Let p,q in ZZ be district primes. Prove that QQ(sqrtp, sqrtq)=QQ(sqrtp+sqrtq), and that QQ sube QQ (sqrtp+sqrtq) is a degree 4 extension.

naivlingr

naivlingr

Answered question

2020-10-20

Let p,qZ be district primes. Prove that Q(p,q)=Q(p+q), and that QQ(p+q) is a degree 4 extension.

Answer & Explanation

SchepperJ

SchepperJ

Skilled2020-10-21Added 96 answers

p,q distinct prime numbers
Let K=Q(p,q),L=Q(p+q)
Claim:
1) K=L
2) deg(LQ)=4
1) Proof that K is contained in L.
p,q distinct prime numbers
Let K=Q(p,q),L=Q(p+q)
Claim:
1) K=L
Proof:
L sube K. Follow from p,qKp+qK
So, L=Q(p+q)K. So, proved.
2) Proof that L is contained in K. Note that i) abd ii) imply that K=L, as required KL
Now, as p and q are distinct primes, the conjugates of p+q are -p+q,pq,p,iL.
So (p+q)+(pq)=2pLpL
Similarly, qL.
So, K=QQ(p,q) sube L, thus proved rArr K=L
3) Proof ofdepends on claim , whose is proved first
deg(LQ)=4
Proof: Form K=L.
But K=Q(p.q)=Q(p)(q)
qQ(p)
Otherwise, q=a+bp, for some a,bQ
Square both sides to get q=a2+b2p+2abpp=qa2+b2p2abQ
which is absurd, as p is prime pQ
Completing the proof that deg (L)=deg(K)=4.
deg(LQ)=deg(KQ)=deg(Q(p)Q)deg(QpqQ(p))=22=4

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