[[Imaginary quadratic field]]
# $\mathbb{Q}(\sqrt{ -14 })$

Consider the monogenic [[imaginary quadratic field]] $K = \mathbb{Q}(\alpha)$ where $\alpha=\sqrt{ -14 }$. #m/thm/num/alg 

> [!code]- Sage
> ```sage
> K.<α> = QuadraticField(-14)
> ```

## Discriminant

By [[Discriminant of an algebraic integer]], 
$$
\begin{align*}
\Delta_{K} = -56.
\end{align*}
$$

## Group of units

By [[Imaginary quadratic field#^P1]],
$$
\begin{align*}
\mathcal{O}_{K}^\times = \{ 1,-1 \}.
\end{align*}
$$

## Class group

[[Minkowski's bound]] is given by
$$
\begin{align*}
M_{K} = \frac{4\sqrt{ 14 }}{\pi} < 5,
\end{align*}
$$
so applying [[Kummer's factorization theorem]]

| $p$ | $x^2 + 14 \bmod p$ | $\langle p \rangle$                  | norms |
| --- | ------------------ | ------------------------------------ | ----- |
| $2$ | $x^2$              | $\mathfrak{p}_{2}^2$                 | $2$     |
| $3$ | $(x+1)(x-1)$       | $\mathfrak{p}_{3} \mathfrak{p}_{3}'$ | $3,3$      |

Clearly no algebraic integers can have these norms, so we can be satisfied that these are not principal.
Since $\mathfrak{p}_{3}^{-1} = \mathfrak{p}_{3}'$, the ideal class group is generated by $\{[\mathfrak{p}_{2}] ,[\mathfrak{p}_{3}]\}$.
Some algebraic integers of small [[field norm]] are

| $t$     | $\opn N_{K:\mathbb{Q}}(\alpha+t)$ |
| ------- | --------------------------------- |
| $\pm 1$ | $3 \cdot 5$                       |
| $\pm 2$ | $2 \cdot 3^2$                     |
| $\pm3$  | $23$                              |
from which we derive the relation
$$
\begin{align*}
\mathfrak{p}_{2} \mathfrak{p}_{3}^2 = \langle 2,\alpha \rangle \langle 3,\alpha+1 \rangle ^2 = \langle \vartheta - 2 \rangle \sim \langle 1 \rangle 
\end{align*}
$$
whence $\mathfrak{p}_{3}^2 \sim \mathfrak{p}_{2}$, so $\mathfrak{p}_{3}^4 \sim \langle 1 \rangle$.
Therefore
$$
\begin{align*}
\Cl K = \langle [\mathfrak{p}_{3}] \rangle \cong \mathrm{C}_{4}.
\end{align*}
$$


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