A quadratic number field is an algebraic field extension of the form ${\ displaystyle K / \ mathbb {Q}}$

${\ displaystyle K = \ mathbb {Q} ({\ sqrt {d}})}$

with a rational number that is not a square in . These are exactly the extensions from the degree above${\ displaystyle d,}$${\ displaystyle \ mathbb {Q}}$${\ displaystyle 2}$${\ displaystyle \ mathbb {Q}.}$

Quadratic number fields are, besides themselves, the simplest number fields . ${\ displaystyle \ mathbb {Q}}$

## introduction

The theory of the quadratic number field developed from the study of binary quadratic forms . In their investigations into Diophantine equations, Euler and Fermat had compiled many fundamental individual results, which subsequently left room for further research. In his Disquisitiones Arithmeticae linked Gauss in Section V of the works of Fermat, Euler and Lagrange and treated there extensively the theory of binary quadratic forms. Although Gauss is in the range of whole numbers in his representation , it is more elegant from today's point of view to expand the field of rational numbers so that the quadratic forms can be broken down into linear factors . Such a decomposition then sees z. B. as follows:

${\ displaystyle x ^ {2} -5y ^ {2} = (x + y {\ sqrt {5}}) \ cdot (xy {\ sqrt {5}})}$

This makes the theory of quadratic number fields a part of the theory of binary quadratic forms.

The body of rational numbers can be expanded into a comprehensive body in various ways . For example, one examines the ring of whole algebraic numbers. It contains precisely those complex numbers that are the zero of a normalized polynomial with integer coefficients . When expanding, however, it often makes sense to only add as many numbers as are required for a given problem: ${\ displaystyle \ mathbb {Q}}$${\ displaystyle K \ subseteq \ mathbb {C}}$ ${\ displaystyle {\ mathcal {O}}}$

Let be finitely many algebraic numbers and let be the smallest subfield of the field of algebraic numbers that contains all these numbers. Then you write ${\ displaystyle \ alpha _ {1}, \ ldots, \ alpha _ {n}}$${\ displaystyle K}$${\ displaystyle {\ overline {\ mathbb {Q}}}}$

${\ displaystyle K = \ mathbb {Q} (\ alpha _ {1}, \ ldots, \ alpha _ {n})}$

and says the body is an extension field of the by adjunction of the elements of results. The couple and is called body expansion and writes for it${\ displaystyle K}$${\ displaystyle \ mathbb {Q},}$${\ displaystyle \ alpha _ {1}, \ ldots, \ alpha _ {n}}$${\ displaystyle \ mathbb {Q}}$${\ displaystyle K}$${\ displaystyle \ mathbb {Q}}$${\ displaystyle K \ colon \ mathbb {Q}.}$

In particular, is an Abelian group . Because also the multiplication of elements from with the scalars from over ${\ displaystyle (K, +)}$${\ displaystyle K}$${\ displaystyle \ mathbb {Q}}$

{\ displaystyle {\ begin {aligned} \ cdot \ colon \ mathbb {Q} \ times K & \; \ to \; K \\ (\ eta, \ alpha) & \; \ mapsto \; \ eta \ alpha \ end {aligned}}}

is explained, one obtains from the field axioms for directly the vector space axioms, so that a vector space can be understood via . The body has finite degrees , which means that as -vector space is finite dimensional. ${\ displaystyle \ mathbb {Q}}$${\ displaystyle K}$${\ displaystyle \ mathbb {Q}}$${\ displaystyle K}$${\ displaystyle \ mathbb {Q}}$ ${\ displaystyle [K \ colon \ mathbb {Q}]}$${\ displaystyle K}$${\ displaystyle \ mathbb {Q}}$

If generated from an algebraic number , then has a base and hence the dimension${\ displaystyle K = \ mathbb {Q} (\ alpha)}$ ${\ displaystyle \ alpha}$${\ displaystyle K}$ ${\ displaystyle \ {1, \ alpha, \ alpha ^ {2}, \ ldots, \ alpha ^ {n-1} \}}$

