# Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$?

Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$ ?

$$\mathbf{Q}(\sqrt{2},\sqrt{3})=\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \mid a,b,c,d\in\mathbf{Q}\}$$

$$\mathbf{Q}(\sqrt{2}+\sqrt{3}) = \lbrace a+b(\sqrt{2}+\sqrt{3}) \mid a,b \in \mathbf{Q} \rbrace$$

So if an element is in $\mathbf Q(\sqrt{2},\sqrt{3})$, then it is in $\mathbf{Q}(\sqrt{2}+\sqrt{3})$, because $\sqrt{6} = \sqrt{2}\sqrt{3}$.

How to conclude from there?

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$\mathbf{Q}(\sqrt{2}+\sqrt{3}) \not= \{a+b(\sqrt{2}+\sqrt{3})\ | a,b \in \mathbf{Q} \}$ because $\sqrt{2}+\sqrt{3}$ does not have degree 2 over $\mathbf{Q}$. – lhf Dec 22 '11 at 14:28
Hi lhf, then what is it? ? ? ? – Tashi Dec 22 '11 at 14:30
$\alpha=\sqrt{2}+\sqrt{3}$ has degree 4 over $\mathbf{Q}$ and so a basis is $1,\alpha,\alpha^2,\alpha^3$. – lhf Dec 22 '11 at 14:35
@Tashi: This may just be an issue of notation, since your argument in the question seems to suggest that you thougt that $\{a+b(\sqrt{2}+\sqrt{3})\ | a,b \in \mathbf{Q} \}$ also includes $\sqrt2\sqrt3$. (It doesn't.) – joriki Dec 22 '11 at 14:39

$\mathbb{Q}(\sqrt{2} + \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3})$ is clear.

Now note that $$(\sqrt{2} + \sqrt{3})^{-1} = \frac{1}{\sqrt{2} + \sqrt{3}} = \frac{\sqrt{2} - \sqrt{3}}{2 - 3} = \sqrt{3} - \sqrt{2}$$ hence $\sqrt{3} - \sqrt{2} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence $\sqrt{2} + \sqrt{3} + \sqrt{3} - \sqrt{2} = 2 \sqrt{3} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence $\sqrt{3} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$. Note that by a similar argument you get $\sqrt{2} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence $\mathbb{Q}(\sqrt{2}, \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2} + \sqrt{3})$.

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Thank you, I can understand this answer! – Tashi Dec 22 '11 at 14:46
@Tashi You're welcome : ) – Matt N. Dec 22 '11 at 14:50
@Tashi With only slightly more effort, you can understand my answer too (which I have elaborated). That will enable you to understand not only this special case, but much more general cases - many of which arise frequently in practice. – Gone Dec 22 '11 at 15:06

HINT $\$ If a field F has two F-linear independent combinations of $\rm\ \sqrt{a},\ \sqrt{b}\$ then you 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's simpler to notice $\rm\ E = \mathbb Q(\sqrt{a} + \sqrt{b})\$ contains the independent $\rm\ \sqrt{a} - \sqrt{b}\$ since

$$\rm \sqrt{a}\ -\ \sqrt{b}\ =\ \dfrac{a-b}{\sqrt{a}+\sqrt{b}}\ \in\ E = \mathbb Q(\sqrt{a}+\sqrt{b})$$

To be explicit, notice that $\rm\ u = \sqrt{a}+\sqrt{b},\ v = \sqrt{a}-\sqrt{b}\in E\$ so solving the linear system for the roots yields $\rm\ \sqrt{a}\ =\ (u+v)/2,\ \ \sqrt{b}\ =\ (v-u)/2\:,\$ both of which are clearly $\rm\:\in E\:,\:$ since $\rm\:u,\:v\in E\:$ and $\rm\:2\ne 0\:$ in $\rm\:E\:,\:$ so $\rm\:1/2\:\in E\:.\:$ 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.

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 Thank you very much. – Tashi Dec 22 '11 at 14:46 @Matt My answer is excerpted from one of my old sci.math posts. [1] was a link to PlanetMath's proof of the Primitive Element Theorem. But PlanetMath seems to have problems currently (or the link has rotted) – Gone Dec 22 '11 at 15:18 Thanks Bill! Now that my comment is obsolete, I'll delete it. – Matt N. Dec 22 '11 at 15:19 I can't seem to access PlanetMath either. Some alternative references: Theorem 3.6 of this handout of Keith Conrad's and Theorem 5.1 of Milne's notes. Oddly enough, they both use Tashi's question as a first example. – Dylan Moreland Dec 22 '11 at 15:22 @BillDubuque In your last sentence, did you mean to write: "...where the determinant of the linear system is non-zero..."? (rather than "invertible") – Matt N. May 21 '12 at 10:25
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If it's allowed to use the Galois theory, it can be proved as following. Since the subgroup of the Galois group of the field extension $\mathbb{Q} (\sqrt2,\sqrt 3)$ over $\mathbb{Q}$ which the subfield $\mathbb{Q}(\sqrt 2+\sqrt 3)$ is trivial, therefor we know the result by the Galois theory. I admit it is not too trivial since one has to verify something as said above.

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