# Algebraic Field Extension of recursive subfields

Let $p_n$ be the n$^{th}$ prime and define subfields $F_n$ of $\mathbb{R}$ recursively by $F_n$=$F_{n-1}(\sqrt[p_n]{p_n})$. Show that $$[F_n:\mathbb{Q} ]=\prod_{i=1}^n p_i$$ and deduce that an algebraic extension need not be finite.

Any help would be greatly appreciated. Best regards.

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@BrandonCarter: I'm not entirely sure where they want us to go with this. Using Eisenstein Criteria, we know that $x^p-p$ is irreducible over $\mathbb{Q}$ for any prime $p$, which I think may help, though I haven't been able to tie it in yet. – Mathmo Feb 8 '12 at 21:56

Let's do it by induction and let me put $P_n=\prod_{i=1}^n p_i$
The statement $[F_n:\mathbb Q]=P_n$ is clear for $n=1$.
Assume it is true for for $n-1$ and let's prove it for $n$.
Let $K=\mathbb Q(\sqrt[p_n](p_n))$.
It is clear that $[F_n:F_{n-1}]\leq p_n$ since $X^{p_n}-p_n$ kills $\sqrt[p_n](p_n)$ and has coefficients in $F_{n-1}$. Hence
$$[F_n:\mathbb Q]= [F_n:F_{n-1}][F_{n-1}:\mathbb Q] \leq p_n\cdot P_{n-1}$$

On the other hand we have $F_{n-1}\subset F_n$ implying that $P_{n-1}|[F_n:\mathbb Q]$ and also $K\subset F_n$ implying $p_n|[F_n:\mathbb Q]$.
Since $P_{n-1}$ and $p_n$ are relatively prime (this is the key point !) we deduce $$p_n\cdot P_{n-1}|[F_n:\mathbb Q]$$
The two displayed equations prove that $[F_n:\mathbb Q]=p_n\cdot P_{n-1}=P_n$

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Wow, thank you very much. I'll spend some time working through this and may get back to you if my understanding fails. Great answer though. – Mathmo Feb 8 '12 at 22:33

Check that $[\mathbb{Q}(\sqrt[p_n]{p_n}):\mathbb{Q}]=p_n$. Suppose by induction $[F_{n-1}:\mathbb{Q}]=\prod_{k=1}^{n-1}p_k$. Notice $[\mathbb{Q}(\sqrt[p_n]{p_n}):\mathbb{Q}] \mid [F_n:\mathbb{Q}]$ and that $[\mathbb{Q}(\sqrt[p_n]{p_n}):\mathbb{Q}]\geq [F_n:F_{n-1}]$.

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Thanks, I see how (when verified) this all ties together. I'm really struggling to show that $[\mathbb{Q}(\sqrt[p_n]{p_n}):\mathbb{Q}]=p_n$, having not really worked with field extensions of this nature before. Conceptually, how is $\mathbb{Q}(\sqrt[p_n]{p_n})$ constructed, and why is the relation true? – Mathmo Feb 8 '12 at 22:14
Georges Elencwajg has submitted an answer, but many thanks for your response, it got me thinking along the right lines. Best. – Mathmo Feb 8 '12 at 22:33