discriminant of $ X^3 - 12 $ is $ -2^4 \cdot 3^5 $, and this polynomial is Eisenstein at $ 3 $; which implies that $ \mathbf Z[12^{1/3}] $ has index $ 2 $ in the full ring of integers, and the discriminant of the ring of integers is $ -2^2 \cdot 3^5 $. (A basis for the actual ring of integers is contained in Dietrich Burde's answer, one can arrive at this by looking at the Newton polygon.) To deduce that the extensions are not isomorphic, just find a prime $ p > 3 $ such that $ 6 $ is a perfect cube modulo $ p $ but $ 12 $ is not, and look at its splitting in both fields. $ p = 7 $ does the trick, for instance.