I want know if my logic in showing that $\Bbb Q(\sqrt 2,\sqrt 3) = \Bbb Q (\sqrt 2 + \sqrt 3)$ is correct I want to know if my logic in showing that $\Bbb Q(\sqrt 2,\sqrt 3) = \Bbb Q (\sqrt 2 + \sqrt 3)$ is correct
Now firstly, it seems that $\Bbb Q(\sqrt 2,\sqrt 3)=(\Bbb  Q (\sqrt 2))(\sqrt3){\overset{{\huge\color\red?}}{=}}\{a+b\sqrt2+c\sqrt 3:a,b,c\in \Bbb Q\}$
But from my last question it seems that we need to check this for multiplicative closure, $(a+b\sqrt2 + c\sqrt3)(d+e\sqrt 2 + f\sqrt 3)=ad+ae\sqrt 2+ af\sqrt 3 + 2eb+bf\sqrt 3+cd\sqrt 3+ ce\sqrt2\sqrt3+ 3cf$

So we actually find I believe that:
$\Bbb Q(\sqrt2,\sqrt3)=\{a+b\sqrt 2+c\sqrt 3+ d\sqrt2\sqrt3:a,b,c,d\in \Bbb Q\}$?


I believe that $\Bbb Q(\sqrt 2 +\sqrt 3)\overset{\huge{\color\red ?}}=\{a+b(\sqrt 2+ \sqrt 3):a,b\in \Bbb Q\}$
So let's verify multiplicative closure $(a+b\sqrt 2+ b\sqrt 3)(c+d\sqrt 2+ d\sqrt 3)$
$$=ac+ad\sqrt 2+ad\sqrt 3+bc\sqrt2+2bd+bd\sqrt3+bc\sqrt3+bd\sqrt2\sqrt3+3bd$$
$$=(ac+3bd+2bd)+(ad+bc)\sqrt 2 + (ad+bd+bc)\sqrt 3 +(bd)\sqrt2\sqrt3$$

Since $\sqrt2$ and $\sqrt 3$ and $\sqrt2\sqrt3$ don't share common coefficients, they are linearly independent and hence $\Bbb Q(\sqrt2+\sqrt 3)=\Bbb Q(\sqrt2,\sqrt3)$(from above, i.e we have deduced that:) $\Bbb Q(\sqrt{2}+\sqrt{3})=\{a+b\sqrt2+c\sqrt3:a,b,c\in\Bbb Q\}$.


Is that the correct way to show this? Also this means we have:
$$\begin{matrix}&&\left(\Bbb Q(\sqrt{2}+\sqrt{3})=\Bbb Q(\sqrt2,\sqrt3)\right)\\\\&{\huge\diagup}&&{\huge\diagdown}\\\Bbb Q(\sqrt2)&&&&\Bbb Q(\sqrt 3)\end{matrix}$$

The "usual" proof that $\Bbb Q(\sqrt{2},\sqrt{3}) \subseteq \Bbb Q(\sqrt{2}+ \sqrt{3})$ ), the other inclusion is obvious.
It suffices to show that $\sqrt{2} \in \Bbb Q(\sqrt{2}+ \sqrt{3})$, since then $\sqrt{3} = (\sqrt{2} + \sqrt{3}) - \sqrt{2}$ must be as well.
Now $(\sqrt{2} + \sqrt{3})^3 = 2\sqrt{2} + 6\sqrt{3} + 9\sqrt{2} + 3\sqrt{3}$
$= 11\sqrt{2} + 9\sqrt{3}$.
Hence $\dfrac{1}{2}[(\sqrt{2} + \sqrt{3})^3 - 9(\sqrt{2} + \sqrt{3})] = \sqrt{2}$, QED.
