In my analysis class these questions have kept coming up over and over again about sums of radicals of primes being irrational. I wanted to just prove the general case so I never had to worry again. The argument should extend to the case where $p_i$ are allowed to be any non-squares in $\mathbb{N}$. Could anyone tell me if my argument holds? Sorry that it's a little long.

First we prove that $\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_{n}})$ is a degree $2^n$ extension of $\mathbb{Q}$, and $\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_{n}})$ is the splitting field for $\{x^2-p_1,...,x^2-p_{n}\}$


Choose $p_1,p_2$ primes. Notice that if deg$(\sqrt{p_1},\mathbb{Q})=2$ for $x^2-p_1$ is irreducible by Eisenstein's Criterion, similarly deg$(\sqrt{p_2},\mathbb{Q})=2$. If $\sqrt{p_2}=q_0+q_1\sqrt{p_1}$, then since $\sqrt{p_2}\not\in\mathbb{Q}$ we find $q_1\not=0$. We compute $$p_1=q_0^2+2q_0q_1\sqrt{p_1}-p_1^2,$$ since $p_2\not=-p_1%2$ we must have $q_0\not=0$. By field operations this contradicts $\sqrt{p_1}\not\in\mathbb{Q}$, therefore $\sqrt{p_2}\not\in\mathbb{Q}(\sqrt{p_1})$. It follows that $\mathbb{Q}(\sqrt{p_1},\sqrt{p_2})=2^2$. Moreover, it is a splitting field for the $\{x^2-p_1,x^2-p_2\}$.

Suppose that $n\geq 1$, and for all $1\leq i\leq n$ it follows that $\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_i})=2^i$ for distinct primes $p_1,...,p_i$. Moreover, suppose $\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_i})=2^i$ is the spliting field for $\{x^2-p_1,...,x^2-p_i\}$. Suppose $p_1,...,p_{n+1}$ are distinct primes. We wish to show that $\sqrt{p_{n+1}}\not\in\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n})$.

Suppose that $$\sqrt{p_{n+1}}=q_1p'+q_2p''+r$$ where $p',p''\in\{\sqrt{\prod_{i=1}^jp_{\sigma{i}}}|\sigma\in\text{S}_n,1\leq j\leq n\}$ are distinct, $r\in\text{Span}(\{\sqrt{\prod_{i=1}^jp_{\sigma{i}}}|\sigma\in\text{S}_n,0\leq j\leq n\}\backslash\{p',p''\})$, and $q_1,q_2\in\mathbb{Q}\backslash\{0\}$. By Eisenstein's $x^2-(p'')^2\in\mathbb{Q}[x]$ is irreducible, therefore $\mathbb{Q}(p'')$ is a degree 2 extension, and is the splitting field for $\{x^2-{p''}^2\}$ over $\mathbb{Q}$. By linear independence $p'\not\in\mathbb{Q}(p'')$, and by Eisenstein's $x^2-(p')^2\in\mathbb{Q}[x]$ is irreducible, therefore $\mathbb{Q}(p',p'')$ is a degree 4 extension of $\mathbb{Q}$, and is the splitting field for $\{x^2-{p''}^2,x^2-(p')^2\}$ over $\mathbb{Q}$. By the isomorphism extension and conjugate isomorphism theorems we can find $\sigma_{(-1)^n,(-1)^m}\in\text{Gal}(\mathbb{Q}(p',p'')/\mathbb{Q})$ such that $\sigma_{(-1)^n,(-1)^m}(p')=(-1)^np'$ and $\sigma_{(-1)^n,(-1)^m}(p'')=(-1)^mp''$ for all $n,m\in\{0,1\}$. By the isomorphism extension theorem these can be extended to isomorphism, $\tau_{(-1)^n,(-1)^m}$, of $\mathbb{Q}(p',p'',r)$ such that $\tau_{(-1)^n,(-1)^m}|_{\{p',p''\}}=\sigma{(-1)^n,(-1)^m}$ and $\tau_{(-1)^n,(-1)^m}(r)=r$. From the conjugate isomorphism theorems it follows that $(-1)^nq_1p'+(-1)^mq_2p''+r$ is a conjugate of $\sqrt{p_{n+1}}$, but $\sqrt{p_{n+1}}$ has two conjugates, a contradiction. Therefore $\sqrt{p_{n+1}}\not=q_1p'+q_2p''+r$.

