Proving that $\mathbb{Q}[\sqrt{2} + \sqrt{3}] = \mathbb{Q}[\sqrt{2},\sqrt{3}].$ Here is the question I want to answer:
Find polynomials $f(x), g(x) \in \mathbb{Q}[x]$ such that $\sqrt{2} = f(\sqrt{2} + \sqrt{3})$ and $\sqrt{3} = g(\sqrt{2} + \sqrt{3}).$ Deduce the equality of fields: $\mathbb{Q}[\sqrt{2} + \sqrt{3}] = \mathbb{Q}[\sqrt{2},\sqrt{3}].$
Here is what I managed to prove so far:
I managed to write $\sqrt{2}$ as a function in $\sqrt{2} + \sqrt{3}$ and $\sqrt{3}$ as a function in $\sqrt{2} + \sqrt{3}$ as can seen below:
$\textbf{Finding f(x).}$
Since $(\sqrt{2} + \sqrt{3})^2 = 5 + 2 \sqrt{2.3}.$ Then we have that \begin{align*}
(\sqrt{2} + \sqrt{3})^2 (\sqrt{2} + \sqrt{3}) &= (5 + 2 \sqrt{2.3})  (\sqrt{2} + \sqrt{3} )\\
&= 5 \sqrt{2} + 5 \sqrt{3} + 4 \sqrt{3} + 6 \sqrt{2} \\
&= 11 \sqrt{2} + 9 \sqrt{3}  \\
&= 9 (\sqrt{2} + \sqrt{3}) +  2 \sqrt{2}
\end{align*}
Then if we put $x = \sqrt{2} + \sqrt{3},$ we will get that $$x^3 = 9x + 2 \sqrt{2}.$$ And so we have that $$\sqrt{2} = \frac{x^3}{2} - \frac{9x}{2} = f(x).$$\
$\textbf{Finding g(x).}$
\begin{align*}
(\sqrt{2} + \sqrt{3})^3  
&= 11 \sqrt{2} +  9 \sqrt{3}\\
&= 11 \sqrt{2} +  9 \sqrt{3} + 2 \sqrt{3} -  2 \sqrt{3}\\
&= 11 (\sqrt{2} + \sqrt{3}) -  2 \sqrt{3}
\end{align*}
Then if we put $x = \sqrt{2} + \sqrt{3},$ we will get that $$x^3 = 11x - 2 \sqrt{3}.$$ And so we have that $$\sqrt{3} = \frac{- x^3}{2} + \frac{11x}{2} = g(x).$$\
And I understand that this proves this inclusion $\mathbb{Q}[\sqrt{2},\sqrt{3}] \subseteq \mathbb{Q}[\sqrt{2} + \sqrt{3}] $
My question is how can I prove the other inclusion $\mathbb Q(\sqrt2+\sqrt3)\subseteq \mathbb Q(\sqrt2,\sqrt3)$? could anyone help me in showing this please pointing out for me what exactly the definition of  $\mathbb Q(\sqrt2,\sqrt3)$ for me.
 A: You are asking about definition for  $\Bbb{Q}(\sqrt{2},\sqrt{3})$.
Let me add some more.
Suppose $F$ is a field of a subfield $K$ and $\alpha$ is an element of $K$. Then the  collection of all subfields  of $K$ containing both $F$ and $\alpha$ is nonempty(why!).
Since the intersection of subfields is again a subfield, it implies that there is a  unique smallest subfiles containing both  $F$ and $\alpha$. Similarly, you can replace $\alpha$ by a collection $\alpha,\beta,\dots$ of elements of $K$.
We conclude a definition from above,
Definition Let $K$ be an extension of the filed $F$ and let  $a,b,c,\dots\in K$ be collection of elements in $K$. Then the smallest field containing both $F$ and the elements $a,b,c\dots$ denoted $F(a,b,c,...)$ is called the field generated by $a,b,c\dots$ over $F$.
Clearly $\Bbb{Q}(\sqrt{2},\sqrt{3})$ is the smallest field containing $\Bbb{Q},\sqrt{2}$ and $\sqrt{3}$.
i.e., $\sqrt{2},\sqrt{3}\in \Bbb{Q}(\sqrt{2},\sqrt{3})$.
Note that $\Bbb{Q}(\sqrt{2},\sqrt{3})$ is a field. So by the closure property w.r.t. addition,
$\sqrt{2}+\sqrt{3}\in\Bbb{Q}(\sqrt{2},\sqrt{3})$
$\Rightarrow \Bbb{Q}(\sqrt{2}+\sqrt{3})\subseteq \Bbb{Q}(\sqrt{2},\sqrt{3})$
