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I have this problem in a exercise list:

"Prove that $K=\mathbb{Q}(2^{\frac{1}{2}},2^{\frac{1}{3}}, 2^{\frac{1}{4}}, \ldots)$ is an algebraic extension, but not a finite extension of $\mathbb{Q}$."

What I did:

Let $\alpha \in K$, then $\alpha = a_0 + a_12^{\frac{1}{n_1}}+a_22^{\frac{1}{n_2}}+\cdots+a_m2^{\frac{1}{n_m}}$.

Now, since each factor of the sum above is algebraic over $\mathbb{Q}$, it follows that $\alpha$ is indeed algebraic over $\mathbb{Q}$ (because the set of algebraic numbers is a field).

Suppose now that $K$ is a finite extension of $\mathbb{Q}$. Then, by Steinitz's theorem, there is $u \in K$ such that $K=\mathbb{Q}(u)$. Let $p(x)$ be the minimal polynomial of $u$, and let $n$ be $p$'s degree. Then, K is an extension os degree $n$. However, the minimal polynomial of $2^{\frac{1}{n+1}}$ has degree $n+1$, so K is an extension of degree at least $n+1$, contradicting the hypothesis that $K$ is an extension of degree $n$. So, $K$ is an infinite extension of $\mathbb{Q}$.

My doubts: Can $\alpha$ be an infinite sum? If yes, can I use the same argument?

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  • $\begingroup$ No, it cant.... $\endgroup$ Sep 10, 2014 at 3:38
  • $\begingroup$ No, unless you add some topology in which the sequence converges. $\endgroup$
    – user40276
    Sep 10, 2014 at 3:41
  • $\begingroup$ @GastónBurrull, so is what I did correct? Or did I make some mistake somewhere? (: $\endgroup$
    – Anna
    Sep 10, 2014 at 3:42
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    $\begingroup$ Yes but Steinitz's theorem is unnecesary, if the extension is finite of degree $n$, by tower law you have that $k|n$ for every $k\in\mathbb{N}$ a contradiction. $\endgroup$ Sep 10, 2014 at 3:45
  • $\begingroup$ @GastónBurrull what would be this k? The degree of the extension $\mathbb{Q}(2^{\frac{1}{k}})$? $\endgroup$
    – Anna
    Sep 10, 2014 at 3:47

1 Answer 1

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As has been mentioned, the result is a direct consequence of the fact that $[\mathbf{Q}(2^{1/n}):\mathbf{Q}]$ = n. This is because we then have, for every $n$, the inequality $$n = [\mathbf{Q}(2^{1/n}):\mathbf{Q}] \leq [K:\mathbf{Q}].$$ Since $[K:\mathbf{Q}]$ is larger than every natural number, it must be infinite.

The trickiest point is why $2^{1/n}$ has degree $n$ over $\mathbf{Q}$. For this it is enough to establish that the polynomial $X^n - 2$ is irreducible. This follows immediately from Eisenstein's criterion for the prime $p = 2$.

There is an error in your proof that every element of $K$ is algebraic over $\mathbf{Q}$. It is not clear at all why you claim any $\alpha \in K$ can be written in the form of a linear combination of various roots of $2$. The only thing that is obvious is that $\alpha$ can be written as a polynomial expression of these numbers, not a linear combination.

The proof that $K$ is algebraic is as follows. The field $K$ is generated by the set $S = \mathbf{Q} \cup \{2^{1/2}, 2^{1/3}, \dots\} \subseteq \overline{\mathbf{Q}}$, where $\overline{\mathbf{Q}}$ is the field of algebraic numbers over $\mathbf{Q}$. It follows that $K$ itself is contained in $\overline{\mathbf{Q}}$. Thus $K$ is algebraic over $\mathbf{Q}$.

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  • $\begingroup$ Why exactly does the equality you mention work? I don't understand why you use $\leq$, when they are equal? could you explain me, please? $\endgroup$
    – Hopmaths
    Oct 15, 2021 at 2:13
  • $\begingroup$ Is algebraic closure of $K$ and $Q$ are same? $\endgroup$
    – Tony
    Sep 4, 2022 at 4:14

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