${\ displaystyle \ dim _ {\ mathbb {Q}} K = [K \ colon \ mathbb {Q}] = n,}$

where is equal to the degree of the minimal polynomial that has as zero. It can be shown that over has degree 2 if the minimal polynomial of is quadratic. Thus is a quadratic number field . ${\ displaystyle n}$ ${\ displaystyle f _ {\ alpha}}$${\ displaystyle \ alpha}$${\ displaystyle K}$${\ displaystyle \ mathbb {Q}}$${\ displaystyle \ alpha}$${\ displaystyle K}$

For a number field called ${\ displaystyle K}$

${\ displaystyle {\ mathcal {O}} _ {K} = K \ cap {\ mathcal {O}}}$

the wholeness ring of or the whole closure of in Thus consists of all elements that are in all algebra; that is, the following applies: ${\ displaystyle K}$${\ displaystyle \ mathbb {Z}}$${\ displaystyle K.}$${\ displaystyle {\ mathcal {O}} _ {K}}$${\ displaystyle K}$

${\ displaystyle {\ mathcal {O}} _ {K} = \ left \ {\ alpha \ in K \; | \; f _ {\ alpha} \ in \ mathbb {Z} [X] \ right \}}$

## definition

A quadratic number field is a quadratic extension of the rational numbers. Quadratic number fields thus arise from the adjunction of the square root . ${\ displaystyle \ mathbb {Q}}$${\ displaystyle {\ sqrt {d}}}$

In the following, let be a square-free integer other than 0 and 1. Then the amount is called ${\ displaystyle d}$

${\ displaystyle \ mathbb {Q} ({\ sqrt {d}}): = \ {x + y {\ sqrt {d}} \ in \ mathbb {C} \; | \; x, y \ in \ mathbb {Q} \}}$

a square number field .

Is , then is called real-square number field , otherwise imaginary- square number field . There is an arbitrary but fixed complex solution to the equation . The second solution to this equation leads to the same number field. ${\ displaystyle d> 0}$${\ displaystyle K}$ ${\ displaystyle {\ sqrt {d}} \ in \ mathbb {C}}$${\ displaystyle X ^ {2} = d}$

## properties

### Conjugation mapping

It is true that every element of the root of a polynomial is of degree . So every element of is algebraic. So you get a tower of bodies: ${\ displaystyle K = \ mathbb {Q} ({\ sqrt {d}})}$${\ displaystyle f \ in \ mathbb {Q} [X]}$${\ displaystyle \ leq 2}$${\ displaystyle K}$

${\ displaystyle \ mathbb {Q} \; \ subsetneq \; K \; \ subsetneq \; {\ overline {\ mathbb {Q}}} \; \ subsetneq \; \ mathbb {C}}$

In particular, is a -base of , that is, it is ${\ displaystyle \ {1, {\ sqrt {d}} \}}$${\ displaystyle \ mathbb {Q}}$${\ displaystyle K}$

${\ displaystyle K = \ mathbb {Q} \ oplus \ mathbb {Q} {\ sqrt {d}}}$

Now the body has exactly two body automorphisms , on the one hand the identical mapping${\ displaystyle K}$

{\ displaystyle {\ begin {aligned} \ operatorname {id} _ {K} \ colon \ qquad K & \; \ to \; K \\ x + y {\ sqrt {d}} & \; \ mapsto \; x + y {\ sqrt {d}} \ end {aligned}}}

and on the other hand the conjugation map :

{\ displaystyle {\ begin {aligned} \ sigma \ colon \ qquad K & \; \ to \; K \\ x + y {\ sqrt {d}} & \; \ mapsto \; xy {\ sqrt {d}} \ end {aligned}}}

In particular, a Galois group of order 2. is the conjugated element about . ${\ displaystyle \ operatorname {Aut} (K) = \ {id_ {K}, \ sigma \}}$${\ displaystyle \ alpha \ in K}$${\ displaystyle \ sigma (\ alpha)}$${\ displaystyle \ alpha}$

### Norm and trace

The two quantities norm and trace of a quadratic number field can be represented as follows using its nontrivial field automorphism : ${\ displaystyle K}$${\ displaystyle \ sigma}$

