# Prove $\mathbb Q (\sqrt2 + \sqrt3 ) = \mathbb Q (\sqrt2 , \sqrt3 )$ [duplicate]

So correct if I am wrong, but these sets are:

$\mathbb Q (\sqrt2 + \sqrt3 )=\{a+b\sqrt2 + c\sqrt3 +d \sqrt6 : a,b,c,d \in \mathbb Q \}$

$\mathbb Q (\sqrt2 , \sqrt3 )=\{a+b\sqrt2 + c\sqrt3 : a,b,c \in \mathbb Q \}$

Was trying to see if the LHS is a subset of RHS and vise versa.

Let $A \in \mathbb Q (\sqrt2 , \sqrt3 )$ with $A=a+b\sqrt2 + c\sqrt3= a+b\sqrt2 + c\sqrt3 + d \sqrt6 - d \sqrt6 = a+b\sqrt2 + c\sqrt3 + d \sqrt6 - d \sqrt3 \sqrt2$ but i can't really do anything after this. If you factor the root $3$ or $2$, it doesn't seem to help.

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• A field that contains $\sqrt{2}+\sqrt{3}$ contains its reciprocal. Compute the reciprocal. – André Nicolas Feb 18 '16 at 17:40
• It suffices to show that $\sqrt{2}$ and $\sqrt{3}$ are in $\mathbb{Q}(\sqrt{2}+\sqrt{3})$. (The other direction is trivial.). If $\sqrt{2}+\sqrt{3}$ is in the field, its reciprocal $\sqrt{3}-\sqrt{2}$ is in, therefore the sum $2\sqrt{3}$ is in, therefore $\sqrt{3}$ is in. Note this is the same idea as a couple of the answers, except I am using division instead of squaring and manipulating. – André Nicolas Feb 18 '16 at 17:59
• If $\sqrt{2}$ and $\sqrt{3}$ are in a certain field, their sum is in the field. – André Nicolas Feb 18 '16 at 20:13
• We are proving that each of the fields is a subfield of the other. The easier part is to show that $Q(\sqrt{2}+\sqrt{3})$ is a subfield of $Q(\sqrt{2},\sqrt{3}$. For clearly $\sqrt{2}$ and $\sqrt{3}$ are in the field on the right, so their sum is, so all of $Q(\sqrt{2}+\sqrt{3})$ is. – André Nicolas Feb 18 '16 at 20:35
• Why are we allowed to take the inverse? $\sqrt{2}+\sqrt{3}$ is in the field $Q(\sqrt{2}+\sqrt{3})$, so its inverse is. – André Nicolas Feb 18 '16 at 20:53

It will suffice to show $\sqrt{2}, \sqrt{3} \in \mathbb{Q(\sqrt{2}+\sqrt{3})}$ and that $\sqrt{2}+\sqrt{3} \in \mathbb{Q}(\sqrt{2}, \sqrt{3})$. The latter is obvious the former is a bit trickier.
Observe that $\sqrt2+\sqrt3\in\mathbb{Q}(\sqrt2,\sqrt3)$ so by minimality $\mathbb{Q}(\sqrt2+\sqrt3)\subseteq\mathbb{Q}(\sqrt2,\sqrt3)$. For the reverse inclusion consider $(\sqrt2+\sqrt3)^2$ for example and so on.
First we take an algebraic closure of $\mathbb{Q}$, say $\mathbb{C}$. Then $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is the smallest subfield of $\mathbb{C}$ containing $\mathbb{Q}$, $\sqrt{2}$ and $\sqrt{3}$. Similarly, $\mathbb{Q}(\sqrt{2},+\sqrt{3})$ is the smallest subfield of $\mathbb{C}$ containing $\mathbb{Q}$ and $\sqrt{2}+\sqrt{3}$. Well, since $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is a field and contains $\sqrt{2}$ and $\sqrt{3}$, the sum $\sqrt{2}+\sqrt{3}$ is in there and so $\mathbb{Q}(\sqrt{2}+\sqrt{3})\subseteq\mathbb{Q}(\sqrt{2},\sqrt{3})$. To show reverse inclusion, it suffices to show that $\sqrt{2}$ and $\sqrt{3}$ are in $\mathbb{Q}(\sqrt{2}+\sqrt{3})$ (why?). Now, $\sqrt{2}+\sqrt{3}$ is in there, so its square $5+2\sqrt{6}$ is in there, so its square $-5$= $2\sqrt{6}$ is in there. Then $2\sqrt{6}(\sqrt{2}+\sqrt{3})=2(2\sqrt{3}+3\sqrt{2})$ is in there. Subtracting $\sqrt{2}+\sqrt{3}$ from this a few times and scaling you get what you want. Play around and have fun!