# root of prime numbers are linearly independent over $\mathbb{Q}$ [duplicate]

How can we prove by mathematical induction that $1,\sqrt{2}, \sqrt{3}, \sqrt{5},\ldots, \sqrt{p_n}$ ($p_n$ is the $n^{\rm th}$ prime number) are linearly independent over the rational numbers ?

$\underline{\text{base case (n=1)}}$: $1,\sqrt{2}$ are linearly independent over the field $\mathbb{Q}$ otherwise $a1+b\sqrt{2}=0 \Leftrightarrow \sqrt{2}=-\frac{a}{b}$ which is absurd.

Then I am stuck.

## marked as duplicate by Magdiragdag, Zain Patel, Vladhagen, dxiv, ShaileshMay 4 '17 at 0:19

• Induction is fine. $\sqrt{p_n} = \sum a_i \sqrt{p_i} => p = (\sum a_i p_i)^2$ implie $\sqrt{p_1} ... \sqrt{p_{n-1}$ are linearly dependent. – fleablood May 3 '17 at 20:53
This is an overkill but it is an interesting one. By quadratic reciprocity and Dirichlet's theorem, given some primes $p_1,p_2,\ldots,p_k$ and a sequence of signs (i.e. elements of $\{-1,+1\}$) $s_1,s_2,\ldots,s_k$, there is an infinite number of primes $q$ such that the Legendre symbol $\left(\frac{p_i}{q}\right)$ equals $s_i$ for any $i\in[1,k]$. If $1,\sqrt{2},\ldots,\sqrt{p_n}$ were linearly dependent over $\mathbb{Q}$, $p_n$ would be a quadratic residue for any large enough prime $q$ admitting $1,2,\ldots,p_{n-1}$ as quadratic residues$^{(*)}$. That contradicts the previous "independency" result.
$(*)$ Let explain this point more clearly. If we assume that $$\eta_0 + \eta_1\sqrt{p_1}+\eta_2\sqrt{p_2}+\ldots+\eta_n\sqrt{p_n}=0$$ with $\eta_k\in\mathbb{Q}$ and $\eta_n\neq 0$, we may take some prime $q$ that is greater than the denominator (in absolute value) of any $\eta_k$ and such that $p_1,p_2,\ldots,p_{n-1}$ are quadratic residues $\!\!\pmod{q}$. It follows that $\pm\sqrt{p_1},\pm\sqrt{p_2},\ldots,\pm\sqrt{p_{n-1}}$ are elements of $\mathbb{F}_q$, as well as $\eta_0,\eta_1,\ldots,\eta_n$. By the previous identity it follows that $\sqrt{p_n}$ is also an element of $\mathbb{F}_q$, hence $p_n$ is a quadratic residue $\!\!\pmod{q}$.