# Finding a basis for $\Bbb{Q}(\sqrt{2}+\sqrt{3})$ over $\Bbb{Q}$.

I have to find a basis for $\Bbb{Q}(\sqrt{2}+\sqrt{3})$ over $\Bbb{Q}$.

I determined that $\sqrt{2}+\sqrt{3}$ satisfies the equation $(x^2-5)^2-24$ in $\Bbb{Q}$.

Hence, the basis should be $1,(\sqrt{2}+\sqrt{3}),(\sqrt{2}+\sqrt{3})^2$ and $(\sqrt{2}+\sqrt{3})^3$.

However, this is not rigorous. How can I be certain that $(x^2-5)^2-24$ is the minimal polynomial that $\sqrt{2}+\sqrt{3}$ satisfies in $\Bbb{Q}$? What if the situation was more complicated? In general, how can we ascertain thta a given polynomial is irreducible in a field?

Moreover, checking for linear independence of the basis elements may also prove to be a hassle. Is there a more convenient way of doing this?

Thanks.

One easy way: $\rm\ F = \Bbb Q(\sqrt{3}+\sqrt{2}) \supseteq \Bbb Q(\sqrt{3},\sqrt{2})\,$ (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.\$ QED $\,$ Below is further discussion from some of my older posts.

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 $\rm\:Q\:.$

In the present case you can argue as follows: clearly

$$\Bbb Q(\sqrt2)\subsetneqq\Bbb Q(\sqrt2+\sqrt3)\implies \dim_{\Bbb Q}\Bbb Q(\sqrt2+\sqrt3)>\dim_{\Bbb Q}\Bbb Q(\sqrt2)=2$$

But since you already found a quartic that has $\;\sqrt2+\sqrt3\;$ as a root and also

$$2\mid\dim_{\Bbb Q}\Bbb Q(\sqrt2+\sqrt3)\;\;\text{(by the above!)}$$

then the dimension must be exactly four and your polynomial is the minimal one and thus irreducible.

In the general case I don't think there's a general algorithm by which to tell whether a given polynomial is irreducible or not.

• Actually there are very good algorithms to factor univariate polynomials over the rationals, including a polynomial-time algorithm based on LLL. link.springer.com/article/10.1007%2FBF01457454 Feb 12 '14 at 17:35

A basis of $\mathbb Q\big[\sqrt{2},\sqrt{3}\big]$ consists of the elements $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$, and hence its dimension over $\mathbb Q$ is equal to $4$.

Clearly, all the above elements belong to $\mathbb Q\big[\sqrt{2},\sqrt{3}\big]$, and hence it remains to show that they are independent over $\mathbb Q$.

There is a rather elegant way to show this, and in fact something more general:

If $n_1,n_2,\ldots,n_k$ are distinct square-free positive integers, then $\sqrt{n_1},\sqrt{n_2},\ldots,\sqrt{n_k}$ are linearly independent over $\mathbb Q$.

(A number $n$ is said to be square free if $k^2\mid n$, implies that $k=1$.)

• Regarding generalizations, if you follow the fnal link in my answer you will find a proof of a more general result of Besicovitch, along with references to even further generalizations by Mordell and Siegel. Feb 17 '14 at 21:42