# How to convert $\Bbb Q(\sqrt 2,\sqrt 3)$ to $\Bbb Q(\alpha)?$

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 $K = \mathbb{Q}(\alpha_1 + \alpha_2 + ... + \alpha_n)$ ?

• See the primitive element theorem. – Git Gud May 1 '14 at 15:48
• To your last question: Generally this is not the case, but it is the case for square roots of primes. – B.Swan Jun 13 '19 at 15:59

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 $\rm\:Q\:.$

• 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 $K = \mathbb{Q}(\alpha_1 + \alpha_2 + ... + \alpha_n)$ ? – Andrei Kh May 1 '14 at 16:09
• @AndreiKh No, generally you need a linear combination of the $\,\alpha_i.\,$ In practice usually the coefficients are small, even though not necessarily $1.\,$ See the linked (standard) proof to better understand how it works. – Bill Dubuque May 1 '14 at 16:14
• Thank you :) . I was initially working on implementing Trager's algorithm for factorization of polynomials over algebraic number fields. The algorithm provided was designed only for elementary extensions. It's sad that it isn't just the sum of the $\alpha_i$. – Andrei Kh May 1 '14 at 16:38
• @AndreiKh Indeed. Coincidentally, Barry Trager was my office-mate when I was an undergrad member of the MIT Macsyma Group (he was working on his thesis on the algebraic case of the Risch integration algorithm). – Bill Dubuque May 1 '14 at 16:52
• @Downvoter If something is not clear, please feel welcome to ask questions. – Bill Dubuque May 1 '14 at 22:45