proof of prime ideal decomposition in quadratic extensions of


Much of the proof of this theorem is given in Marcus’ Number FieldsMathworldPlanetmath (http://planetmath.org/NumberField); however, all of the details will be filled in here, and some aspects of the proof here will differ from those of Marcus.

Note that gcd(a,b) refers to the greatest common divisorMathworldPlanetmathPlanetmath in Z of a and b (which must necessarily be rational integers).

Proof.

Let d be a squarefreeMathworldPlanetmath integer with d1 and K=(d).

If p is a rational prime that divides d, then

p,d2=p2,pd,d=gcd(p2,d),pd=p,pd=p=p𝒪K.

Note that p,d𝒪K. (If they were equal, then p,d2 would equal 𝒪K.)

If d3mod4, then disc(K)=4d. Note that 2 divides disc(K). Thus, 2 ramifies in 𝒪K. Therefore, 2𝒪K=P2 for some prime idealPlanetmathPlanetmathPlanetmath P of 𝒪K. Moreover, P is the unique ideal of 𝒪K of norm (http://planetmath.org/IdealNorm) 2. Since d-1mod2,1+d, then

𝒪K/2,1+d={a+bd+2,1+d:a,b}={a-b+2,1+d:a,b}={0+2,1+d,1+2,1+d}.

Since 2,1+d has 2, it follows that P=2,1+d and 2𝒪K=2,1+d2.

If d1mod8, then disc(K)=d. Note that 2 does not divide disc(K). Thus, 2 does not ramify in 𝒪K. Since

2,1+d22,1-d2=4,1+d,1-d,1-d4=4,2,2(1-d2),1-d4=2=2𝒪K,

we have that 2,1+d2 and 2,1-d2 must be distinct. Proving that these ideals are indeed given below.

If d5mod8, then consider the minimal polynomialPlanetmathPlanetmath f(x)[x] for 1+d2. Since 1+d2, it must be the case that degf2.

α=1+d22α-1=d(2α-1)2=d4α2-4α+1=d4α2-4α+1-d=0α2-α+1-d4=0

Thus, f(x)=x2-x+1-d4.

Let P be a lying over 2 in 𝒪K. Note that f(x) has a root (http://planetmath.org/Root) in 𝒪K and thus in 𝒪K/P. On the other hand, since f(x)x2+x+1mod2, f(x) considered as an element of 𝔽2[x] has no root in 𝔽2. Thus, 𝒪K/P and 𝔽2 are not isomorphic. Therefore, [𝒪K/P:𝔽2]>1. Since 1<[𝒪K/P:𝔽2]=f(P|2)[K:]=2, we have that f(P|2)=2. Thus, 2 is inert in 𝒪K. It follows that 2𝒪K is in 𝒪K.

If p is an odd prime (http://planetmath.org/Prime) that does not divide d and dn2modp, then p does not divide disc(K) (which equals either d or 4d). Thus, p does not ramify in 𝒪K. Also, p does not divide n. Since

p,n+dp,n-d=p2,pn+pd,pn-pd,n2-d=p2,2pn,pn-pd,n2-d=gcd(p2,2pn),pn-pd,n2-d=p,pn-pd,n2-d=p=p𝒪K,

we have that p,n+d and p,n-d must be distinct. It will be proven that these ideals are indeed .

Let I denote the norm of the ideal I (http://planetmath.org/IdealNorm) of 𝒪K and σGal(K/) with σ(d)=-d. Then

p,n+d=σ(p,n+d)=σ(p,n+d)=σ(p),σ(n+d)=p,n-d.

Note that p2=p𝒪K=p,n+dp,n-d=p,n-d2. Therefore, p,n+d=p,n-d=p. It follows that the indicated ideals are .

Finally, if p is an odd prime that does not divide d and d is not a square modp, then consider the minimal polynomial g(x)=x2-d for d over . Let P be a lying over p in 𝒪K. Note that g(x) has a root in 𝒪K and thus in 𝒪K/P. On the other hand, since discg(x)=-4(-d)=4d, which is not a square in 𝔽p, then g(x) considered as an element of 𝔽p[x] has no root in 𝔽p. Thus, 𝒪K/P and 𝔽p are not isomorphic. Therefore, [𝒪K/P:𝔽p]>1. Note that 1<[𝒪K/P:𝔽p]=f(P|p)[K:]=2. Thus, f(P|p)=2. Therefore, p is inert in 𝒪K. It follows that p𝒪K is in 𝒪K. ∎

References

  • 1 Marcus, Daniel A. Number Fields. New York: Springer-Verlag, 1977.
Title proof of prime ideal decomposition in quadratic extensions of
Canonical name ProofOfPrimeIdealDecompositionInQuadraticExtensionsOfmathbbQ
Date of creation 2013-03-22 15:59:06
Last modified on 2013-03-22 15:59:06
Owner Wkbj79 (1863)
Last modified by Wkbj79 (1863)
Numerical id 20
Author Wkbj79 (1863)
Entry type Proof
Classification msc 11R11