Proving that $\left(\mathbb Q[\sqrt p_1,\dots,\sqrt p_n]:\mathbb Q\right)=2^n$ for distinct primes $p_i$.

I have read the following theorem:

If $p_1,p_2,\dots,p_n$ are distinct prime numbers, then$$\left(\mathbb Q\left[\sqrt p_1,\dots,\sqrt p_n\right]:\mathbb Q\right)=2^n.$$

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 $n\geq 1$. The set $B_n:=\left\{\sqrt {p_1^{\epsilon_1}}\sqrt {p_2^{\epsilon_2}}\cdots\sqrt {p_n^{\epsilon_n}}\,|\,(\epsilon_1,\epsilon_2,\cdots,\epsilon_n)\in\{0,1\}^n\right\}$ has $2^n$ elements and is a $\mathbb Q-$basis of $\mathbb Q\left[\sqrt p_1,\sqrt p_2,\cdots,\sqrt p_n\right].$

The proof will be by induction.

For $n=1,$ we have $B_n=\left\{1,\sqrt {p_1}\right\}.$ It is clear that $\sqrt{p_1}\neq 1,$ so the set has $2=2^1$ elements. It is the basis of $\mathbb Q[\sqrt{p_1}]$ because the minimal polynomial of $\sqrt {p_1}$ over $\mathbb Q$ has degree $2,$ and there is a theorem that $K[a]$ has $a^0,\cdots,a^{d-1}$ 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 $n\geq 2.$ We have

$$ \left(B_n=B_{n-1}\cup\sqrt{p_n}B_{n-1}\right)\text { and } \left(B_{n-1}\cap\sqrt{p_n}B_{n-1}=\emptyset\right), $$

which is easy to see. It is also easy to see that $\operatorname{card}(B_{n-1})=\operatorname{card}(\sqrt{p_n}B_{n-1}),$ and therefore

$$ \operatorname{card}B_{n}=2^n. $$

Let

$$ \sum_{x\in B_{n}}q_xx=0 $$

for some $\{q_x\}_{x\in B_n}\subset\mathbb Q.$ Let $p(x):=\sqrt{p_n}x$ for all $x\in B_{n-1}.$ We have

$$ \sum_{x\in B_{n}}q_xx=\sum_{x\in B_{n-1}} q_xx+\sum_{x\in \sqrt{p_n}B_{n-1}} q_xx=\sum_{x\in B_{n-1}} q_xx+\sum_{x\in B_{n-1}} q_{p(x)}\sqrt{p_n}x. $$

Therefore

$$ \sum_{x\in B_{n-1}} q_xx=-\sqrt{p_n}\sum_{x\in B_{n-1}} q_{p(x)}x,\tag1 $$

and we can make the following division iff $q_{p(x)}\neq 0$ for all $x\in B_{n-1}$ (because $B_{n-1}$ is linearly indepentent over $\mathbb Q$):

$$ \sqrt{p_n}=-\frac{\sum_{x\in B_{n-1}} q_xx}{\sum_{x\in B_{n-1}} q_{p(x)}x}, $$

The right-hand side belongs to $\mathbb Q\left[\sqrt p_1,\sqrt p_2,\cdots,\sqrt p_{n-1}\right],$ so we have

$$ \sqrt{p_n}\in \mathbb Q\left[\sqrt p_1,\sqrt p_2,\cdots,\sqrt p_{n-1}\right]. $$

Therefore we can write $\sqrt{p_n}$ uniquely in the basis $B_{n-1}$.

$$ \sqrt{p_n}=\sum_{y\in B_{n-1}}c_yy $$

for some $\{c_y\}_{y\in B_{n-1}}\subset \mathbb Q.$

After squaring this equation we will obtain

$$ p_n=\sum_{y\in B_{n-1}}c_y^2y^2+2\sum_{y,z\in B_{n-1}}c_yc_zyz. $$

The last sum must be zero because it is not in $\mathbb Q$ and because after reducing it, we obtain a representation of $p_n$ in the basis $B_{n-1},$ which is unique. Thus

$$p_n=\sum_{y\in B_{n-1}}c_y^2y^2.$$

Unfortunately, I can't prove that $c_yc_z$ is always zero. This was my first thought, but clearly there's trouble with the possibility of reductions in $$ \sum_{y,z\in B_{n-1}}c_yc_zyz. $$

