I have read the following theorem:
If p1,p2,…,pn are distinct prime numbers, then(Q[√p1,…,√pn]:Q)=2n.
I have tried to prove a more general statement but I have a problem at one point. (I still don't know how to prove the theorem above, too, because I don't know how not to use linear independence, which I do in the more general statement below.) Could you please help me overcome the obstacle I've encountered? I will post the intended proof and make it clear where I'm having trouble.
I want to prove the following statement:
Let n≥1. The set Bn:={√pϵ11√pϵ22⋯√pϵnn|(ϵ1,ϵ2,⋯,ϵn)∈{0,1}n} has 2n elements and is a Q−basis of Q[√p1,√p2,⋯,√pn].
The proof will be by induction.
For n=1, we have Bn={1,√p1}. It is clear that √p1≠1, so the set has 2=21 elements. It is the basis of Q[√p1] because the minimal polynomial of √p1 over Q has degree 2, and there is a theorem that K[a] has a0,⋯,ad−1 as a basis, where d is the degree of the minimal polynomial of a over K.
Suppose the statement is true for n−1, where n≥2. We have
(Bn=Bn−1∪√pnBn−1) and (Bn−1∩√pnBn−1=∅),
which is easy to see. It is also easy to see that card(Bn−1)=card(√pnBn−1), and therefore
cardBn=2n.
Let
∑x∈Bnqxx=0
for some {qx}x∈Bn⊂Q. Let p(x):=√pnx for all x∈Bn−1. We have
∑x∈Bnqxx=∑x∈Bn−1qxx+∑x∈√pnBn−1qxx=∑x∈Bn−1qxx+∑x∈Bn−1qp(x)√pnx.
Therefore
∑x∈Bn−1qxx=−√pn∑x∈Bn−1qp(x)x,
and we can make the following division iff qp(x)≠0 for all x∈Bn−1 (because Bn−1 is linearly indepentent over Q):
√pn=−∑x∈Bn−1qxx∑x∈Bn−1qp(x)x,
The right-hand side belongs to Q[√p1,√p2,⋯,√pn−1], so we have
√pn∈Q[√p1,√p2,⋯,√pn−1].
Therefore we can write √pn uniquely in the basis Bn−1.
√pn=∑y∈Bn−1cyy
for some {cy}y∈Bn−1⊂Q.
After squaring this equation we will obtain
pn=∑y∈Bn−1c2yy2+2∑y,z∈Bn−1cyczyz.
The last sum must be zero because it is not in Q and because after reducing it, we obtain a representation of pn in the basis Bn−1, which is unique. Thus
pn=∑y∈Bn−1c2yy2.
Unfortunately, I can't prove that cycz is always zero. This was my first thought, but clearly there's trouble with the possibility of reductions in
∑y,z∈Bn−1cyczyz.
Different pairs y,z may yield the same element of Bn−1 in the product yz. This happens for example when y=√5√3, z=√5√2, and y′=√11√2, z′=√11√3.
If it were true that cycz is always zero, I would be able to continue my proof as follows. We would have only one y0 such that cy0≠0 and we'd get
pn=c2y0y20.
Let cy0=kl. We can write
l2pn=k2y20.
But y20 is the product of some primes different from pn. Therefore the greatest power of pn that divides the right-hand side is even. However, the greatest power of pn that divides the left-hand side is odd. A contradiction.
The contradiction proves that qp(x)=0 for all x∈Bn−1. Hence (1) gives us that
∑x∈Bn−1qxx=0
and linear independence of Bn−1 gives us that qx=0 for all x∈Bn−1.
This gives us that Bn is linearly independent. It generates the whole Q[√p1,√p2,⋯,√pn] because
Q[√p1,√p2,⋯,√pn]=(Q[√p1,√p2,⋯,√pn−1])[√pn].
This would end the proof.
Answer
HINT An inductive proof follows easily from this
LEMMA [K(√a,√b):K]=4 if √a, √b, √ab all are not in K and 2≠0 in K.
Proof Let L=K(√b). Then [L:K]=2 via √b∉K, so it is sufficient to prove [L(√a):L]=2. It fails only if √a∈L=K(√b) and then √a = r+s √b for r,s∈K. But that is impossible since squaring yields (1): a = r2+b s2+2rs √b, which contradicts hypotheses as follows:
rs≠0 ⇒ √b ∈ K by solving (1) for √b, using 2≠0
s=0 ⇒ √a ∈ K via √a = r∈K
r=0 ⇒ √ab∈K via √a = s √b, times √b QED
Using the above as the inductive step one easily proves the following result of Besicovic.
THEOREM Let Q be a field with 2≠0, and L=Q(S) be an extension of Q generated by n square roots S={√a,√b,…} of elts a,b,…∈Q.
If every nonempty subset of S has product not in Q then each successive
adjunction Q(√a), Q(√a,√b),… doubles the degree over Q, so, in total, [L:Q] = 2n. Hence the 2n subproducts of the product of S comprise a basis of L over Q.
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