{\ displaystyle {\ begin {aligned} N \ colon \; K & \; \ to \; \ mathbb {Q} \\\ alpha & \; \ mapsto \; \ alpha \ sigma (\ alpha) \ end {aligned} }}

and

{\ displaystyle {\ begin {aligned} \ operatorname {Sp} \ colon \; K & \; \ to \; \ mathbb {Q} \\\ alpha & \; \ mapsto \; \ alpha + \ sigma (\ alpha) \ end {aligned}}}

Since the embedding forms a ring homomorphism , the norm becomes multiplicative and the trace additive. By inserting you get: ${\ displaystyle \ sigma}$

{\ displaystyle {\ begin {aligned} N (\ alpha) & = (x + y {\ sqrt {d}}) (xy {\ sqrt {d}}) = x ^ {2} -dy ^ {2} \\ Sp (\ alpha) & = (xy {\ sqrt {d}}) + (x + y {\ sqrt {d}}) = 2x \ end {aligned}}}

The norm is thus a square shape . Due to the fact that the whole algebraic numbers form a ring , is obviously a ring too. This takes on a role analogous to that of the ring in and it is therefore a sub-ring of Thus all elements of the form are always whole algebraic, and one obtains an inclusion of rings: ${\ displaystyle K.}$${\ displaystyle {\ mathcal {O}}}$${\ displaystyle {\ mathcal {O}} _ {K}}$${\ displaystyle K}$${\ displaystyle \ mathbb {Z}}$${\ displaystyle \ mathbb {Q},}$${\ displaystyle {\ mathcal {O}} _ {K} \ cap \ mathbb {Q} = \ mathbb {Z}.}$${\ displaystyle {\ mathcal {O}} _ {K}}$${\ displaystyle K.}$${\ displaystyle x + y {\ sqrt {d}}, x, y \ in \ mathbb {Z}}$

${\ displaystyle \ mathbb {Z} [{\ sqrt {d}}] \ subseteq {\ mathcal {O}} _ {K}}$

The following shows that equality does not necessarily apply here

example
Let us consider the third unit root. This is a zero of the normalized polynomial which, by the way, is not its minimal polynomial, and thus an integer algebraic number. So the so-called Eisenstein numbers , however${\ displaystyle \ zeta _ {3} = {\ tfrac {-1 + {\ sqrt {-3}}} {2}} \ in {\ mathbb {Q} ({\ sqrt {-3}})}. }$${\ displaystyle X ^ {3} -1 \ in \ mathbb {Z} [X],}$${\ displaystyle \ zeta _ {3} \ in {\ mathcal {O}} _ {\ mathbb {Q} ({\ sqrt {-3}})},}$${\ displaystyle \ zeta _ {3} \ notin \ mathbb {Z} [{\ sqrt {-3}}].}$

There is a very simple way of identifying the whole algebraic numbers in a square number field, because a number lies in if and only if its norm and trace are whole numbers. ${\ displaystyle \ alpha \ in K}$${\ displaystyle {\ mathcal {O}} _ {K},}$

Since countable is infinite , countable is also infinite, because each only has a finite number of zeros. Hence the set of algebraic numbers is countably infinite. ${\ displaystyle \ mathbb {Q}}$ ${\ displaystyle \ mathbb {Q} [X]}$${\ displaystyle f \ in \ mathbb {Q} [X]}$

There remains the question of the form of the integral algebraic elements from Here hang the many variants of the elements and from the congruence from modulo fourth As a square-free number, modulo 4 can only be congruent to 1, 2 or 3 from the outset . The following applies now: ${\ displaystyle {\ mathcal {O}} _ {K}.}$${\ displaystyle x}$${\ displaystyle y}$${\ displaystyle d}$${\ displaystyle d}$

