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