Gender theory

The gender theory is a term from the mathematical branch of algebraic number theory . In many cases, gender theory gives a satisfactory answer to the question of how prime numbers are represented by non-equivalent, binary quadratic forms with the same discriminant . That is, it makes it possible to decide whether or not a prime number is represented by a square form in two variables. However, it does not generally make any statements about the representation of general forms.

Genders

In his Disquisitiones Arithmeticae , Carl Friedrich Gauß developed the gender theory as the theory of the sexes of square forms. One of the greatest achievements that Gauss made on gender theory is the calculation of the number of genders of forms with a given discriminant . He was finally able to show that their number is the same , where denotes the number of prime factors contained in. In addition, he proved that there is always a divisor of the (real) equivalence classes of primitive, positively definite forms with discriminants . In the following denote the totality ring of the square number fields . The gender theory can be treated with ideal classes in addition to the square forms. If the division into ideal classes in the narrower sense is finer than that in the usual, then the division into sexes is very rough. Two ideals that are different from one another are called similar (in the symbol ) if their norms apply, with one of the totally different ones being totally positive . The associated equivalence classes are called genders. ${\displaystyle D}$ ${\displaystyle 2^{r-1}}$ ${\displaystyle r}$ ${\displaystyle D}$ ${\displaystyle 2^{r-1}}$ ${\displaystyle D}$ ${\displaystyle {{\rm {\mathcal {O}}}_{{\mathbb {Q} }({\sqrt {d}})}}}$ ${\displaystyle {\mathbb {Q} }({\sqrt {d}})}$ ${\displaystyle {(0)}}$ ${\displaystyle {\rm {\mathfrak {a}}},{\rm {\mathfrak {b}}}{\rm {\trianglelefteq }}{\rm {\mathcal {O}}}_{{\mathbb {Q} }({\sqrt {d}})}}$ ${\displaystyle {\rm {\mathfrak {a}}}{\overset {+}{\approx }}{\rm {\mathfrak {b}}}}$ ${\displaystyle N({\rm {\mathfrak {a}}})=N(\lambda )N({\rm {\mathfrak {b}}})}$ ${\displaystyle 0}$ ${\displaystyle \lambda \in {\mathbb {Q} }({\sqrt {d}})^{\times }}$

In particular, the set of all genders forms an Abelian group , the so-called gender class group . The element of is called the main sex . It is that which contains the main ideals in the strict sense. Ideals, which are equivalent in the narrower sense, obviously belong to the same sex if they are too prime. One can now show that ideals are similar if and only if they belong to the same gender, i.e. if their ideal classes in the narrower sense differ by a square , that is, it applies ${\displaystyle Cl_{gen}^{+}({\mathbb {Q} }({\sqrt {d}}))}$ ${\displaystyle Cl_{gen}^{+}({\mathbb {Q} }({\sqrt {d}}))}$ ${\displaystyle d}$ ${\displaystyle {\rm {\mathfrak {a}}},{\rm {\mathfrak {b}}}{\rm {\trianglelefteq }}{\rm {\mathcal {O}}}_{{\mathbb {Q} }({\sqrt {d}})}}$

${\displaystyle {\rm {\mathfrak {a}}}{\overset {+}{\approx }}{\rm {{\mathfrak {b}}\;\;}}\Leftrightarrow {\rm {\;\;{\mathfrak {a}}}}{\overset {+}{\sim }}{\rm {{\mathfrak {b}}{\mathfrak {c}}}}^{2}}$

for an ideal . ${\displaystyle {\rm {\mathfrak {c}}}{\rm {\trianglelefteq }}{\rm {\mathcal {O}}}_{{\mathbb {Q} }({\sqrt {d}})}}$

This means that the gender class group is isomorphic to , with the ideal class group being referred to in the narrower sense. One can show that in a quadratic number field with a discriminant the number of genders is the same . It then immediately follows that the number of classes is in each gender , where denotes the number of different prime divisors of . ${\displaystyle {Cl_{gen}^{+}({\mathbb {Q} }({\sqrt {d}}))}}$ ${\displaystyle C_{+}/C_{+}^{2}}$ ${\displaystyle C_{+}}$ ${\displaystyle D}$ ${\displaystyle 2^{r-1}}$ ${\displaystyle {\rm {\mathfrak {K}}}(D)={\frac {h(D)}{2^{r-1}}}}$ ${\displaystyle r}$ ${\displaystyle D}$

Correspondence

In algebraic number theory there is a correspondence theorem that makes a statement about the connection between real equivalence classes of primitive quadratic forms and the equivalence classes of ideals in the narrower sense.