Let it be square-free and the associated square number field, then:${\ displaystyle d \ in \ mathbb {Z} \ setminus \ {0,1 \}}$${\ displaystyle K = \ mathbb {Q} ({\ sqrt {d}})}$
${\ displaystyle {\ mathcal {O}} _ {K} = {\ begin {cases} {\ mathbb {Z} + \ mathbb {Z} {\ sqrt {d}}}, \; & {\ text {if }} d \ equiv 2,3 {\ text {mod}} 4 \\\ mathbb {Z} + \ mathbb {Z} {\ frac {1 + {\ sqrt {d}}} {2}}, \; & {\ text {falls}} d \ equiv 1 {\ text {mod}} 4 \ end {cases}}}$
example
The third root of unity is because of in and is of the form In contrast, the whole Gaussian numbers in because of congruence have the form${\ displaystyle \ zeta _ {3} = {\ tfrac {-1 + {\ sqrt {-3}}} {2}}}$${\ displaystyle d = -3 \ equiv 1 {\ text {mod}} 4}$${\ displaystyle {\ mathcal {O}} _ {\ mathbb {Q} ({\ sqrt {-3}})}}$${\ displaystyle x + y {\ tfrac {1 + {\ sqrt {-3}}} {2}}.}$${\ displaystyle \ mathbb {Q} [i]}$${\ displaystyle d = -1 \ equiv 3 {\ text {mod}} 4}$${\ displaystyle x + y {\ sqrt {-1}} = x + yi.}$

### units

A first essential difference between real and imaginary square number fields is in terms of their units . So is z. B. the unit group of the ring the cyclic group of the order The description of the unit group of the entirety ring depends on whether it is real or imaginary quadratic. So the unit group for imaginary quadratic number fields is finite and we can describe it as follows: ${\ displaystyle \ mathbb {Z} ^ {\ times} = \ {- 1,1 \}}$${\ displaystyle \ mathbb {Z}}$${\ displaystyle 2.}$${\ displaystyle {\ mathcal {O}} _ {K} ^ {\ times}}$${\ displaystyle {\ mathcal {O}} _ {K}}$${\ displaystyle K}$

Let and the associated (imaginary) square number field. The following applies to his unit group :${\ displaystyle d <0}$${\ displaystyle K = \ mathbb {Q} ({\ sqrt {d}})}$${\ displaystyle {\ mathcal {O}} _ {K} ^ {\ times}}$
${\ displaystyle {\ mathcal {O}} _ {K} ^ {\ times} = {\ begin {cases} \ {\ pm 1, \ pm i \} \ cong \ mathbb {Z} / 4 \ mathbb {Z }, & {\ text {if}} d = -1 \\\ left \ {\ pm 1, {\ frac {1 \ pm {\ sqrt {-3}}} {2}}, {\ frac {- 1 \ pm {\ sqrt {-3}}} {2}} \ right \} \ cong \ mathbb {Z} / 6 \ mathbb {Z}, & {\ text {if}} d = -3 \\\ {-1,1 \} \ cong \ mathbb {Z} / 2 \ mathbb {Z}, & {\ text {otherwise}} \ end {cases}}}$

In the case of a real square number field, the description of the unit group is more complex. It turns out that every real square number field has an infinite number of units. The determination of the unit group amounts to solving Pell's equation . One can now show by means of Dirichlet's drawer principle that this equation provides an infinite number of units (solutions). Since the pigeonhole principle is not constructive, it is used to determine the units, the continued fraction expansion of${\ displaystyle x ^ {2} -dy ^ {2} = \ pm 1}$${\ displaystyle {\ sqrt {d}}.}$

## Construction of quadratic number fields

A classic example of the construction of a quadratic field is the uniquely determined square intermediate body is one of a primitive formed th root of unity cyclotomic field to take an odd prime. The uniqueness follows from the fact that the Galois group is isomorphic to and therefore cyclic. By looking at the branch you can see that the square intermediate body is equal to ; namely, the discriminant of is a power, and therefore this must also apply to the discriminant of the square intermediate field. According to the above statement, it must be , otherwise there is also branching. The same also applies to any powers of an odd prime number. ${\ displaystyle p}$${\ displaystyle p}$${\ displaystyle \ mathbb {Q} (\ zeta _ {p}) / \ mathbb {Q}}$${\ displaystyle (\ mathbb {Z} / p \ mathbb {Z}) ^ {\ times}}$${\ displaystyle \ mathbb {Q} ({\ sqrt {p ^ {*}}})}$${\ displaystyle p ^ {*} = (- 1) ^ {\ frac {p-1} {2}} p}$${\ displaystyle \ mathbb {Q} (\ zeta _ {p}) / \ mathbb {Q}}$${\ displaystyle p}$${\ displaystyle p ^ {*} \ equiv 1 {\ text {mod}} 4}$${\ displaystyle 2}$

