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?

**Answer**

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 $K=\mathbb{Q}(\alpha)$.

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

**Attribution***Source : Link , Question Author : Community , Answer Author : Community*