Different pairs $y,z$ may yield the same element of $B_{n-1}$ in the product $yz.$ This happens for example when $y=\sqrt 5\sqrt 3,$ $z=\sqrt 5\sqrt 2,$ and $y'= \sqrt 11\sqrt 2,$ $z'=\sqrt 11\sqrt 3$.

If it were true that $c_yc_z$ is always zero, I would be able to continue my proof as follows. We would have only one $y_0$ such that $c_{y_0}\neq 0$ and we'd get

$$p_n=c_{y_0}^2y_0^2.$$

Let $c_{y_0}=\frac kl$. We can write $$l^2p_n=k^2y_0^2.$$

But $y_0^2$ is the product of some primes different from $p_n$. Therefore the greatest power of $p_n$ that divides the right-hand side is even. However, the greatest power of $p_n$ that divides the left-hand side is odd. A contradiction.

The contradiction proves that $q_{p(x)}=0$ for all $x\in B_{n-1}.$ Hence $(1)$ gives us that

$$ \sum_{x\in B_{n-1}} q_xx=0 $$

and linear independence of $B_{n-1}$ gives us that $q_x=0$ for all $x\in B_{n-1}.$

This gives us that $B_n$ is linearly independent. It generates the whole $\mathbb Q\left[\sqrt p_1,\sqrt p_2,\cdots,\sqrt p_n\right]$ because

$$ \mathbb Q\left[\sqrt p_1,\sqrt p_2,\cdots,\sqrt p_n\right]=\left(\mathbb Q\left[\sqrt p_1,\sqrt p_2,\cdots,\sqrt p_{n-1}\right]\right)\left[\sqrt{p_n}\right]. $$

This would end the proof.

HINT $\ $ An inductive proof follows easily from this

LEMMA $\rm\ \ [K(\sqrt{a},\sqrt{b}) : K] = 4\ $ if $\rm\ \sqrt{a},\ \sqrt{b},\ \sqrt{a\:b}\ $ all are not in $\rm\:K\:$ and $\rm\: 2 \ne 0\:$ in $\rm\:K\:.$

Proof $\ \ $ Let $\rm\ L = K(\sqrt{b})\:.\:$ Then $\rm\: [L:K] = 2\:$ via $\rm\:\sqrt{b} \not\in K\:,\:$ so it is sufficient to prove $\rm\: [L(\sqrt{a}):L] = 2\:.\:$ It fails only if $\rm\:\sqrt{a} \in L = K(\sqrt{b})\ $ and then $\rm\ \sqrt{a}\ =\ r + s\ \sqrt{b}\ $ for $\rm\ r,s\in K\:.\:$ But that is impossible since squaring yields $\rm(1):\ \ a\ =\ r^2 + b\ s^2 + 2\:r\:s\ \sqrt{b}\:,\: $ which contradicts hypotheses as follows:

$\rm\qquad\qquad rs \ne 0\ \ \Rightarrow\ \ \sqrt{b}\ \in\ K\ \ $ by solving $(1)$ for $\rm\sqrt{b}\:,\:$ using $\rm\:2 \ne 0$

$\rm\qquad\qquad\ s = 0\ \ \Rightarrow\ \ \ \sqrt{a}\ \in\ K\ \ $ via $\rm\ \sqrt{a}\ =\ r \in K$

$\rm\qquad\qquad\ r = 0\ \ \Rightarrow\ \ \sqrt{a\:b}\in K\ \ $ via $\rm\ \sqrt{a}\ =\ s\ \sqrt{b}\:,\: \ $times $\rm\:\sqrt{b}\quad\quad$ QED

Using the above as the inductive step one easily proves the following result of Besicovic.

