| Version 2 |
Version 1 |
| Let $K$ be a quadratic number field, i.e. $K=\Rats(\sqrt{d})$ for |
Let $K$ be a quadratic number field, i.e. $K=\Rats(\sqrt{d})$ for |
| some square-free integer $d$. The discriminant of the extension is |
some square-free integer $d$. The discriminant of the extension is |
| \begin{eqnarray*} |
$$ |
|
D_{K/\Rats}=\begin{cases} d, & \text{ if } d\equiv 1 \ \operatorname{mod}\ 4,\\
|
D_{K/\Rats}=\begin{cases} d, \text{ if } d\equiv 1 \ \operatorname{mod}\ 4,\\
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4d, & \text{ if } d\equiv 2,3 \operatorname{mod}\ 4.\end{cases}
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4d, \text{ if } d\equiv 2,3 \operatorname{mod}\ 4.\end{cases}
|
| \end{eqnarray*} |
$$ |
| Let $\mathcal{O}_K$ denote the ring of integers of $K$. We have: |
Let $\mathcal{O}_K$ denote the ring of integers of $K$. We have: |
|
\begin{eqnarray*}\mathcal{O}_K\cong \begin{cases}
|
$$\mathcal{O}_K\cong \begin{cases}
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\Ints\oplus \frac{1+\sqrt{d}}{2}\Ints, & \text{ if } d\equiv 1 \ \operatorname{mod}\ 4,\\
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\Ints\oplus \frac{1+\sqrt{d}}{2}\Ints, \text{ if } d\equiv 1 \ \operatorname{mod}\ 4,\\
|
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\Ints\oplus \sqrt{d}\Ints, & \text{ if } d\equiv 2,3
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\Ints\oplus \sqrt{d}\Ints, \text{ if } d\equiv 2,3
|
| \operatorname{mod}\ 4. \end{cases} |
\operatorname{mod}\ 4. \end{cases} |
| \end{eqnarray*} |
$$ |
| Prime ideals of $\Ints$ decompose as follows in $\mathcal{O}_K$: |
Prime ideals of $\Ints$ decompose as follows in $\mathcal{O}_K$: |
|
|
| \begin{thm} Let $p\in \Ints$ be a prime. |
\begin{thm} Let $p\in \Ints$ be a prime. |
| \begin{enumerate} |
\begin{enumerate} |
| \item If $p\mid d$ (divides), then |
\item If $p\mid d$ (divides), then |
| $p\mathcal{O}_K=(p,\sqrt{d})^2$; |
$p\mathcal{O}_K=(p,\sqrt{d})^2$; |
|
|
| \item If $d$ is odd, then |
\item If $d$ is odd, then |
|
\begin{eqnarray*}2\mathcal{O}_K=\begin{cases}
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$$2\mathcal{O}_K=\begin{cases}
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(2,1+\sqrt{d})^2, & \text{ if } d\equiv 3\ \operatorname{mod}\ 4,\\
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(2,1+\sqrt{d})^2, \text{ if } d\equiv 3\ \operatorname{mod}\ 4,\\
|
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\left(2,\frac{1+\sqrt{d}}{2}\right)\left(2,\frac{1-\sqrt{d}}{2}\right), & \text{ if } d\equiv 1\ \operatorname{mod}\ 8,\\
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\left(2,\frac{1+\sqrt{d}}{2}\right)\left(2,\frac{1-\sqrt{d}}{2}\right), \text{ if } d\equiv 1\ \operatorname{mod}\ 8,\\
|
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\text{prime}, & \text{ if } d\equiv 5\ \operatorname{mod}\ 8.
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\text{prime}, \text{ if } d\equiv 5\ \operatorname{mod}\ 8.
|
|
\end{cases} \end{eqnarray*}
|
\end{cases} $$
|
|
|
| \item If $p\neq 2$, $p$ does not divide $d$, then |
\item If $p\neq 2$, $p$ does not divide $d$, then |
| \begin{eqnarray*} |
$$ |
|
p\mathcal{O}_K=\begin{cases} (p,n+\sqrt{d})(p,n-\sqrt{d}), & \text{ if } d\equiv n^2\ \operatorname{mod}\ p,\\
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p\mathcal{O}_K=\begin{cases} (p,n+\sqrt{d})(p,n-\sqrt{d}), \text{ if } d\equiv n^2\ \operatorname{mod}\ p,\\
|
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\text{prime}, & \text{ if $d$ is not prime } \operatorname{mod}\ p.
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\text{prime}, \text{ if $d$ is not prime } \operatorname{mod}\ p.
|
| \end{cases} |
\end{cases} |
| \end{eqnarray*} |
$$ |
| \end{enumerate} |
\end{enumerate} |
| \end{thm} |
\end{thm} |
|
|
| \begin{thebibliography}{9} |
\begin{thebibliography}{9} |
| \bibitem{marcus} Daniel A.Marcus, {\em Number Fields}. Springer, New York. |
\bibitem{marcus} Daniel A.Marcus, {\em Number Fields}. Springer, New York. |
| \end{thebibliography} |
\end{thebibliography} |