It follows that if $\sqrt{p_{n+1}}\in\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n})$, then $\sqrt{p_{n+1}}=q_0+q_1p'$ where $p'\{\sqrt{\prod_{i=1}^jp_{\sigma{i}}}|\sigma\in\text{S}_n,1\leq j\leq n\}$, and $q_0,q_1\in\mathbb{Q}$ with $q_1\not=0$. We compute $$\sqrt{p_{n+1}}=q_0+q_1p'$$ $$p_{n+1}=q_0^2+2q_0q_1p'+q_1^2(p')^2.$$ Since $p'$ is irrational we have $q_0=0$ and $p_{n+1}=(q_1p')^2$. Note, $(p')^2$ is a product of distinct primes, so since the denominator of $q_1^2$ is a perfect square, and $q_1^2(p')^2\in\mathbb{N}$ we have $q_1^2\in\mathbb{N}$. Then every prime dividing $(p')^2$ divides $p_{n+1}$, a contradiction, so $\sqrt{p_{n+1}}\not=q_0+q_1p'$, and $\sqrt{p_{n+1}}\not\in\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n})$. Since $x^2-p_{n+1}$ is irreducible of degree 2 with roots $\pm\sqrt{p_{n+1}}$ it follows that $\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_{n+1}})$ is a degree $2^{n+1}$ extension of $\mathbb{Q}$. Moreover, $\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_{n+1}})$ is the splitting field for $\{x^2-p_1,...,x^2-p_{n+1}\}$. By P.M.I. this completes the proof. $\square$

Now we prove $\sum_{i=1}^nq_i\sqrt{p_i}+q_0$ is irrational for $q_0,...,q_n\in\mathbb{Q}$, $q_1,...,q_n\not=0$, and $p_i$ distinct primes.


Suppose $r_1,...,r_n$ are distinct products of primes, and $q_1,...,q_n\in\mathbb{Q}$ are non-zero. Moreover, suppose no square divides any $r_i$. Assume $p_1,...,p_t$ are all distinct primes dividing $r_1\cdot...\cdot r_n$. It follows that $\sqrt{r_1},...,\sqrt{r_n}$ are distinct elements of $\{\sqrt{\prod_{i=1}^jp_{\sigma{i}}}|\sigma\in\text{S}_n,0\leq j\leq n\}$, which is a basis for $\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n})$ over $\mathbb{Q}$, therefore by linear independence $\sum_{i=1}^nq_1\sqrt{r_i}\not=q$ for any $q\in\mathbb{Q}$. Our claim is a special case of what we've just proven, therefore this completes the proof.$\square$

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    $\begingroup$ It's surely time to learn Kummer theory. $\endgroup$ – Lord Shark the Unknown Oct 20 '18 at 6:51
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    $\begingroup$ This kind of question has been asked many times. The complications in your proof come from the mix of multiplicative and additive structures. Using (elementary) Kummer theory, you can appeal only to the multiplicative structure, see e.g. math.stackexchange.com/a/1609061/300700 $\endgroup$ – nguyen quang do Oct 21 '18 at 15:07
  • $\begingroup$ I don't fully understand the linked proof. Thank you both for mentioning Kummer Theory. I will plan on studying it when I get the time, or, obviously if it comes up in any of my courses. It seems the Kummer Theory approach, while being simpler, also has much broader applications. Which is awesome! $\endgroup$ – Melody Oct 21 '18 at 16:48
  • $\begingroup$ Kummer theory is great for this, but you can get away with more elementary reasoning. See here for some ways assuming various levels of machinery. $\endgroup$ – Jyrki Lahtonen Oct 24 '18 at 6:00
  • $\begingroup$ But I really wonder in what kind of classes on analysis questions like this come up over and over again? I can see things like $\sqrt2\notin\Bbb{Q}$ playing a role, but the more general variants? Of course, if the teacher really wanted to teach algebra but is stuck with this course.... (whistles innocently). $\endgroup$ – Jyrki Lahtonen Oct 24 '18 at 6:04

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