# How to show two field extensions are equal.

How do I go about showing that $\mathbb{Q}(\sqrt{3}+\sqrt{5})=\mathbb{Q}(\sqrt{3},\sqrt{5})$?

Since $\mathbb{Q}(\sqrt{3}+\sqrt{5}) \subset\mathbb{Q}(\sqrt{3},\sqrt{5})$, we need to show the reverse inclusion. Does it suffice to prove that $\sqrt{2} \in \mathbb{Q}(\sqrt{3}+\sqrt{5})$?

Thank you

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Where did $\sqrt{2}$ come from? –  Tobias Kildetoft Dec 5 '12 at 18:56
My argument could be completely flawed but I was thinking that $\sqrt{3}\sqrt{5}=2\sqrt{2}$ is in $\mathbb{Q}(\sqrt{3},\sqrt{5})$ so I need to show it is also in $\mathbb{Q}(\sqrt{3}+\sqrt{5})$. –  user39280 Dec 5 '12 at 19:01
Your first equality is not correct. –  Tobias Kildetoft Dec 5 '12 at 19:03
I am saying that $\sqrt{3}\sqrt{5}$ is not $2\sqrt{2}$. –  Tobias Kildetoft Dec 5 '12 at 19:06
Oh my god, sorry. I am so tired I cannot even think. –  user39280 Dec 5 '12 at 19:09

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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We have

$$\frac{(\sqrt{3} + \sqrt{5})^3 - 14(\sqrt{3} + \sqrt{5})}{4} = \frac{18\sqrt{3} + 14\sqrt{5} - 14\sqrt{3} -14 \sqrt{5}}{4} = \frac{4\sqrt{3}}{4} = \sqrt{3}$$

So $\sqrt{3} \in \mathbb{Q}(\sqrt{3} + \sqrt{5})$, and hence so is $\sqrt{5}$.

The general trick in these situations is to write one of the irrational numbers as a polynomial in their sum, with coefficients from $\mathbb{Q}$.

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Thank you. I don't know why I thought the answer was much more complicated. –  user39280 Dec 5 '12 at 19:06

Note that $$\frac{1}{\sqrt{5}+\sqrt{3}}=\frac{\sqrt{5}-\sqrt{3}}2$$ So $\sqrt{5}-\sqrt{3}\in\mathbb Q[\sqrt{5}+\sqrt{3}]$ and therefore $\sqrt{3},\sqrt 5\in\mathbb Q[\sqrt{5}+\sqrt{3}]$

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