calculation on the discriminant of the field extension This is a homework problem:
Suppose $m,n$ are natural numbers such that $0<m<n$ and $m,n$ do not contain square factors. Suppose $x=\sqrt[3]{m^2n}$ and $K=\mathbb{Q}(x)$. Then show that $y=\sqrt[3]{mn^2}$ is in $\mathcal{O}_K$, and $\{1,x,y\}$ is a basis $\mathcal{B}$ of $K$. Calculate the discriminant of $\mathcal{B}$.
The fiest 2 questions are easy, for the discriminant, the lecture note says that:
For a number field $K$, suppose $f_1,...,f_n$ are the embeddings of $K$ into $\mathbb{C}$, and $\{b_1,...,b_n\}$ a $\mathbb{Q}$-basis, then $\Delta_K(b_1,...,b_n)=$det$(f_i(b_i))^2$.
So I calculated the discriminant of the matrix \begin{pmatrix}
1 & x & y \\
x & x^2 & xy \\
y & xy & y^2
\end{pmatrix}
But the solution calculated two matrices
\begin{equation}
\delta_x=\begin{pmatrix}
0 & 0 & mn \\
1 & 0 & 0 \\
0 & m & 0
\end{pmatrix}
\end{equation}
\begin{equation}
\delta_y=\begin{pmatrix}
0 & mn & 0 \\
0 & 0 & n \\
1 & 0 & 0
\end{pmatrix}
\end{equation}
Then calculate the determinant of
\begin{pmatrix}
1 & Tr(x) & Tr(y) \\
Tr(x) & Tr(x^2) & Tr(xy) \\
Tr(y) & Tr(xy) & Tr(y^2)
\end{pmatrix}
I dont really understand this, can anyone give an explanation?
 A: EDIT: You should determine what is integral basis of $\mathcal{O}_{K}$. When $m^2\equiv n^2(\mod 9)$, $\{x,y,\frac{1+mx+ny}{3}\}$ is integral basis. Otherwise, $\{1,x,y\}$ is an integral basis. This can be shown via Dedekind's factorization theorem, I will elaborate this later.
When $m^2\not \equiv n^2(\mod 9)$, it is straightforward to calculate the discriminant via trace form $\Delta_K = \det(\text{Tr}(\alpha_i \alpha_j))$, where $\{\alpha_i\}$ are integral basis of $K$. So in your case, it suffices to calculate
$$
\det \left(\begin{array}{ccc}
\operatorname{Tr}(1) & \operatorname{Tr}(x) & \operatorname{Tr}(y) \\
\operatorname{Tr}(x) & \operatorname{Tr}\left(x^{2}\right) & \operatorname{Tr}(x y) \\
\operatorname{Tr}(y) & \operatorname{Tr}(x y) & \operatorname{Tr}\left(y^{2}\right)
\end{array}\right)
$$
where $\operatorname{Tr}(\alpha)=\operatorname{Tr}_{K/\mathbb{Q}}(\alpha)$.
By the definition of trace, $\operatorname{Tr}_{K/\mathbb{Q}}(\alpha)$=Trace of left multiplication of $\alpha$. So you can check we have the following: $\operatorname{Tr}_{K/\mathbb{Q}}(1)=3, \operatorname{Tr}_{K/\mathbb{Q}}(x)=\operatorname{Tr}_{K/\mathbb{Q}}(y)=\operatorname{Tr}_{K/\mathbb{Q}}(x^2)=\operatorname{Tr}_{K/\mathbb{Q}}(y^2)=0$ and $\operatorname{Tr}_{K/\mathbb{Q}}=3mn$. Thus $\Delta_K = -27m^2n^2$. And I will leave other case to you.
