# Discriminant of $\Bbb Q(\sqrt[3]{2})$

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?

• Is the class using a textbook or did the teacher just pull that out of thin air? – Robert Soupe Mar 11 '17 at 18:59

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 $N(3\alpha^2) = N(3)N(\alpha)^2 = 3^3 \cdot 2^2$.

• what's the discriminant of $\Bbb Q(\sqrt[3] 4)$ ? – mercio Mar 15 '17 at 14:01
• This formula works only if powers of $\alpha$ form an integral basis. In case you didn't get mercio's hint :-) – Jyrki Lahtonen Mar 15 '17 at 14:07
• Thanks for the correction; I edited my answer accordingly. – Ricardo Buring Mar 17 '17 at 17:05

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.

• 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

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 $$$$det\big(\theta_i^{j-1}\big)^2 = (\theta_1-\theta_2)^2(\theta_1-\theta_3)^2(\theta_2-\theta_3)^2 = -108.$$$$ Now we obtain the discriminant.