Square roots of primes are rationally independent This problem was proposed by Vitaly Bergelson as a seminar problem.
I was wondering if anyone could give me some feedback on proof-writing/style - the idea of analyzing modulo a large prime is not mine, but the writeup is.
Problem: Call a set $S \subset \mathbb{R}$ rationally independent if no nontrivial linear combination of its elements with rational coefficients is zero. Prove $\{\sqrt{p}: p \text{ prime}\}$ is rationally independent.
Solution: Clearing denominators, assume that $\sum_{i=1}^{n}{a_i}\sqrt{p_i} = 0$ for $a_i \in \mathbb{Z}$; we claim that each $a_k = 0$. For each $k$, we construct a large prime $P_k$ such that $\left(\frac{p_i}{P_k}\right)$ for each $1 \leq i \leq n$ such that $i \neq k$, and $\left(\frac{p_k}{P_k}\right) = -1$. To do this, let $P_k$ be such that
$$ P_k \equiv 1 \bmod{p_i} \text{ for }1 \leq i \leq n, i \neq k $$
$$ P_k \equiv b \bmod{p_k} \text{ where } \left(\frac{b}{p_k}\right) = -1 $$
$$ P_k \equiv 1 \bmod{4} $$
This gives us a solution $P_k$ modulo $4p_1p_2\dots p_n$, so by Dirichlet's theorem on arithmetic progressions, we can choose $P_k$ to be an arbitrarily large prime. Now, by quadratic reciprocity, we have $\left(\frac{p_i}{P_k}\right) = \left(\frac{P_k}{p_i}\right)= 1$. As $a = \sum_{1 \leq i \leq n, i\neq k}{a_i}\sqrt{p_i}$ is an element of $\mathbb{F}_{P_k}$, using the fact that $\sum_{i=1}^{n}{a_i}\sqrt{p_i} = 0$, we have
$$
    a = -a_k\sqrt{p_k} \implies a^2 \equiv a_k^2 p_k \pmod{P_k}
$$
If $a_k \not \equiv 0 \pmod{P_k}$, then we have $p_k \equiv (aa_k^{-1})^2 \pmod{P_k}$, contradicting $\left(\frac{p_k}{P_k}\right) = 1$. Thus, $a_k \equiv 0 \bmod{P_k}.$ The same argument for all $k$ between $1$ and $n$ shows that arbitrarily large primes divide each $a_k$, so in particular, each $a_k = 0$. $\blacksquare$
 A: I reccomend you to distinguish clearly between the element of $\mathbb{Z}$ (or its extension) and that of $\mathbb{F}_{P_k}$.
(Some readers may be not familiar with the term "ring" or "ring homomorphism" (as I was so before), but they are the term to describe many things precisely.)
So instead of stating "$a = \sum_{1 \leq i \leq n, i\neq k}{a_i}\sqrt{p_i}$ is an element of $\mathbb{F}_{P_k}$", the formal description is that:
There exist a ring homomorphism $f:\mathbb{Z} [ \sqrt{p_1},..\sqrt{p_{k-1}},\sqrt{p_{k+1}},\sqrt{p_n}]→\mathbb{F}_{P_k}$.
(The notation $\mathbb{Z}[\sqrt{p_1},..\sqrt{p_{k-1}},\sqrt{p_{k+1}},\sqrt{p_n}]$ means the set of linear combination of $\sqrt{p_1},..\sqrt{p_{k-1}},\sqrt{p_{k+1}},\sqrt{p_n}$ with coefficients in $\mathbb{Z}$.)
Then the discussion will go as:
$f$ sends $a^2$ to a square element in $\mathbb{F}_{P_k}$ while it sends $a_k^2p_k$ to non-square element unless $a_k=0$, (which contradict to $a^2=a_k^2p_k$ and the property of $f$ being ring homomorphism).
・Minor point:
It is better to declare $1\leq k \leq n$ when $k$ first appears.
