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

The degree of $[K:\mathbb{Q}]=3$ since $x^3-6$ is the minimal polynomial and is irreducible. Similarly $[L:\mathbb{Q}]=3$.

  1. Show that $\mathbb{Z}_K=\mathbb{Z}[\alpha]$ and calculate disc $K$.

disc $K= $disc$(x^3-6)=-2^2.3^5$. Not sure how I can show the first part without using the discriminant.

  1. Work out disc $L$ in the form $-2^a.3^5$ where $a \in \mathbb{N}$ What are the possible values for $a$..

I have disc $L=$disc$(x^3-12)=-2^4.3^5$. $a=1,2,4$ I think.

  1. Prove that $L \cong \mathbb{Q}(18^{1/3})$. Hint look at $\frac{12^{2/3}}{2}$.

  2. Deduce disc $L=$disc $K$.

There is more but I think I need to understand all this first before I move on.

Community
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Walter
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  • My instinct here would be to show that after localising at $2$ or $3$ they are the same extensions. – mercio Feb 15 '17 at 15:11

2 Answers2

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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.

Ege Erdil
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  • We know the exact power of $ 3 $ in the discriminant in both cases because of the lemma I mention in the answer. Since $ X^3 - 12 $ is Eisenstein at $ 3 $, the index of $ \mathbf Z[12^{1/3}] $ in the ring of integers is coprime to $ 3 $; and since the discriminant scales by the square of submodule index, the discriminant of the ring of integers can only differ from the discriminant of $ X^3 - 12 $ by the factor of a power of $ 2 $. – Ege Erdil Feb 16 '17 at 13:43
  • Well, you also noted that the degree of $ \beta^2 / 2 $ is the same as the degree of the extension, so yes; you can. – Ege Erdil Feb 16 '17 at 14:01
  • Equal number fields have the same discriminant; in fact, they have the same anything. That's what it means for two number fields in $ \mathbf C $ to be equal. – Ege Erdil Feb 16 '17 at 14:03
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The cubic number field $K=\mathbb{Q}(\sqrt[3]{6})$ has discriminant $-2^23^5=-972$. This can be indeed seen by showing that $\{1,\alpha,\alpha^2\}$ is an integral power basis for the ring of integers in $K$, where $\alpha =\sqrt[3]{6}$. Then we have \begin{align*} D(1,\alpha,\alpha^2) & =\det \begin{pmatrix} tr (1) & tr (\alpha) & tr (\alpha^2) \\ tr (\alpha) & tr (\alpha ^2) & tr (\alpha ^3) \\ tr (\alpha ^2) & tr (\alpha ^3) & tr(\alpha^4)\\ \end{pmatrix} \\[0.2cm] & = \det \begin{pmatrix} 3 & 0 & 0 \\ 0 & 0 & 18 \\ 0 & 18 & 0 \end{pmatrix} =-3\cdot 18^2=-972. \end{align*} Here we have used the properties of the trace, and that the discriminant can be computed by the matrix of traces with respect to an integral basis. For $L=\mathbb{Q}(\sqrt[3]{12})$ an integral basis is given by $$ \{1,\beta,\frac{\beta^2}{2}\}, $$ where $\beta=\sqrt[3]{12}$. Again with the trace matrix we see that the discriminant of $L$ is $-972$, too.

Dietrich Burde
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  • how do you show it's an integral power basis ? – mercio Feb 15 '17 at 15:10
  • Since $\alpha$ satisfies the polynomial equation $x^3-6=0$ it is integral, so that of course $\mathbb{Z}[\alpha]\subseteq \mathcal{O}_K$. For the other direction one has to do more, see for example the arguments here, and the notes of Keith Conrad. It goes like the case $\mathbb{Q}[\sqrt[3]{2}]$. – Dietrich Burde Feb 15 '17 at 15:12
  • @Walter Which isomorphism? The number fields are not isomorphic, because the rings of integers are different. One is monogenic, the other not. – Dietrich Burde Feb 15 '17 at 15:54
  • The ring $\mathcal O_L$ is monogenic: it contains $\gamma := \sqrt[3]{12} + \sqrt[13]{18} = \beta + \beta^2/2$ and the minimal poly. of $\gamma$ over $\mathbf Q$ is $x^3-18x-30$, which has discriminant $-972 = {\rm disc}(L)$, so $\mathcal O_L = \mathbf Z[\gamma]$. To see $K$ and $L$ are not isomorphic, look at primes splitting completely: $7$ splits completely in $K$ and is inert in $L$. – KCd May 29 '21 at 02:48