The body , on the other hand, has exactly the three bodies , and as square intermediate bodies ; this is because the Galois group of the extension is no longer cyclic ( see prime residual class group ). ${\ displaystyle \ mathbb {Q} (\ zeta _ {8}) / \ mathbb {Q}}$${\ displaystyle \ mathbb {Q} ({\ sqrt {-1}})}$${\ displaystyle \ mathbb {Q} ({\ sqrt {2}})}$${\ displaystyle \ mathbb {Q} ({\ sqrt {-2}})}$${\ displaystyle (\ mathbb {Z} / 8 \ mathbb {Z}) ^ {\ times} \ simeq \ mathbb {Z} / 2 \ mathbb {Z} \ times \ mathbb {Z} / 2 \ mathbb {Z} }$

For the special case one obtains the whole ring of the Gaussian numbers , for the whole ring of the Eisenstein numbers . These two wholeness rings are the only wholeness rings of square number fields that are also circular divisions . ${\ displaystyle d = -1}$${\ displaystyle d = -3}$

## Ambiguity of prime factorization

In 1843 Peter Dirichlet drew Ernst Eduard Kummer's attention to the ambiguity of the prime factorization in certain number rings. In his supposed proof of the Fermatschen conjecture , which included the algebraic numbers, Kummer had considered the fundamental theorem of number theory to be proven for all algebraic numbers, so that these also have a clear decomposition like the ordinary whole numbers. That this is no longer given in the ring can easily be shown for the number 21. ${\ displaystyle {\ rm {\ mathcal {O}}} _ {{\ mathbb {Q}} ({\ sqrt {-5}})}}$

So is on the one hand and on the other . That the numbers are all irreducible and not associated with one another can be seen with the help of the norm as follows. Suppose the number 3 is separable. For example with , where there are no units. Then is and consequently must be. Now are of the form with and hence it follows that is the norm . Now the equation is obviously unsolvable in integers, which contradicts our assumption. So the number in is irreducible and one proves analogously that the numbers are too . It is clear that the numbers and are not associated with each other. Likewise, and as conjugates cannot be associated with one another. Assuming the numbers and are too associated, then the fractions would be . However, since both the trace from and from are not integers, the elements cannot lie in. So the numbers are not associated with each other. Consequently, there are two different prime factorizations in for the number . ${\ displaystyle 21 = 3 \ cdot 7}$${\ displaystyle 21 = (1 + 2 {\ sqrt {-5}}) (1-2 {\ sqrt {-5}})}$${\ displaystyle 3, {\ rm {\;}} 7 {\ rm {\;}}, 1 \ pm 2 {\ sqrt {-5}}}$${\ displaystyle {\ rm {\ mathcal {O}}} _ {{\ mathbb {Q}} ({\ sqrt {-5}})}}$${\ displaystyle 3 = \ alpha \ cdot \ beta}$${\ displaystyle \ alpha, \ beta \ in {\ rm {\ mathcal {O}}} _ {{\ mathbb {Q}} ({\ sqrt {-5}})}}$${\ displaystyle N _ {{\ mathbb {Q}} ({\ sqrt {-5}})} (3) = N _ {{\ mathbb {Q}} ({\ sqrt {-5}})} (\ alpha ) N _ {{\ mathbb {Q}} ({\ sqrt {-5}})} (\ beta) = 9}$${\ displaystyle N _ {{\ mathbb {Q}} ({\ sqrt {-5}})} (\ alpha) = N _ {{\ mathbb {Q}} ({\ sqrt {-5}})} (\ beta) = \ pm 3}$${\ displaystyle \ alpha, \ beta}$${\ displaystyle x + y {\ sqrt {-5}}}$${\ displaystyle x, y \ in {\ mathbb {Z}}}$${\ displaystyle N _ {{\ mathbb {Q}} ({\ sqrt {-5}})} (x + y {\ sqrt {-5}}) = x ^ {2} + 5y ^ {2} \ in {\ mathbb {Z}}}$${\ displaystyle x ^ {2} + 5y ^ {2} = \ pm 3}$${\ displaystyle 3}$${\ displaystyle {\ rm {\ mathcal {O}}} _ {{\ mathbb {Q}} ({\ sqrt {-5}})}}$${\ displaystyle 7 {\ rm {\;}}, 1 \ pm 2 {\ sqrt {-5}} {\ rm {\;}}}$${\ displaystyle 3}$${\ displaystyle 7}$${\ displaystyle 1 + 2 {\ sqrt {-5}} {\ rm {\;}}}$${\ displaystyle 1-2 {\ sqrt {-5}}}$${\ displaystyle 3}$${\ displaystyle 7}$${\ displaystyle 1 \ pm 2 {\ sqrt {-5}}}$${\ displaystyle {\ frac {1 \ pm 2 {\ sqrt {-5}}} {3}}, {\ frac {1 \ pm 2 {\ sqrt {-5}}} {7}} \ in {\ rm {\ mathcal {O}}} _ {{\ mathbb {Q}} ({\ sqrt {-5}})}}$${\ displaystyle {\ frac {1 \ pm 2 {\ sqrt {-5}}} {3}}}$${\ displaystyle {\ frac {1 \ pm 2 {\ sqrt {-5}}} {7}}}$${\ displaystyle {\ frac {1 \ pm 2 {\ sqrt {-5}}} {3}}, {\ frac {1 \ pm 2 {\ sqrt {-5}}} {7}}}$${\ displaystyle {\ rm {\ mathcal {O}}} _ {{\ mathbb {Q}} ({\ sqrt {-5}})}}$${\ displaystyle 21}$${\ displaystyle {\ rm {\ mathcal {O}}} _ {{\ mathbb {Q}} ({\ sqrt {-5}})}}$

