Let $p_1,\ldots,p_n\in\mathbb{N}$ be different prime numbers, it can be shown that $[\mathbb{Q}(\sqrt{p_1},\ldots,\sqrt{p_n}):\mathbb{Q}]=2^n$ and in any case it is clearly finite since $[\mathbb{Q}(\sqrt{p_1},\ldots,\sqrt{p_n}):\mathbb{Q}]\leq2^n$.

Since $char(\mathbb{Q})=0$ then $\mathbb{Q}$ is perfect hence every field extension is separable, in particular $\mathbb{Q}(\sqrt{p_{1}},\sqrt{p_{2}},\ldots,\sqrt{p_{n}})/\mathbb{Q}$ is separable.

Since $\mathbb{Q}(\sqrt{p_{1}},\sqrt{p_{2}},\ldots,\sqrt{p_{n}})/\mathbb{Q}$ is a finite and separable field extension, by the primitive element theorem, it holds that there exist $\alpha\in\mathbb{Q}(\sqrt{p_{1}},\sqrt{p_{2}},\ldots,\sqrt{p_{n}})$ s.t $\mathbb{Q}(\alpha)=\mathbb{Q}(\sqrt{p_{1}},\sqrt{p_{2}},\ldots,\sqrt{p_{n}})$.

I wish to find such element $\alpha$ (i.e. a primitive element, that we know exist).

I know how to do this in the case $n=2$, I tried to generalize and prove this claim by induction, in the induction step I need to prove:

  1. $\sqrt{p_{1}}+\cdots+\sqrt{p_{n-1}}\in\mathbb{Q}(\sqrt{p_{1}}+\sqrt{p_{2}}+\sqrt{p_{n}})$

  2. $\sqrt{p_{n}}\in\mathbb{Q}(\sqrt{p_{1}}+\sqrt{p_{2}}+\sqrt{p_{n}})$

What I tried to do is to look at :

$$ \begin{align} & (\sqrt{p_{1}}+\sqrt{p_{2}}+\sqrt{p_{n}})(\sqrt{p_{1}}+\sqrt{p_{2}}-\sqrt{p_{n}}) \\ & =((\sqrt{p_{1}}+\sqrt{p_{2}}+\sqrt{p_{n-1}})+\sqrt{p_{n}})((\sqrt{p_{1}}+\sqrt{p_{2}}+\sqrt{p_{n-1}})-\sqrt{p_{n}}) \\ & =(\sqrt{p_{1}}+\cdots+\sqrt{p_{n-1}})^{2}-p_{n} \end{align} $$

If $n=2$ then this product is in $\mathbb{Q}$ hence in $\mathbb{Q}(\sqrt{p_{1}}+\sqrt{p_{2}})$ hence $\sqrt{p_{1}}-\sqrt{p_{2}}\in\mathbb{Q}(\sqrt{p_{1}}+\sqrt{p_{2}})$ so adding we get $\sqrt{p_1}\in\mathbb{Q}(\sqrt{p_1}+\sqrt{p_2})$ hence $\sqrt{p_2}\in\mathbb{Q}(\sqrt{p_1}+\sqrt{p_2})$ and we have proven $(2)$

So the reason I fail here is that I can't manage to show $$\sqrt{p_{1}}+\sqrt{p_{2}}-\sqrt{p_{n}}\in\mathbb{Q}(\sqrt{p_{1}}+\sqrt{p_{2}}+\sqrt{p_{n}}).$$

Can someone please help me find a primitive element, or help complete the proof I am trying to do here ? help is very much appriciated!

  • $\begingroup$ It seems easier to just guess the answer and show that it has $2^n$ Galois conjugates. [I haven't worked out the details—this is just something I've seen used on this site for similar questions.] $\endgroup$ – Dylan Moreland Jul 9 '12 at 21:31
  • $\begingroup$ @DylanMoreland - what do you mean by "something with $2^n$ Galois conjugates" ? $\endgroup$ – Belgi Jul 9 '12 at 21:32
  • $\begingroup$ This question has been answered here before, or something very similar:math.stackexchange.com/questions/93453 $\endgroup$ – Geoff Robinson Jul 9 '12 at 21:36
  • $\begingroup$ @GeoffRobinson - I tried reading the solution suggested in the link you gave and I could not understand them. I am hoping for a more elementary solution $\endgroup$ – Belgi Jul 9 '12 at 21:44
  • $\begingroup$ OK. I think you need some Galois theory, at least implicitly, as also sugested by Dylan Moreland. $\endgroup$ – Geoff Robinson Jul 9 '12 at 21:54

I believe this is the solution Dylan was hinting at. To show that $\alpha=\sum \sqrt{p_i}$ generates $E=\mathbb Q(\{\sqrt{p_i}\})$ it suffices to show that $\alpha$ is not fixed by any automorphism of $E$. Notice that any automorphism of $E$ maps $\sqrt{p_i}$ to $\pm \sqrt{p_i}$. So it suffices to demonstrate that

$$\sum \sqrt{p_i} \neq \sum s_i\sqrt{p_i}$$

for any choice of $s_i$ such that at least one $s_i$ is $-1$. But this is immediate because by cancelling the positive $s_i$ we would have

$$\sum \sqrt{p_j}=-\sum \sqrt{p_j}.$$

It follows that $\alpha$ is not fixed by any automorphism of $E$ and so $E=\mathbb Q(\alpha)$.

This is pretty much the same argument as Geoffs, but maybe it's a little bit clearer.

  • $\begingroup$ Can you please explian why it suffices to show that $\alpha$ is not fixed by any automorphism of $E$ ? I am tring to justify it to myself and I say that if there was an automorphism of $E$ that would of fixed $\alpha$ than there is a proper subgroup of the Galois group of $E$ that fixed $\alpha$. this subgroup corresponds to a proper subfield of $E$ (but I don't know if/why it is $\mathbb{Q}(\alpha)$ and I think that what I do need is the other direction of what I am writing here...) thanks for the help! $\endgroup$ – Belgi Jul 9 '12 at 22:13
  • $\begingroup$ @Belgi If $\mathbf Q(\alpha)\subsetneq E$, then there is a nontrivial automorphism of $E$ fixing $Q(\alpha)$. It's a simple fact, no need to bring the full strength of Galois theorem here. $\endgroup$ – tomasz Jul 9 '12 at 22:20
  • $\begingroup$ OK, I understand. Can we also use Galois theory to give a simple argument about why the degree of the extension is $2^n$ ? In the spirit of this answer I wanted to show that every such map ($p_i$ maps to plus or minus itself) is an automorphism, but it seems that we can't prove that this map is even $1-1$ without knowing that $\sqrt{p_i}$'s are independent over $\mathbb{Q}$ $\endgroup$ – Belgi Jul 9 '12 at 22:48
  • $\begingroup$ @Belgi I don't think so. You really have to show that the primes are independent, I think Bill Dubuque's argument is as good as it gets. $\endgroup$ – JSchlather Jul 10 '12 at 2:25

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