## You are here

Homelemma for imaginary quadratic fields

## Primary tabs

# lemma for imaginary quadratic fields

For determining the imaginary quadratic fields whose ring of integers has unique factorization, one can use the following

Lemma. Let $d$ be a negative integer with $d\equiv 1\;\;(\mathop{{\rm mod}}4)$, $p$ the greatest odd irreducible integer with $p\leqq\sqrt{\frac{1}{3}|d|}$ and $q=\frac{1}{4}(1\!-\!d)$. In the imaginary quadratic field $\mathbb{Q}(\sqrt{d})$, the factorization of integers is unique if and only if the integers

$\displaystyle t^{2}\!-\!t\!+\!q\quad\;\left(t=1,\,2,\,\ldots,\,\frac{p\!+\!1}{% 2}\right)$ | (1) |

are irreducible in the field of the rational numbers.

The lemma yields the below table:

$q$ | $d=1-4q$ | $p$ | $\frac{1}{2}(p\!+\!1)$ | the numbers (1) |
---|---|---|---|---|

$1$ | $-3$ | $1$ | $1$ | 1 |

$2$ | $-7$ | $1$ | $1$ | 2 |

$3$ | $-11$ | $1$ | $1$ | 3 |

$5$ | $-19$ | $1$ | $1$ | 5 |

$11$ | $-43$ | $3$ | $2$ | 11, 13 |

$17$ | $-67$ | $3$ | $2$ | 17, 19 |

$41$ | $-163$ | $7$ | $4$ | 41, 43, 47, 53 |

# References

- 1 K. Väisälä: Lukuteorian ja korkeamman algebran alkeet. Tiedekirjasto No. 17. Kustannusosakeyhtiö Otava, Helsinki (1950).

Related:

ListOfAllImaginaryQuadraticPIDs, ClassNumbersOfImaginaryQuadraticFields

Type of Math Object:

Theorem

Major Section:

Reference

Parent:

Groups audience:

## Mathematics Subject Classification

11R11*no label found*11R04

*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections

## Comments

## squarefree?

does d have to be squarefree in this lemma?

## Re: squarefree?

V\"ais\"al\"a's proof is quite long, and partly rests on former theorems where one usually supposes the squarefreeness. Of course, one never needs to think that when speaking of quadratic field Q(\sqrt{d}), d were other than square-free.

Regards,

Jussi