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 $\mathbb{Q}$?

  • $\begingroup$ I don't understand what the field you are talking about is. If you mean a prime as in a prime number, then those are all contained in $\mathbb{Z}$, so in particular in $\mathbb{Q}$ and the extension you're looking at is $\mathbb{Q}/\mathbb{Q}$: $\endgroup$
    – Thorgott
    Dec 17, 2019 at 18:41
  • $\begingroup$ Sorry, my fault, I forgot the sqrt $\endgroup$
    – Gabrielek
    Dec 17, 2019 at 18:43
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    $\begingroup$ Why do you think that $r \in \mathbb{Q}[\sqrt{p} \ : p \textrm{ prime}]$? I don't see why this should hold. $\endgroup$ Dec 17, 2019 at 18:45
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    $\begingroup$ $r$ is not an element of the field: there are no infinite sums in fields. $\endgroup$ Dec 17, 2019 at 18:46
  • 3
    $\begingroup$ @Gabrielek: Sum is a BINARY operation. You can only add finitely many things. There is no infinitary sum operation on the field. $\endgroup$ Dec 17, 2019 at 18:51

2 Answers 2


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.

  • $\begingroup$ (+1) Great didactic point. However, you cannot a priori claim that $r$ is not an element of the extension, even though OPs rationale appears to be mistaken. Saying anything about the nature of $r$ seems to be non-trivial. $\endgroup$
    – Thorgott
    Dec 17, 2019 at 19:15
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    $\begingroup$ @Thorgott: I can claim that the assertion that $r$ lies in the field is unwarranted, and so I need not consider $r$ at all unless and until the OP establishes that it does lie in the field. It's certainly not a "counterexample" to the extension given being algebraic. $\endgroup$ Dec 17, 2019 at 19:16
  • $\begingroup$ Arturo thank you for your answer, but thank to this can I conclude that it is a non primitive extention? $\endgroup$
    – Gabrielek
    Dec 17, 2019 at 19:41
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    $\begingroup$ @Gabrielek: It's a nonprimitive extension because algebraic primitive extensions are of finite degree, and this is an algebraic extension of infinite degree. $\endgroup$ Dec 17, 2019 at 19:43

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


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