I have a basic question about algebraic field extensions:
How can I convert a multiple extension like $\mathbb{Q}(\sqrt{2},\sqrt{3})$ to a single (elementary) field extension (like $\mathbb{Q}(\sqrt{5})$) ?
Is it even possible to do this in general?
Is it always the case, that if we got a field extension $K = \mathbb{Q}(\alpha_1, \alpha_2, ..., \alpha_n)$ and $\alpha_i$'s are algebraic over $\mathbb{Q}$ then $K = \mathbb{Q}(\alpha_1 + \alpha_2 + ... + \alpha_n)$ ?
Generally one can use the Primitive Element Theorem (see below). But here there is a simple optimization: $\rm\ F = \Bbb Q(\sqrt{3}+\sqrt{2}) \supseteq \Bbb Q(\sqrt{3},\sqrt{2})\,$ (and reverse is clear), since $\rm\,F\,$ contains not only $\, u = \sqrt{3}+\sqrt{2}\, $ but also $\,v = \sqrt{3}-\sqrt{2} = (3-2)/(\sqrt{3}+\sqrt{2}), \, $ thus $\,\sqrt{3},\sqrt{2} = (u\pm v)/2 \in\rm F.$
If field F has $2\,$ F-linear independent combinations of $\rm\, \sqrt{a},\ \sqrt{b}\, $ then we can solve for $\rm\, \sqrt{a},\ \sqrt{b}\, $ in F. For example, the Primitive Element Theorem works that way, obtaining two such independent combinations by Pigeonholing the infinite set $\rm\ F(\sqrt{a} + r\ \sqrt{b}),\ r \in F,\ |F| = \infty,\,$ into the finitely many fields between F and $\rm\ F(\sqrt{a}, \sqrt{b}),\,$ e.g. see PlanetMath's proof.
In this case it is simpler to notice $\rm\ F = \mathbb Q(\sqrt{a} + \sqrt{b})\ $ contains the independent $\rm\ \sqrt{a} - \sqrt{b}\ $ since
$$\rm \sqrt{a}\ -\ \sqrt{b}\ =\ \dfrac{\ a\,-\,b}{\sqrt{a}+\sqrt{b}}\ \in\ F = \mathbb Q(\sqrt{a}+\sqrt{b}) $$
To be explicit, notice that $\rm\ u = \sqrt{a}+\sqrt{b},\ v = \sqrt{a}-\sqrt{b}\in F\ $ so solving the linear system for the roots yields $\rm\ \sqrt{a}\ =\ (u+v)/2,\ \ \sqrt{b}\ =\ (u-v)/2,\ $ both of which are clearly $\rm\,\in F,\:$ since $\rm\:u,\:v\in F\:$ and $\rm\:2\ne 0\:$ in $\rm\:F,\:$ so $\rm\:1/2\:\in F.\:$ This works over any field where $\rm\:2\ne 0\:,\:$ i.e. where the determinant (here $2$) of the linear system is invertible, i.e. where the linear combinations $\rm\:u,v\:$ of the square-roots are linearly independent over the base field.
More generally, one may use the following lemma (which is the basis of a general result on linear independence of square roots due to Besicovitch, see below).
Lemma $\rm\ \ [K(\sqrt{a},\sqrt{b}) : K] = 4\ $ if $\rm\ \sqrt{a},\ \sqrt{b},\ \sqrt{a\:b}\, $ are all $\rm\,\not\in K,\:$ and $\rm\: 2 \ne 0\:$ in $\rm\,K.$
Proof $\ \ $ Let $\rm\ L = K(\sqrt{b})\:.\:$ Then $\rm\: [L:K] = 2\:$ via $\rm\:\sqrt{b} \not\in K,\:$ thus it suffices to show $\rm\: [L(\sqrt{a}):L] = 2\:.\:$ It fails only if $\rm\:\sqrt{a} \in L = K(\sqrt{b})\ $ and then $\rm\ \sqrt{a}\ =\ r + s\ \sqrt{b}\ $ for $\rm\ r,s\in K.\:$ But that's impossible, since squaring yields $\rm(1):\ \ a\ =\ r^2 + b\ s^2 + 2\:r\:s\ \sqrt{b}\:,\: $ contra, hypotheses, as follows
$\rm\quad\quad\quad\quad\quad\quad\quad\quad rs \ne 0\ \ \Rightarrow\ \ \sqrt{b}\ \in\ K\ \ $ by solving $(1)$ for $\rm\sqrt{b}\:,\:$ using $\rm\:2 \ne 0$
$\rm\quad\quad\quad\quad\quad\quad\quad\quad\ s = 0\ \ \Rightarrow\ \ \ \sqrt{a}\ \in\ K\ \ $ via $\rm\ \sqrt{a}\ =\ r \in K$
$\rm\quad\quad\quad\quad\quad\quad\quad\quad\ r = 0\ \ \Rightarrow\ \ \sqrt{a\:b}\in K\ \ $ via $\rm\ \sqrt{a}\ =\ s\ \sqrt{b}\:,\: \ $times $\rm\:\sqrt{b}\quad$ QED
Using the above as the inductive step one easily proves the following result of Besicovic.
Theorem $\ $ Let $\rm\:Q\:$ be a field with $2 \ne 0\:,\:$ and $\rm\ L = Q(S)\ $ be an extension of $\rm\:Q\:$ generated by $\rm\: n\:$ square roots $\rm\ S = \{ \sqrt{a}, \sqrt{b},\ldots \}$ of elts $\rm\ a,\:b,\:\ldots \in Q\:.\:$ If every nonempty subset of $\rm\:S\:$ has product not in $\rm\:Q\:$ then each successive adjunction $\rm\ Q(\sqrt{a}),\ Q(\sqrt{a},\:\sqrt{b}),\:\ldots$ doubles the degree over $\rm\,Q,\,$ so, in total, $\rm\: [L:Q] \ =\ 2^n\:.\:$ So the $\rm\:2^n\:$ subproducts of the product of $\rm\:S\:$ comprise a basis of $\rm\:L\:$ over $\rm\:Q\:.$
