On non primitive fields extension

Question

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 $\mathbb{Q}$?

Answer 1

By definition, an extension $F/K$ is algebraic if and only if every element of $F$ is algebraic over $K$. We also have:

Theorem. Let $S$ be a subset of the algebraic closure of $K$. Then $K(S)$ is an algebraic extension of $K$.

Therefore, since $S=\{\sqrt{p}\mid p\text{ is a prime}\}$ is a set of algebraic numbers, $\mathbb{Q}(S)$ is algebraic over $\mathbb{Q}$.

As to your element $r$, it's not an element of the field. Sum is a binary operation on the field: it takes two arguments. Inductively, you may define a sum with $n$ terms. But you cannot obtain an operation of infinite arity this way. There is no infinite sum in the field.

Even in the real numbers you don't truly have an infinite sum. Instead, what you have is an analytic notion of limit, and you define an "infinite sum" as a limit of finite sums. But in the algebraic side, you do not have this notion, because there is no native notion of convergence. You cannot define this infinite sum as a limit, and you cannot add infinitely many things in a field. So your element $r$ is not an element of the field, even if it is a real number.

Think about it: if you could actually do this, then $\pi$ would be a rational number, because it is equal to $$3 + \frac{1}{10} + \frac{4}{10^2} + \frac{1}{10^3}+\frac{5}{10^4} +\cdots$$ But this "sum", really a limit, is not in the field of rational numbers.

That means that your $r$ cannot function as a "counterexample" to the extension being algebraic: you have no warrant whatsoever to assert that it lies in the extension.

Answer 2

$\mathbb{Q}[\sqrt{p} \text{ : p is prime}]$ is an algebraic extension of $\mathbb{Q}$, since it is generated by the $\sqrt{p}$, which are all algebraic over $\mathbb{Q}$.