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I want to understand a way of computing the discriminant of the number field $K=\mathbb{Q}(\sqrt[3]{2})$. The degree of $K|\mathbb{Q}$ is $n=3$ and we have $3=1+2\cdot 1$, so there are one real and two complex embeddings.

Now my teacher concludes that the absolute value of the discriminant is equal to $2^2\cdot 3^3$. Why that?

3 Answers3

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I'm not sure that this is what your teacher had in mind, but it allows a quick calculation.

Let $f \in \mathbb{Q}[x]$ monic of degree $n$ be the minimum polynomial of $\alpha$ and let $K=\mathbb{Q}(\alpha)$.

  1. The discriminant of $\mathbb{Z}[\alpha]$ is $(-1)^{\frac{1}{2}n(n-1)}N^K_{\mathbb{Q}}(f'(\alpha))$.
  2. The field norm $N^K_{\mathbb{Q}}$ is multiplicative.
  3. $N^K_{\mathbb{Q}}(b) = b^n$ for $b \in \mathbb{Q}$.
  4. $N^K_{\mathbb{Q}}(\alpha)$ is $(-1)^n$ times the constant term of $f$.

Putting these together for $f = x^3 - 2$ (using $\mathcal{O}_K = \mathbb{Z}[\alpha]$ here, which is nontrivial) yields

The absolute value of the discriminant of $\mathbb{Q}(\sqrt[3]{2})$ is $N(3\alpha^2) = N(3)N(\alpha)^2 = 3^3 \cdot 2^2$.

Community
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Ricardo Buring
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Concerning the necessary background, the article The splitting field of $x^3-2$ by K. Conrad does explain this in detail, see Theorem $2$ on page $4$. Taking the determinant of the matrix $(\sigma_i(x_j))^2_{i,j}$ gives the discriminant, where $\{x_1,x_2,\ldots, x_n\}$ is an integral basis of the number field $K$ over $\mathbb{Q}$, and the $\sigma_i$ the embeddings.Here $n=3$. Another method is explained here; and using the trace matrix for the discriminant of a cubic number field has been computed here.

hbghlyj
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Dietrich Burde
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    Emphasis: ${x_1,x_2,\ldots,x_n}$ should be an integral basis. Of course, it may be that some texts imply with the phrase a basis of a number field that the basis is to be integral. – Jyrki Lahtonen Mar 15 '17 at 14:06
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Let $L=\mathbb{Q}(\sqrt[3]{2})$ and let $f(x)=x^3-2$ be the minimal polynomial of $\sqrt[3]{2}$ over $\mathbb{Q}$. Consider $\{e_1,e_2,e_3\}:=\{1,x,x^2\}$ be a basis of $L$ over $\mathbb{Q}$. Let $\theta_1 = \sqrt[3]{2},\theta_2 = w\sqrt[3]{2}$ and $\theta_3=w^2\sqrt[3]{2}$ be the roots of $f$ (in $\mathbb{C}$ as a splitting field of $f$) where $w$ is a complex cube root of $1$. Then the discriminant is the square of the determinant of the matrix $\big(\sigma_i(e_j)\big)_{ij}$ where $\sigma_i$ are $\mathbb{Q}$-embeddings.

We know that the $i$-th embedding takes $x$ to $\theta_i$, i.e. $\sigma_i(1)=1$, $\sigma_i(x)=\theta_i$ and $\sigma_i(x^2)=\theta_i^2$. Clearly the matrix $\big(\sigma_i(e_j)\big)_{ij} = \big(\theta_i^{j-1}\big)$ is Vandermonde whose determinant satisfies \begin{equation} \det\big(\theta_i^{j-1}\big)^2 = (\theta_1-\theta_2)^2(\theta_1-\theta_3)^2(\theta_2-\theta_3)^2 = -108. \end{equation} Using $\mathcal{O}_{\mathbb{Q}(\sqrt[3]{2})} = \mathbb{Z}[\sqrt[3]{2}]$, we obtain the discriminant.

Yibo Zhang
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    That does not prove the discriminant of $L$ is $-108$ because the discriminant of a number field is defined to be the discriminant of a $\mathbf Z$-basis of its ring of integers. Your calculation shows ${\rm disc}(\mathbf Z[\sqrt[3]{2}]) = {\rm disc}({1,\sqrt[3]{2},\sqrt[3]{4}})= -108$, but you didn't show $\mathcal O_L =\mathbf Z[\sqrt[3]{2}]$. We can say $-108={\rm disc}(\mathbf Z[\sqrt[3]{2}]) = [\mathcal O_L:\mathbf Z[\sqrt[3]{2}]]^2{\rm disc}(L)$, so $[\mathcal O_L:\mathbf Z[\sqrt[3]{2}]]$ is $1$, $2$, $3$, or $6$. Explain why it is $1$. – KCd Jan 18 '24 at 19:46
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    You're right. I should emphasise that $\mathcal{O}_{\mathbb{Q}(\sqrt[3]{2})} = \mathbb{Z}[\sqrt[3]{2}]$. – Yibo Zhang Jan 18 '24 at 21:04