Prove $\mathbb Q (\sqrt2 + \sqrt3 ) = \mathbb Q (\sqrt2 , \sqrt3 )$ 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.
 A: 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.
A: 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.
A: I myself had this problem a while ago, but this is because I didn't quite get the definitions. Instead of trying to explicitly describe these sets, remember the definitions.
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!
EDIT: André Nicolas gave a way simpler argument for showing the "hard part" in the comments. Anyways, it is always good to play around and discover things for yourself