Be ( not a square) a fundamentally discriminant . Then there is a bijective correspondence between the real equivalence classes of primitive quadratic forms with discriminants and the equivalence classes in the narrower sense of ideals of . In particular, the number of equivalence classes of ideals in the narrower sense is equal to the class number ${\displaystyle D\equiv 0,1\mod {4},}$ ${\displaystyle D}$ ${\displaystyle D}$ ${\displaystyle {\rm {\mathcal {O}}}_{{\mathbb {Q} }({\sqrt {d}})}}$ ${\displaystyle h_{{\mathbb {Q} }({\sqrt {d}})}^{+}}$ ${\displaystyle h(D)}$ .

In the proof, the correspondence between ideals and quadratic forms is explicitly constructed, see correspondence theorem of algebraic number theory .

Note that in general there is no bijection between the equivalence classes of primitive, positively definite quadratic forms and the ideal classes in the ordinary sense. Is about , then and . In contrast is . The reason for this is that the fundamental unit of totally is positive. ${\displaystyle D=-303}$ ${\displaystyle h(-303)=10}$ ${\displaystyle h_{{\mathbb {Q} }({\sqrt {-303}})}^{}=5}$ ${\displaystyle h_{{\mathbb {Q} }({\sqrt {-303}})}^{+}=h(-303)=10}$ ${\displaystyle \mathbb {Q} ({\sqrt {-303}})}$

Division into gender classes

If there is a square shape with a discriminant and any two numbers represented by the shape (it doesn't matter whether the numbers are prime numbers or not), then the product can always be brought into the shape . ${\displaystyle f=\left\langle a,b,c\right\rangle }$ ${\displaystyle D}$ ${\displaystyle z,w}$ ${\displaystyle f}$ ${\displaystyle zw}$ ${\displaystyle x^{2}-dy^{2},{\rm {\;\;}}x,y\in {\mathbb {Z} }}$

example

${\displaystyle z=a\alpha ^{2}+2b\alpha \gamma +c\gamma ^{2}{\rm {\;\;}},{\rm {\;\;{\textit {w}}=a}}\beta ^{2}+2b\beta \delta +c\delta ^{2}}$where , ${\displaystyle \alpha ,\beta ,\gamma ,\delta \in {\mathbb {Z} }}$

then the mold is by a unimodular transformation with ${\displaystyle \left\langle a,b,c\right\rangle }$

${\displaystyle \left({\begin{array}{cc}{\alpha }&{\beta }\\{\gamma }&{\delta }\end{array}}\right)\in {\mathbb {Z} }^{2\times 2}}$ and ${\displaystyle \alpha \delta -\gamma \beta =1}$

into the shape over. Then its discriminant is of the form , that is, the product of the form . ${\displaystyle \left\langle z,x,w\right\rangle }$ ${\displaystyle x^{2}-zw}$ ${\displaystyle dy^{2}}$ ${\displaystyle zw}$ ${\displaystyle x^{2}-dy^{2}}$

For the division of the square shapes into gender classes, this results in:

1. Let be for odd in rising prime numbers, then for every natural number which can be represented by the form and which is not a divisor of , has the Legendre symbol ${\displaystyle p_{1},\ldots ,p_{t}}$ ${\displaystyle t\geq 0}$ ${\displaystyle D}$ ${\displaystyle m}$ ${\displaystyle f}$ ${\displaystyle p_{i}{\rm {\;}},{\rm {\;}}1\leq i\leq t}$ ${\displaystyle m}$

${\displaystyle \chi _{i}(m)=\left({\frac {m}{p_{i}}}\right)}$

one and the same value. Because if there are any two prime numbers that can be represented by, then it follows that ${\displaystyle m_{1},m_{2}}$ ${\displaystyle p}$ ${\displaystyle f}$

${\displaystyle m_{1}m_{2}\equiv x^{2}\mod {p}}$and with it , so . This is called a Dirichlet character modulo . ${\displaystyle {\frac {m_{1}m_{2}}{p}}=+1}$ ${\displaystyle {\frac {m_{1}}{p}}={\frac {m_{2}}{p}}}$ ${\displaystyle \chi _{i}}$ ${\displaystyle p}$

