Rigorously prove $\mathbb{Q}(\sqrt2,\sqrt3) = \mathbb{Q}(\sqrt2) (\sqrt3)$ As stated, I want to argue that the identity holds i.e. the smallest field containing $\sqrt2$, $\sqrt3$ and $\mathbb{Q}$ is indeed the smallest field containing $\sqrt3$ and $\mathbb{Q}(\sqrt2)$. 
 A: There is no smallest field containing a field $F$ and elements $a_1,\dotsc,a_n$. This is also not the correct definition of $F(a_1,\dotsc,a_n)$. Here is the proper definition:
Let $L/F$ be an extension and $a_1,\dotsc,a_n \in L$. Then $F(a_1,\dotsc,a_n)$ is the smallest subfield of $L$ which contains $F$ and the elements $a_1,\dotsc,a_n$. That means:
$F(a_1,\dotsc,a_n)$ is a subfield of $L$ containing $F,a_1,\dotsc,a_n$. If $E$ is another subfield of $L$ containing $F,a_1,\dotsc,a_n$, then $F(a_1,\dotsc,a_n) \subseteq E$.
This definition is quite abstract, but you can also check that the elements of $F(a_1,\dotsc,a_n)$ are just fractions of polynomial expressions in $L$ with variables from $a_1,\dotsc,a_n$and coefficients from $F$. So an example would be $\dfrac{a_1 a_2^3 - u}{ a_3 - a_1}$ (if $u \in F$ and $a_1 \neq a_3$).
Claim. If $L/F$ is a field extension and $a,b \in L$, then $F(a)(b)=F(a,b)$ as subfields of $L$.
Proof.


*

*$F(a)(b)$ contains $F(a)$. Thus, it contains $F$ and $a$. It also contains $b$. Thus, $F(a,b) \subseteq F(a)(b)$.

*$F(a,b)$ contains $F$ and $a$. Thus, $F(a) \subseteq F(a,b)$. But $F(a,b)$ also contains $b$. Thus, $F(a)(b) \subseteq F(a,b)$. $\checkmark$


In your example, $L=\mathbb{R}$ (or $\mathbb{C}$ perhaps), $F=\mathbb{Q}$, $n=2$, $a_1=\sqrt{2}$, $a_2 = \sqrt{3}$.
A: Put a=√2+√3 ; we note K=Q(√2) , L=Q(√3) , M=Q(a) and show K=M=L.
x∈M↔x=r_4
 a^4+r_3 a^3+r_2 a^2+r_1 a+r_0  ;〖  r〗_i∈Q
The calculation gives x=A√6+B√2+C√3+D where A, B, C, D are rational. Therefore x=(A√3+B) √2+(C√3+D)∈L(√2) ;  x=(A√2+C) √3+(B√2+D)∈K(√3) .
Then K=M=L since clearly K and L are contained in M.
