Showing that a cubic extension of an imaginary quadratic number field is unramified. Let $\alpha^3-\alpha-1=0$, $K=\mathbb Q(\sqrt{-23})$, $K'=\mathbb Q(\alpha)$, and $L=\mathbb Q(\sqrt{-23},\alpha)$. 
Then I am asked to show that the field extension $L/K$ is unramified.
I know that if $\mathfrak p\in\operatorname{Max}(\mathcal O_K)$ ramifies in $L$ then $\mathfrak p\mid\mathfrak d$ where $\mathfrak d$ is the discriminant ideal, i.e. the ideal of $\mathcal O_L$ generated by all $\operatorname{disc}(x_1,x_2,x_3)$ such that $(x_1,x_2,x_3)$ is a $K$-basis of $L$ and $x_1,x_2,x_3\in\mathcal O_L$. 
I know that $(1,\alpha,\alpha^2)$ is one such basis with discriminant $-23$, so if $\mathfrak p$ ramifies in $L$ then $23\in\mathfrak p$, and in $\mathcal O_K$ we have $(23)=(\sqrt{-23})^2$. Therefore the only candidate for $\mathfrak p$ is $(\sqrt{23})$. If I could find a basis with discriminant not divisible by $23$ then I would be done, but that quickly turns messy.
Now factoring $23$ in $\mathcal O_{K'}$ gives $(23)=(23,\alpha-3)(23,\alpha-10)^2$, so I need would like to have two different prime ideals of $\mathcal O_L$ containing $(23,\alpha-10)$, then I will have three different prime ideals of $\mathcal O_L$ containing $\sqrt{-23}$ and I will be done. 
Alternatively I need to show that no prime ideal $\mathfrak q$ of $\mathcal O_L$ has $\sqrt{-23}\in\mathfrak q^2$.
Any ideas?

Observe that $$\begin{align}23 &= \alpha^3-\alpha-24=(\alpha-3)(\alpha^2+3\alpha+8) \\&=(\alpha-3)(\alpha-(-\frac{3}2+\sqrt {-23}))(\alpha-(-\frac{3}2-\sqrt {-23}))\end{align}$$
So $$(23,\alpha-10)\subset(\alpha-(-\frac 32\pm \sqrt{-23}),\alpha-10)\subset\mathcal O_L$$