2. Be . Then for all the odd numbers represented by this form has the expression ${\displaystyle D\equiv 3\mod {4}}$ ${\displaystyle m}$

${\displaystyle \varsigma (m)=(-1)^{\frac {m-1}{2}}}$

one and the same value. Because if any two numbers are odd, then and since the product is odd, one of the two numbers must be even and the other odd. So that also implies and with it . ${\displaystyle m_{1},m_{2}}$ ${\displaystyle m_{1}m_{2}=x^{2}-dy^{2}\equiv x^{2}+y^{2}\mod {4}}$ ${\displaystyle m_{1}m_{2}}$ ${\displaystyle x,y}$ ${\displaystyle m_{1}m_{2}\equiv 1\mod {4}}$ ${\displaystyle m_{1}\equiv m_{2}\mod {4}}$ ${\displaystyle (-1)^{\frac {m_{1}-1}{2}}=(-1)^{\frac {m_{2}-1}{2}}}$

3. Be . Then for all the odd numbers represented by this form has the expression ${\displaystyle D\equiv 2\mod {8}}$ ${\displaystyle m}$

${\displaystyle \varepsilon (m)=(-1)^{\frac {m^{2}-1}{8}}}$

one and the same value.

4. Is . Then for all odd numbers represented by this form has the expression ${\displaystyle D\equiv 6\mod {8}}$ ${\displaystyle m}$

${\displaystyle \varsigma (m)\cdot \varepsilon (m)=(-1)^{{\frac {m-1}{2}}+{\frac {m^{2}-1}{8}}}}$

one and the same value.

5. Be . Then for all the odd numbers represented by this form has the expression ${\displaystyle D\equiv 4\mod {8}}$ ${\displaystyle m}$

${\displaystyle \varsigma (m)=(-1)^{\frac {m-1}{2}}}$

one and the same value.

6. Be . Then for all odd numbers that can be represented by the same form has each of the two expressions ${\displaystyle D\equiv 0\mod {8}}$ ${\displaystyle m}$

${\displaystyle \varsigma (m)=(-1)^{\frac {m-1}{2}}}$ and ${\displaystyle \varepsilon (m)=(-1)^{\frac {m^{2}-1}{8}}}$

an unchangeable value in itself. Because it follows . ${\displaystyle m_{1}m_{2}=x^{2}-dy^{2}\equiv x^{2}\equiv 1\mod {8}}$ ${\displaystyle m_{1}\equiv m_{2}\mod {8}}$

With this the division of binary quadratic forms with given discriminant into genders is found and one obtains: ${\displaystyle D}$

Discriminant	Associated characters
${\displaystyle D\equiv 1\mod {4}}$	${\displaystyle \chi _{1},\ldots ,\chi _{t}}$
${\displaystyle D=4D',{\rm {\;}}D'\equiv 1\mod {4}}$	${\displaystyle \chi _{1},\ldots ,\chi _{t}}$
${\displaystyle D=4D',{\rm {\;}}D'\equiv 3\mod {4}}$	${\displaystyle \chi _{1},\ldots ,\chi _{t},\varsigma }$
${\displaystyle D=4D',{\rm {\;}}D'\equiv 2\mod {8}}$	${\displaystyle \chi _{1},\ldots ,\chi _{t},\varepsilon }$
${\displaystyle D=4D',{\rm {\;}}D'\equiv 6\mod {8}}$	${\displaystyle \chi _{1},\ldots ,\chi _{t},\varsigma \varepsilon }$
${\displaystyle D=4D',{\rm {\;}}D'\equiv 4\mod {8}}$	${\displaystyle \chi _{1},\ldots ,\chi _{t},\varsigma }$
${\displaystyle D=4D',{\rm {\;}}D'\equiv 0\mod {8}}$	${\displaystyle \chi _{1},\ldots ,\chi _{t},\varsigma ,\varepsilon }$