THEOREM $\ $ Let $\rm\:Q\:$ be a field with $2 \ne 0\:,\:$ and $\rm\ L = Q(S)\ $ be an extension of $\rm\:Q\:$ generated by $\rm\: n\:$ square roots $\rm\ S = \{ \sqrt{a}, \sqrt{b},\ldots \}$ of elts $\rm\ a,\:b,\:\ldots \in Q\:.\:$ If every nonempty subset of $\rm\:S\:$ has product not in $\rm\:Q\:$ then each successive adjunction $\rm\ Q(\sqrt{a}),\ Q(\sqrt{a},\:\sqrt{b}),\:\ldots$ doubles the degree over $\rm\:Q\:,\:$ so, in total, $\rm\: [L:Q] \ =\ 2^n.\:$ Hence the $\rm2^n$ subproducts of the product of $\rm\:S\:$ comprise a basis of $\rm L$ over $\rm\:Q\:.$

Nothing wrong with the other answers. I just want to try my hand at this. I will prove by induction on the number of primes $n$ that

1. For $K_n=\Bbb{Q}(\sqrt{p_1},\sqrt{p_2},\ldots,\sqrt{p_n})$ we have $[K_n:\Bbb{Q}]=2^n$ AND
2. The extension $K/\Bbb{Q}$ is Galois with Galois group isomorphic to the $n$-fold Cartesian product $(C_2)^n$ generated by the automorphisms $\tau_i, i=1,2,\ldots,n,$ such that $\tau_i(\sqrt{p_j})=(-1)^{\delta_{ij}}\sqrt{p_j}$.

The base case $n=1$ is easy. Skipping that. Assume that the claim holds when we have $k$ primes.

Claim 1. $\sqrt{p_{k+1}}\notin K_k$.

Proof. Assume contrariwise that $\sqrt{p_{k+1}}\in K_k$. We know that $\Bbb{Q}(\sqrt{p_{k+1}})$ is a quadratic extension of $\Bbb{Q}$. By Galois theory the quadratic subfields of $K_k$ are exactly the fixed fields of index two subgroups of $\operatorname{Gal}(K_k/\Bbb{Q})$. By part two of the induction hypothesis, this Galois group is isomorphic to the additive group of a $k$-dimensional vector space $V$ over $\Bbb{F}_2$. The non-degenerate bilinear form, $B:V\times V\to\Bbb{F}_2, B\big((x_1,x_2,\ldots,x_k),(y_1,y_2,\ldots,y_k)\big)=\sum_{i=1}^kx_iy_i$, shows that the maximal subgroups are exactly the duals of the minimal subgroups of $V$. There are $2^k-1$ of those, namely the ones containing a single non-zero vector of $V$. Therefore $K_k$ has exactly $2^k-1$ quadratic subfields. But if $S$ is the product of any non-empty subset of $\{p_1,p_2,\ldots,p_k\}$, then $\Bbb{Q}(\sqrt{S})$ is a quadratic subfield of $K_k$. Such quadratic fields are easily seen to be distinct (only need the analogue of the argument showing $\sqrt2\notin\Bbb{Q}(\sqrt3)$ for this). Similarly we see that $\sqrt{p_{k+1}}$ is not in any of those quadratic subfields. The claim follows.

Claim 2. The inductive step is valid.

Proof. By Claim 1. $[K_{k+1}:K_k]=2$, so part 1 one of the induction hypothesis implies part 1 of the inductive step. Because $K_{k+1}$ is the splitting field of $\prod_{i=1}^{k+1}(x^2-p_i)\in\Bbb{Q}[x]$, it follows that $K_{k+1}$ is Galois over $\Bbb{Q}.$ Any automorphism $\tau\in \operatorname{Gal}(K_{k+1}/\Bbb{Q})$ is fully determined if we know the images $\tau(\sqrt{p_i}), i=1,2,\ldots,k+1$. There are two choices for each of those images (up to sign) - a total of $2^{k+1}$ combinations. Because the extension is Galois, we know that there will be exactly $2^{k+1}$ automorphisms, so all those sign combinations must occur. Q.E.D.