# 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 $$(23,\alpha-10)\subset(\alpha-(-\frac 32\pm \sqrt{23}),\alpha-10)\subset\mathcal O_L$$ Dec 2, 2015 at 18:37
• I wish, @Mathmo123, that you had put this into an answer (after correcting the minor typos). Dec 5, 2015 at 3:50
• @user : related: math.stackexchange.com/questions/492888 Jan 8, 2017 at 18:44

The field $K$ is ramified only at $23$, and since the $\Bbb Q$-discriminant of $k=\Bbb Q(\alpha)$ is of absolute value $23$ as well, the only possibility for ramification of $K'=Kk$ over $K$ is above the prime $23$, in other words at the unique prime of $K$ lying over the $\Bbb Z$-prime $23$. So we may think of localizing and completing.
Calling $\,f(X)=X^3-X-1$, we have $f(3)=23$ and $f(10)=989=23\cdot43$. Indeed, over $\Bbb F_{23}$, we have $f(X)\equiv(X-3)(X-10)^2$. Let’s examine $g(X)=f(X+10)=X^3+30X^2+299X+989=X^3+30X+13\cdot23X+43\cdot23$, which shows that $g$ has two roots of $23$-adic valuation $1/2$ (additive valuation, that is). Now let’s go farther, and, calling $\sqrt{-23}=\beta$, look at \begin{align} h(X)=g(X+4\beta)&=X^3+(30+12\beta)X^2+(-805 + 240\beta)X+(-10051 - 276\beta)\\ &=X^3+(30+12\beta)X^2+(-35\cdot23+240\beta)X+(-19\cdot23^2-12\cdot23\beta)\,. \end{align} Look at the $23$-adic valuations of the coefficients: $0$ for the degree-$3$ and degree-$2$ terms, $1/2$ for the linear term, and $3/2$ for the constant term. So the Newton polygon of $h$ has three segments of width one, slopes $0$, $1/2$, and $1$. Thus $h(X)=f(X+10+4\beta)$ has three roots all in $\Bbb Q_{23}(\sqrt{-23}\,)$, and therefore the unique prime of $K$ above $23$ splits completely in $K'$, and the extension is unramified.
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 $$(23,\alpha-10)\subset(\alpha-(-\frac 32\pm \sqrt{-23}),\alpha-10)\subset\mathcal O_L$$