So we see that the fundamental theorem of number theory and thus the uniqueness of the prime factorization can generally no longer be assumed.

Problems of this kind can be got under control today with Kummer's ideal theory . Guided by the complex numbers, Kummer's intention was to create a wider range of new ideal numbers so that they could be uniquely decomposed into the product of ideal prime numbers . The theory of ideal numbers developed by Kummer was systematized by the German mathematician Richard Dedekind and today the ideal numbers are simply referred to as the Dedekindian ideals of the ring . The fundamental theorem of Dedekind's ideal theory now provides the generalization of the theorem of unambiguous prime factorization and shows a way to deal with the ambiguity of prime factorization and to restore an analogy to the fundamental theorem of number theory. (See for example Dedekindring ). ${\ displaystyle {\ rm {\ mathcal {O}}} _ {{\ mathbb {Q}} ({\ sqrt {d}})}}$

## Prime ideal decomposition

That the prime ideal decomposition of a main ideal , for a prime number , cannot be arbitrary, follows from the norm . That is, is either a prime ideal or is broken down into the product of two (not necessarily different) prime ideals of the norm . A prime number is called in${\ displaystyle p {\ rm {\ mathcal {O}}} _ {{\ mathbb {Q}} ({\ sqrt {d}})}}$ ${\ displaystyle p}$${\ displaystyle N (p {\ rm {\ mathcal {O}}} _ {{\ mathbb {Q}} ({\ sqrt {d}})}) = p ^ {2}}$${\ displaystyle p {\ rm {\ mathcal {O}}} _ {{\ mathbb {Q}} ({\ sqrt {d}})}}$${\ displaystyle p}$${\ displaystyle p}$${\ displaystyle {\ mathbb {Q}} ({\ sqrt {d}})}$