If the set of all associated characters is given by and their number is given by , again describing the number of different prime numbers rising up in different prime numbers, then the set of certain values that these characters have for a certain form is called the total character of the form. Depending on how the result of the total character turns out, all forms with the same discriminant and type are divided into genders. In other words, two forms each belong to the same sex or to two different sexes, depending on whether the total character of one form corresponds to the other or not. ${\displaystyle \Theta }$ ${\displaystyle r}$ ${\displaystyle r}$ ${\displaystyle D}$ ${\displaystyle \pm 1}$ ${\displaystyle r}$ ${\displaystyle \Theta }$ ${\displaystyle \left\langle a,b,c\right\rangle }$

A gender is the epitome of all original forms of the same discriminant and type, for which each of the characters has the same value. Since all numbers that can be represented by a certain form are also represented by their (real) equivalent forms, all these forms of the same class also belong to the same gender. It turns out that the individual characters of a given primitive form can always be recognized from one of the coefficients . Because as often as there is a prime divisor of , one of the numbers will certainly not be divisible by, because if both were divisible by, then would also merge into and thus also into . But that would not make the shape primitive. ${\displaystyle r}$ ${\displaystyle \Theta }$ ${\displaystyle \left\langle a,b,c\right\rangle }$ ${\displaystyle a,c}$ ${\displaystyle p}$ ${\displaystyle D}$ ${\displaystyle p}$ ${\displaystyle p}$ ${\displaystyle p}$ ${\displaystyle b^{2}=D+ac}$ ${\displaystyle b}$

example

For the discriminant one obtains the two primitive, non-equivalent reduced forms and . The determinant can be broken: . It follows from this . So there are exactly two genders and there is exactly one of the forms in each of the genders. Well is . You therefore get the two characters: ${\displaystyle D=-20}$ ${\displaystyle f_{1}:=\left\langle 1,0,5\right\rangle }$ ${\displaystyle f_{2}:=\left\langle 2,2,3\right\rangle }$ ${\displaystyle D=-20=-2^{2}\cdot 5}$ ${\displaystyle {\rm {\mathfrak {A}}}(-20)=2^{2-1}=2}$ ${\displaystyle -20\equiv 4\mod {8}}$

${\displaystyle \delta (p)=(-1)^{\frac {p-1}{2}}}$and . ${\displaystyle \chi (p)={\frac {p}{5}}}$

Now it is easy to see that the total character has value . So the crowd is the main sex. And since it has total character , it is not main sex. If now is an odd prime number, then is represented by if and only if ${\displaystyle f_{1}}$ ${\displaystyle (-1)^{\frac {p-1}{2}}={\frac {p}{5}}=+1}$ ${\displaystyle \left\{f_{1}\right\}}$ ${\displaystyle f_{2}}$ ${\displaystyle (-1)^{\frac {p-1}{2}}={\frac {p}{5}}=-1}$ ${\displaystyle \left\{f_{2}\right\}}$ ${\displaystyle p\neq 5}$ ${\displaystyle p}$ ${\displaystyle f_{1}}$

${\displaystyle (-1)^{\frac {p-1}{2}}=+1}$and . ${\displaystyle {\frac {p}{5}}=+1}$

This is the case if and only if is. Analogously, one obtains that is represented by if and only if ${\displaystyle p\equiv 1,9\mod {20}}$ ${\displaystyle p}$ ${\displaystyle f_{2}}$

${\displaystyle (-1)^{\frac {p-1}{2}}=-1}$and . ${\displaystyle {\frac {p}{5}}=-1}$

So if is. ${\displaystyle p\equiv 3,7\mod {20}}$

The representation of the prime numbers is thus clearly characterized by the forms and . ${\displaystyle f_{1}}$ ${\displaystyle f_{2}}$

The limits of gender theory

In his work published in 1744, Leonhard Euler deals among other things with the form . If one looks at this form and tries to divide prime numbers into genders, one first finds that the quadratic congruence cannot be solved trivially for precisely those prime numbers for which a square is in the remainder class field . From the quadratic reciprocity law it follows that this applies except for or only for the prime numbers . There are four reduced, primitive forms of the discriminant : ${\displaystyle x^{2}+14y^{2}}$ ${\displaystyle x^{2}+14y^{2}\equiv 0\mod {p}}$ ${\displaystyle -14}$ ${\displaystyle {\rm {\mathbb {F} }}_{p}}$ ${\displaystyle p=2}$ ${\displaystyle p=7}$ ${\displaystyle p\equiv 1,3,5,9,13,15,19,23,25,27,39,45\mod {56}}$ ${\displaystyle D=-56}$

${\displaystyle f_{1}=\left\langle 1,0,14\right\rangle ,{\rm {\;\;}}f_{2}=\left\langle 3,2,5\right\rangle ,{\rm {\;\;}}f_{3}=\left\langle 2,0,7\right\rangle ,{\rm {\;\;}}f_{4}=\left\langle 3,-2,5\right\rangle }$.

