$\mathbb{Q}(\sqrt{2}, \sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$ I want to show that $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$. In my understanding, $\mathbb{Q}(\sqrt{2},\sqrt{3})=\{a\sqrt{2}+b\sqrt{3}+c:a,b,c\in\mathbb{Q} \}$, but $\mathbb{Q}(\sqrt{2}+\sqrt{3})=\{a(\sqrt{2}+\sqrt{3})+b):a,b\in\mathbb{Q} \}=\{a\sqrt{2}+a\sqrt{3}+b:a,b\in\mathbb{Q} \}$. Bu the two sets are actually different. I suspect that I'm not completely getting the notation. Also, I read that $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]=3$, but my interpretation of notation assumes that $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]=2$. Moreover, I don't see why $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$. Isn't it the case that $\mathbb{Q}(\sqrt[3]{2})=\{a+b\sqrt[3]{2}\}$?
Update: I suspect that $\sqrt{2}+\sqrt{3}$ is transcendental over $\mathbb{Q}$, so this is probably why its index is $3$. But anyway, $\sqrt[3]2$ is not transcendental over $\mathbb{Q}$.
 A: I think you are misunderstanding the definition of a field extension.
Given a field $K$, the extension $K(\alpha)$ is the smallest field containing K and $\alpha$. 
To illustrate, $\mathbb Q(\sqrt2,\sqrt3)= \{a + b\sqrt2 + c\sqrt3 + d\sqrt6  \mid a,b,c,d \in \mathbb{Q}\}$ as $\sqrt2\sqrt3 = \sqrt6$ (so that the field is closed). (Remember the definition of $[K(\alpha):K]$ is the dimension of $K(\alpha)$ as a vector space over $K$).
I guess the above should answer why $[\mathbb Q(\sqrt2,\sqrt3):\mathbb Q] = 4$ and $ [\mathbb Q(\sqrt[3]{2}):\mathbb Q] = 3$ as $\mathbb Q(\sqrt[3]{2}) = \{ a + b \sqrt[3]{2} + c \sqrt[3]{4} \mid a,b,c \in \mathbb{Q}\}$
Going back to the original question:
From the above it should be clear (noting that $\mathbb Q(\sqrt2,\sqrt3)$ is a field)  that $\mathbb Q(\sqrt2 + \sqrt3) \subset \mathbb Q(\sqrt2,\sqrt3)$
As for the reverse inclusion notice that $(\sqrt2 + \sqrt3)^2 = 5 + 2 \sqrt{6}$
so we get $\sqrt{6} \in \mathbb Q(\sqrt2 + \sqrt3)$. (again all I am using is closure)
Furthermore, $\sqrt{6}(\sqrt2 + \sqrt3) - 2(\sqrt2 + \sqrt3) = \sqrt{2} \in \mathbb{Q}(\sqrt2 + \sqrt3)$, implying $\sqrt{3} \in \mathbb Q(\sqrt2 + \sqrt3)$.
Hence $\mathbb Q(\sqrt2 + \sqrt3) = \mathbb Q(\sqrt2, \sqrt3)$