• sluggish when a prime ideal is,${\ displaystyle p {\ rm {\ mathcal {O}}} _ {{\ mathbb {Q}} ({\ sqrt {d}})} = {\ rm {\ mathfrak {p}}}}$
• dismantled, if with prime ideals ,${\ displaystyle p {\ rm {\ mathcal {O}}} _ {{\ mathbb {Q}} ({\ sqrt {d}})} = {\ rm {\ mathfrak {p}}} {\ rm { \ mathfrak {p}}} '}$${\ displaystyle {\ rm {\ mathfrak {p}}} \ neq {\ rm {\ mathfrak {p}}} '{\ rm {\ trianglelefteq}} {\ rm {\ mathcal {O}}} _ {{ \ mathbb {Q}} ({\ sqrt {d}})}}$
• branched out if for a prime ideal .${\ displaystyle p {\ rm {\ mathcal {O}}} _ {{\ mathbb {Q}} ({\ sqrt {d}})} = {\ rm {\ mathfrak {p}}} ^ {2} }$${\ displaystyle {\ rm {\ mathfrak {p}}} {\ rm {\ trianglelefteq}} {\ rm {\ mathcal {O}}} _ {{\ mathbb {Q}} ({\ sqrt {d}} )}}$

The third case occurs precisely for the (finitely many) prime divisors of the discriminant . The other two cases occur "equally frequently" in a certain sense; this follows from Chebotarev's theorem of density .

One can now find without much effort that for the discriminant of a square number field: ${\ displaystyle d \ neq 0.1}$

${\ displaystyle \ Delta _ {{\ mathbb {Q}} ({\ sqrt {d}})}: = {\ begin {cases} 4d & {\ text {falls}} d \ equiv 2,3 \, {\ text {mod}} 4 \\ d & {\ text {falls}} \, \, \, d \ equiv 1 \, {\ text {mod}} 4. \ end {cases}}}$

Note that it always holds. ${\ displaystyle {\ mathbb {Q}} ({\ sqrt {d}}) = {\ mathbb {Q}} ({\ sqrt {\ Delta _ {{\ mathbb {Q}} ({\ sqrt {d} })}}})}$

With the help of the discriminant and the Legendre symbol , a clear description of the behavior of odd prime numbers in a square number field can be given:

 Theorem (law of decomposition ): For an odd prime number in the following applies: ${\ displaystyle p}$${\ displaystyle {\ mathbb {Q}} ({\ sqrt {d}})}$ Is , then is and is branched.${\ displaystyle p \ mid \ Delta _ {{\ mathbb {Q}} ({\ sqrt {d}})}}$${\ displaystyle (p) = (p, {\ sqrt {d}}) ^ {2}}$${\ displaystyle p}$ Is then disassembled.${\ displaystyle \ left ({\ frac {\ Delta _ {{\ mathbb {Q}} ({\ sqrt {d}})}} {p}} \ right) = + 1}$${\ displaystyle p}$ Is , then is sluggish.${\ displaystyle \ left ({\ frac {\ Delta _ {{\ mathbb {Q}} ({\ sqrt {d}})}} {p}} \ right) = - 1}$${\ displaystyle p}$

Proof: See: law of decomposition

Note: The prime number was excluded. But it is true that in is sluggish when . It is broken down if and it is branched if . ${\ displaystyle 2}$${\ displaystyle 2}$${\ displaystyle {\ mathbb {Q}} ({\ sqrt {d}})}$${\ displaystyle d \ equiv 5 {\ text {mod}} 8}$${\ displaystyle d \ equiv 1 {\ text {mod}} 8}$${\ displaystyle d \ equiv 2,3 {\ text {mod}} 4}$

The statement about inertia also applies to the decomposition into prime elements ; In general, however, such statements can be extended to prime elements if and only if is the main ideal ring , i.e. has a clear decomposition into prime elements, or equivalently has a class number . ${\ displaystyle {{\ mathbb {Q}} ({\ sqrt {d}})}}$ ${\ displaystyle 1}$

example

If you look at, for example . Then by applying the quadratic reciprocity law several times, we get that the prime number in is inert. Because . ${\ displaystyle \ left ({\ frac {-15} {37}} \ right)}$${\ displaystyle 37}$${\ displaystyle {\ mathbb {Q}} ({\ sqrt {-15}})}$${\ displaystyle \ left ({\ frac {-15} {37}} \ right) = \ left ({\ frac {-1} {37}} \ right) \ left ({\ frac {3} {37} } \ right) \ left ({\ frac {5} {37}} \ right) = (- 1) ^ {18} \ left ({\ frac {1} {8}} \ right) \ left ({\ frac {2} {5}} \ right) = 1 \ cdot 1 \ cdot (-1) = - 1}$