The discriminant can be broken down into the two different prime divisors and . So there are exactly different genders. It also follows that the number of classes in each gender is exact . Well is . So the following three characters are to be considered: ${\displaystyle D=-56=-2^{3}\cdot 7}$ ${\displaystyle 2}$ ${\displaystyle 7}$ ${\displaystyle 2}$ ${\displaystyle h(-56)=4}$ ${\displaystyle {\rm {\mathfrak {K}}}(-56)={\frac {4}{2^{}}}=2}$ ${\displaystyle D=-56\equiv 0\mod {8}}$

${\displaystyle \varsigma (p)=(-1)^{\frac {p-1}{2}}{\rm {\;}},{\rm {\;}}\varepsilon (p)=(-1)^{\frac {p^{2}-1}{8}}{\rm {\;}},{\rm {\;}}\chi (p)={\frac {p}{7}}}$

The main sex consists of the forms . These have the total character . The non-main sex from the forms with the total character . It follows from this that a prime number is represented by or if and only if ${\displaystyle \left\{f_{1},f_{3}\right\}}$ ${\displaystyle \left\{+1,+1\right\}}$ ${\displaystyle \left\{f_{2},f_{4}\right\}}$ ${\displaystyle \left\{-1,-1\right\}}$ ${\displaystyle p\neq 2,7}$ ${\displaystyle f_{1}}$ ${\displaystyle f_{2}}$

${\displaystyle {\rm {\;}}(-1)^{\frac {p^{2}-1}{8}}{\rm {\;}}=+1,{\frac {p}{7}}=+1}$

applies. A simple calculation shows that this is the case if and only if . However, no statement can be made about whether by or represented. Similarly, it is easy to see that a prime number is represented by or if and only if is. Again, can not a requirement derives whether by or represented. ${\displaystyle p\equiv 1,9,15,23,25,39\mod {56}}$ ${\displaystyle p}$ ${\displaystyle f_{1}}$ ${\displaystyle f_{3}}$ ${\displaystyle p\neq 2,7}$ ${\displaystyle f_{2}}$ ${\displaystyle f_{4}}$ ${\displaystyle p\equiv 3,5,13,19,27,45\mod {56}}$ ${\displaystyle p}$ ${\displaystyle f_{2}}$ ${\displaystyle f_{4}}$

This shows that gender theory has reached its limits and cannot answer all questions regarding the representation of prime numbers by binary quadratic forms satisfactorily. Such questions and problems can be treated today with the help of class field theory .

Web links

Wikibooks: Correspondence Theorem of Algebraic Number Theory  - Learning and Teaching Materials

• Franz Lemmermeyer: Quadratic number fields: A taster course.

literature

• Peter Gustav Lejeune Dirichlet : Lectures on number theory. VDM Verlag-Edition Classic, 1863.
• Carl Friedrich Gauss: Disquisitiones Arithmeticae : Investigations over higher arithmetic. Springer, Berlin 1889. (Reprint: Kessel, Remagen 2009, ISBN 978-3-941300-09-5 )
• Paulo Ribenboim : The World of Prime Numbers: Secrets and Records. 1st edition. Springer, Berlin 2006, ISBN 3-540-34283-4 .
• Paulo Ribenboim: My numbers, my friends: highlights of number theory. Springer textbook, 2009, ISBN 978-3-540-87955-8 .
• Don B. Zagier : Zeta functions and quadratic fields: An introduction to higher number theory. 1st edition. Springer, Berlin 1981, ISBN 3-540-10603-0 .
• David A. Cox : Primes of the form : Fermat, Class Field Theory, and Complex Multiplication. ${\displaystyle x^{2}+ny^{2}}$Wiley-Interscience, New York 2013, ISBN 978-1-118-39018-4 .
• Candy Walter : Quadratic number fields and the gender theory. Leibniz University Hannover, 2009, doi: 10.13140 / RG.2.2.24046.84800 .