I was searching for a non primitive extension, where for primitive extension it's meant a field extension $E/F$ such that $E = F(u)$ for some $u \in E \ $ (Here is given an example of non primitive extension).
One of my colleagues suggested me another and more pleasant example: $\mathbb{Q}[\sqrt{p}\mid p\text{ is prime}]$ over $\mathbb{Q}$, where $p$ varies among all primes.
If that extension would be primitive, it should exists an $r$ such that $Q[r]$ is that extension. However the extension proposed is algebric so $[\mathbb{Q}[r]:\mathbb{Q}] < \infty$ but this is absurd since $\mathbb{Q}[\sqrt{p} \mid p\text{ is prime}]$ has infinite degree.
My question is:
Is $\mathbb{Q}[\sqrt{p} \mid p\text{ is prime}]$ an algebraic extension of $\mathbb{Q}$?
My opinion is that it is not. I gave the following counterexample: $$ r = 1 + \Big(\frac{1}{\sqrt{2}}\Big)^3 + \Big(\frac{1}{\sqrt{3}}\Big)^3 + \cdots+ \Big(\frac{1}{\sqrt{p}}\Big)^3+ \cdots$$
(I made the power $3$ just to be sure the series converges).
Is that an algebraic element over $\mathbb{Q}$?