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.

Question

Take $p_1, p_2, \ldots, p_n, p_{n+1}$ be $n+1$ prime numbers in $\mathbb{P} \subseteq \mathbb{N}$. $\sqrt{p_{n+1}} \notin \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots, \sqrt{p_n})$ seems to be quite clear, but still need a proof. I know some proofs are involved with Galois theory, which is not I want.

Answer 1

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 has $b=0$ 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 has $a=0$ 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 has $$\sqrt{p_n}=\frac{q_{1}\cdots q_m-a^2-b^2p_n}{2ab}\in \mathbb{Q}(\sqrt{p_1}, \ldots, \sqrt{p_{n-1}}) $$ 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.