Dgdrxrt

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

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

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

Let $f in mathbbQ[x]$ monic of degree $n$ be the minimum polynomial of $alpha$ and let $K=mathbbQ(alpha)$.

1. The discriminant of $mathbbZ[alpha]$ is $(-1)^frac12n(n-1)N^K_mathbbQ(f'(alpha))$.


2. The field norm $N^K_mathbbQ$ is multiplicative.


3. $N^K_mathbbQ(b) = b^n$ for $b in mathbbQ$.


4. $N^K_mathbbQ(alpha)$ is $(-1)^n$ times the constant term of $f$.

Putting these together for $f = x^3 - 2$ (using $mathcalO_K = mathbbZ[alpha]$ here, which is nontrivial) yields

The absolute value of the discriminant of $mathbbQ(sqrt[3]2)$ is $N(3alpha^2) = N(3)N(alpha)^2 = 3^3 cdot 2^2$.

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 $mathbbQ$, 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.

Let $L=mathbbQ(sqrt[3]2)$ and let $f(x)=x^3-2$ be the minimal polynomial of $sqrt[3]2$ over $mathbbQ$. Consider $e_1,e_2,e_3:=1,x,x^2$ be a basis of $L$ over $mathbbQ$. Let $theta_1 = sqrt[3]2,theta_2 = wsqrt[3]2$ and $theta_3=w^2sqrt[3]2$ be the roots of $f$ (in $mathbbC$ 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 $mathbbQ$-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 $big(sigma_i(e_j)big)_ij = big(theta_i^j-1big)$ is Vandermonde whose determinant satisfies
beginequation
detbig(theta_i^j-1big)^2 = (theta_1-theta_2)^2(theta_1-theta_3)^2(theta_2-theta_3)^2 = -108.
endequation
Now we obtain the discriminant.