# Elementary proof for $\sqrt{p_{n+1}} \notin \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots, \sqrt{p_n})$ where $p_i$ are different prime numbers.

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

#### Solutions Collecting From Web of "Elementary proof for $\sqrt{p_{n+1}} \notin \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots, \sqrt{p_n})$ where $p_i$ are different prime numbers."

(I will prove the following more general statement)

Theorem Let $P(n)$ be the following statement
$$\forall m\in \Bbb N^+\ \ \sqrt{q_1\cdots q_m} \notin \mathbb{Q}(\sqrt{p_1}, \ldots, \sqrt{p_n}) \text{ for any }\text{distinct primes } p_1,\ldots,p_{n},q_1,\ldots,q_m$$
we claim that $P(n)$ is true for any integer $n\in \Bbb N$

Proof by induction

• Basis step: Given a positive integer $m$ and $q_1,\cdots ,q_m$ distinct prime numbers assume that:
$$\sqrt{q_1\cdots q_m}\in \Bbb Q$$
hence there exists $a$ and $b$ integers such that $q_1\cdots q_m=\frac{a^2}{b^2}$ thus $$1=v_{q_1}(q_1\cdots q_m)=2(v_{q_1}(a)-v_{q_1}(b))$$
the first equality holds because $q_1,\cdots q_m$ are distinct, It follows that $1$ is even which is absurd, finally $P(0)$ is true.

• Induction step : Assume that $P(n-1)$ is true we will prove $P(n)$ by contradiction, assume that $P(n)$ is false then there exists an integer $m\geq 1$ and distinct primes $p_1,\cdots,p_n,q_1,\cdots,q_m$ such that:
$$\sqrt{q_1\cdots q_m} \in \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots, \sqrt{p_n})$$
hence there exists $a,b\in \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots, \sqrt{p_{n-1}})$ such that $\sqrt{q_{1}\cdots q_m}=a+b\sqrt{p_n}$ by squaring
either:

• one have $b=0$ then $\sqrt{q_{1}\cdots q_m}\in \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots, \sqrt{p_{n-1}})$ and $p_1,\cdots,p_{n-1},q_1,\cdots,q_m$ are distinct
• or one have $a=0$ in which case $bp_n=\sqrt{q_{1}\cdots q_mp_n}\in \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots, \sqrt{p_{n-1}})$and $p_1,\cdots,p_{n-1},q_1,\cdots,q_m,q_{m+1}=p_n$ are distinct
• or one have :
$$\sqrt{p_n}=\frac{q_{1}\cdots q_m-a^2-b^2p_n}{2ab}\in \mathbb{Q}(\sqrt{p_1}, \ldots, \sqrt{p_{n-1}})$$
and $p_1,\cdots,p_{n-1},q_1=p_n$ are distinct here the new positive integer $m$ is $1$

In all cases there is a contradiction with $P(n-1)$, finally $P(n)$ is true.