Let $2\leq q_1<q_2< \dots<q_r$ be square-free natural numbers such that $mcd(q_i,q_j)=1$. Prove that the field extension $\Bbb{Q}(\sqrt{q_1},\dots,\sqrt{q_r})|\Bbb{Q}$ has degree $$[\Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{r}}):\Bbb{Q}]=2^r$$

I tried to prove the assertion by induction over $r$. For $r=1$, we need to prove that $\Bbb{Q}(\sqrt{q})|\Bbb{Q}$ has degree $2$, but this is clear since the minimal polynomial of $\sqrt{q}$ over $Q$ is

$$P(t)=t^2-q$$ which has degree $2$.

Now let's assume the assertion is true for $r=k$ and prove it for $k+1$. Consider

$$\Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k+1}})=\Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k}})( \sqrt{q_{k+1}})$$ Then we have the following chain of inclusions

$$\Bbb{Q} \subset \Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k}}) \subset \Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k}})( \sqrt{q_{k+1}}) $$


$$[\Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k+1}}):\Bbb{Q}]=[\Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k+1}}):\Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k}})]\cdot [\Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k}}):\Bbb{Q}]=2\cdot 2^k$$

since the minimal polynomial of $\sqrt{q_{k+1}}$ over $ \Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{r}})$ is


which again has degree $2$.

That should conclude the proof. However, something seems to be wrong, since I didn't use the hypothesis that $mcd(q_i,q_j)=1$ and that each $q_i$ is square-free at all. So I'm afraid that I left something unproved, that requires the use of the hypothesis. I think that might be the part where I stated that the minimal polynomial of $\sqrt{q_{k+1}}$ over the "small field" was the one I said it was, but I really don't know how to prove it. Any help would be appreciated. Thanks in advance!

  • $\begingroup$ You have used the square-free part by having the squareroots not being in rational numbers. $\endgroup$ – Zelos Malum Mar 8 '16 at 14:58
  • $\begingroup$ For the "square-free" part this is not really important, because if $q=a^2q'$ with $a,q,q'\geq 2$, then $\mathbb{Q}(\sqrt{q})=\mathbb{Q}(\sqrt{q'})$. So this hypothesis is really just a way to keep only the interesting part. $\endgroup$ – Arnaud D. Mar 8 '16 at 15:04

I would like to give a detailed proof of why $[\Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k+1}}):\Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k}})] = 2$. Then you can see where you have implicitly used what.

Using the inductive hypothesis, $\sqrt{q_k},\sqrt{q_{k+1}},\sqrt{q_{k+1}q_{k}} \notin \Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k-1}})$ (and because $\gcd(q_i, q_j) = 1$).
So suppose $\sqrt{q_{k+1}} \in \Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k}})$.
Then there are $x,y \in \Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k-1}})$ so that $$\sqrt{q_{k+1}} = x + y \cdot \sqrt{q_{k}}$$ $$\text{(inductive hyp. provides } [\Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k}}):\Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k-1}})] = 2) \text{)}$$

(note: $x \neq 0$, because otherwise $q_{k} \mid q_{k+1}$. And also $y \neq 0$, because $\sqrt{q_{k+1}} \notin \Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k-1}})$)

So we get $q_{k+1} = x^2 + y^2q_{k} + 2xy\sqrt{q_{k}}$ and this implies that $$\underbrace{2xy\sqrt{q_{k}}}_{\notin\ L:= \Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k-1}})} = -\underbrace{x^2}_{\in L} - \underbrace{y^2q_{k}}_{\in L} + \underbrace{q_{k+1}}_{\in \Bbb{N} \subset L} $$ which is a contradiction. So $$\sqrt{q_{k+1}} \notin \Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k}}) \implies [\Bbb{Q}(\sqrt{q_1}, \dots, \sqrt{q_{k+1}}):\Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k}})] = 2$$

  • $\begingroup$ what is the argument for $\sqrt{q_{k+1}} \in \Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k}})$. $\implies$ there are $x,y \in \Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k-1}})$ such that $\sqrt{q_{k+1}} = x + y \cdot \sqrt{q_{k}}$ ? shouldn't it be $x,y \in \Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k}})$ ? $\endgroup$ – reuns Mar 8 '16 at 23:05
  • $\begingroup$ @user1952009 This is because the inductive hyp. provides $[\Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k}}):\Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k-1}})] = 2$. That means, that $L:= \Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k}})$ is a 2-dimensional $\Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k-1}})$-vector space. Thus $x, y \in L$, because $(1, \sqrt{q_k})$ is a $L$-basis of $\Bbb{Q}(\sqrt{q_1},\dots, \sqrt{q_{k}})$ $\endgroup$ – johnnycrab Mar 8 '16 at 23:55
  • $\begingroup$ Sorry, there is a typo in the second sentence. Of course $L$ is the smaller field :) everything else is good :) $\endgroup$ – johnnycrab Mar 9 '16 at 7:42
  • $\begingroup$ I've spent hours filling in the 'elementary details' of your detailed proof. Although your inductive/algebraic logic might have some 'jumps or rough spots', it is 'dazzling and delicate'. (+1) $\endgroup$ – CopyPasteIt Apr 1 at 18:06

You have used it, albeit unstated. As $q_i$ is square-free, we know that $\sqrt{q_i}\notin\mathbb{Q},$ which is necessary, as otherwise it's redundant. You have also used that $\gcd(q_i,q_j)=1$ by having that $q_j\notin\mathbb{Q}(\sqrt{q_i}),$ as otherwise that might occur. With them not sharing any prime divisors, we know that the previous statement holds true.


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