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Consider the set $\{\sqrt{p_1},\dots,\sqrt{p_n}\}$, where none of $p_i$'s are perfect squares, and ${\rm gcd}(p_i,p_j)=1$ for every $i \neq j$. Prove that $0$ cannot be expressed as $\sum\limits_{k=1}^n q_k\sqrt{p_k}$, where $q_k\in \mathbb{Q}$ for every $k$, and $(q_1,\dots,q_k)\neq (0,0,\dots,0)$.

TBTD
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    Good catch. Fixed, thanks Thomas. – TBTD Dec 18 '18 at 02:00
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    Show that $[\mathbb{Q}(\sqrt{p_1},\ldots,\sqrt{p_n}):\mathbb{Q}]= 2^n$ so $\mathbb{Q}(\sqrt{p_1},\ldots,\sqrt{p_n})$ is a $2^n$ dimensional $\mathbb{Q}$-vector space. Do you see a basis ? – reuns Dec 18 '18 at 02:04
  • Ah, reuns. So, the thing is I did a brief google search, hence I know that the tools I need are related to Galois fields, but I know close to $0$ abstract algebra, not to even mention Galois theory. Any theorems that you recommend me reading, so that I can tackle this problem? – TBTD Dec 18 '18 at 02:08
  • Or, whether there is an elementary way of solving this? – TBTD Dec 18 '18 at 02:09
  • If $x^2-m$ is irreducible in $K[x]$ then ${ a+b \sqrt{m}, (a,b) \in K}$ is a field (denoted $K(\sqrt{m})$) and a $2$-dimensional vector space (with basis $1,\sqrt{m}$). Here you need to look at $K_0 = \mathbb{Q}, K_{j+1} = K_j(\sqrt{p_{j+1}})$, to obtain that each $K_j$ is a $2^j$-dimensional $ \mathbb{Q}$ vector space. The basis will appear naturally, as well as your result. – reuns Dec 18 '18 at 02:12
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    In this answer I think Bill Dubuque managed to stirp the argument to bare essentials. The inductive step there does take some care. My answer below makes the induction step easier, but at the cost of assuming basic Galois theory. – Jyrki Lahtonen Dec 23 '18 at 11:19

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Oops, I have a semi-rigorous idea. Can someone help me with checking and rigorizing it?

I will prove the result when $p_i$'s are all prime.

Lemma: There exists a large prime $p$ such that, in modulo $p$, $p_1,\dots,p_{n-1}$ are all quadratic residues, while $p_n$ is not.

Proof. Here is the construction. Let $P$ be a large prime. The strategy is to use the law of quadratic reciprocity. Select $p\equiv 1\pmod{4}$, and $p\equiv 1\pmod{p_i}$, for every $i$. Now, the law of quadratic reciprocity implies, $$ \left(\frac{P}{p_i}\right)\left(\frac{p_i}{P}\right)=\left(\frac{p_i}{P}\right)=(-1)^{\frac{p_i-1}{2}\cdot \frac{P-1}{2}}=1, $$ hence, $p_i$ is indeed a quadratic residue, in modulo $P$, for every $i =1,2,\dots,n-1$. Now, let $\alpha_i\in\{0,1,\dots,p_n-1\}$ be a number, such that $\alpha_n$ is a quadratic non-residue, in modulo $p_n$. With this, we select $P\equiv \alpha_n\pmod{p_n}$. One can immediately see, using the law of quadratic reciprocity that, $p_n$ is a quadratic non-residue, in modulo $P$. Finally, the existence of such a prime $P$ obeying the congruences $P\equiv 1\pmod{4}$, $P\equiv 1\pmod{p_i}$ for $i =1,2,\dots,n-1$, and $P\equiv \alpha_n\pmod{p_n}$, follows immediately from the Chinese Remainder Theorem and Dirichlet's theorem on primes.

Now, in this prime $P$ modulus, if $\sum_{i=1}^n q_i\sqrt{p_i}=0$, then by multiplying with a large $M$ everywhere we get $\sum_{i =1}^{n-1} M_i\sqrt{p_i} \equiv -q_i\sqrt{p_n}\pmod{P}$, from here, we deduce that $p_n$ is a quadratic-residue, a contradiction.

TBTD
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  • From quadratic reciprocity and Dirichlet theorem on arithmetic progression, for any $a_1,\ldots,a_n$ coprime and any $b_i = \pm 1$ you can find a prime $p$ such that $(a_i/p) = b_i$. But how does it help here since you don't know anything on the $q_i$ ? – reuns Dec 18 '18 at 03:00
  • Very naively, on one side you get $\sum_{i=1}^{n-1}q_i\sqrt{p_i}$, where since $p_i$ is a quadratic residue, $\sqrt{p_i}$ exists modulo $P$, from here, you'd deduce that $q_n\sqrt{p_n}=c$ for some $c$ in modulo $P$. Hence, $\sqrt{p_n}$ must also exist in modulo $P$, but $p_n$ is not a quadratic residue. Note $q_i$'s are all rational, hence by taking P large enough (in fact, assuming the existence of a non-trivial combo, looking at all the denominators of involved $q_i$'s and taking $P$ sufficiently large), we can carry out modulo $p$ computation, no? – TBTD Dec 18 '18 at 03:04
  • The only thing is that, how to make this rigorous? – TBTD Dec 18 '18 at 03:04
  • Ah ok then yes (That's the same as showing that $x^2-p_n$ is irreducible over $\mathbb{Q}(\sqrt{p_1},\ldots,\sqrt{p_{n-1}})$). You can look at the ring $R = \mathbb{Z}/P \mathbb{Z}[\sqrt{p_1},\ldots,\sqrt{p_{n-1}}]$ where the $\sqrt{p_j}$ are symbols whose square is $p_j$. That ring may be not an integral domain, but there is a prime ideal such that $R/I$ is an integral domain, thus a finite field, containing the square root of each $p_j$. If $\mathbb{Z}/P \mathbb{Z}$ already contained a square root of each $p_j$ then $R/I \cong \mathbb{Z}/P \mathbb{Z}$ so you can lift $\sum_i M_i \sqrt{p_i}$ – reuns Dec 18 '18 at 03:09
  • see here – Jean Marie Sep 19 '22 at 15:33