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