Sheet1 B5

$\DeclareMathOperator{\tr}{tr}\DeclareMathOperator{\disc}{disc}$Suppose that $β$ is a root of $X^3+p X+q=0$, where $X^3+p X+q$ is an irreducible polynomial in $𝐙[X]$, and let $K=𝐐(β)$. Compute $\tr_{K/𝐐}(β^i)$ for $i=0,1, …, 4$. Deduce that $\disc_{K/𝐐}(1, β, β^2)=-4 p^3-27 q^2$. Hence, give an example of a cubic number field $K$ such that $𝒪_K$ has a power integral basis. $\{1,β,β^2\}$ is a 𝐐-basis of $K$ $β^3=-q-pβ$ $\tr_{K/𝐐}(1)=\tr\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}=3$ $\tr_{K/𝐐}(β)=\tr\begin{pmatrix}0&1&0\\0&0&1\\-q&-p&0\end{pmatrix}=0$ $\tr_{K/𝐐}(β^2)=\tr\begin{pmatrix}0&0&1\\-q&-p&0\\0&-q&-p\end{pmatrix}=-2p$ $\tr_{K/𝐐}(β^3)=\tr\begin{pmatrix}-q&-p&0\\0&-q&-p\\pq&p^2&-q\end{pmatrix}=-3q$ $\tr_{K/𝐐}(β^4)=\tr\begin{pmatrix}0&-q&-p\\pq&p^2&-q\\q^2&2pq&p^2\end{pmatrix}=2p^2$ \begin{align*}\disc_{K/𝐐}(1, β, β^2)&=\begin{vmatrix}\tr_{K/𝐐}(1)&\tr_{K/𝐐}(β)&\tr_{K/𝐐}(β^2)\\\tr_{K/𝐐}(β)&\tr_{K/𝐐}(β^2)&\tr_{K/𝐐}(β^3)\\\tr_{K/𝐐}(β^2)&\tr_{K/𝐐}(β^3)&\tr_{K/𝐐}(β^4)\\\end{vmatrix}\\&=\begin{vmatrix}3&0&-2p\\ 0&-2p&-3q\\-2p&-3q&2p^2\end{vmatrix} \\&=-4 p^3 - 27 q^2\end{align*} When $p=q=1$, $\disc_{K/𝐐}(1, β, β^2)=-31$ is squarefree, by 2.18 $K=𝐐(β)$ is a cubic number field such that $𝒪_K$ has a power integral basis $\{1, β, β^2\}$.