<s>[INST] 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:
$$\begin{matrix}&&\left(\Bbb Q(\sqrt{2}+\sqrt{3})=\Bbb Q(\sqrt2,\sqrt3)\right)\\\\&{\huge\diagup}&&{\huge\diagdown}\\\Bbb Q(\sqrt2)&&&&\Bbb Q(\sqrt 3)\end{matrix}$$
 [/INST] Just to add to completeness of the other answers let me find minimal polynomial of $\alpha = \sqrt 2 + \sqrt 3$ over $\mathbb Q$ directly.
\begin{align}
\alpha = \sqrt 2 + \sqrt 3 &\implies \alpha - \sqrt 2 = \sqrt 3\\
&\implies \alpha^2 -2\alpha\sqrt 2 + 2 = 3\\
&\implies \alpha^2 - 1 = 2\alpha\sqrt 2\\
&\implies \alpha^4 - 2\alpha^2 + 1 = 8\alpha^2\\
&\implies \alpha^4 - 10\alpha^2 + 1 = 0
\end{align}
Thus, $\alpha$ is a root of $p(x) = x^4 - 10x^2 + 1$, so if we show that it is irreducible over $\mathbb Q$, it will follow that it is minimal polynomial. By Gauss's lemma, since $p$ is primitive, it is enough to prove that it is irreducible over $\mathbb Z$. Assume the contrary, that $p = fg$ for non-invertible $f, g\in\mathbb Z[x]$. Neither $f$ nor $g$ can be of degree $0$ since $p$ is primitive, nor of degree $1$ since $p$ has no rational roots by Rational root theorem. So we have $\deg f = \deg g = 2$ and thus can write $$ p(x) = (x^2 + ax + b)(x^2 + cx + d),\quad a,b,c,d\in\mathbb Z $$ By expanding we get $$x^4 - 10x^2 + 1 = x^4 + (a+c)x^3 + (b+d+ac)x^2 + (ad + bc)x + bd$$ or equivalently
\begin{align}
a + c &= 0\\
b + d + ac &= -10\\
ad + bc &= 0\\
bd &= 1
\end{align}
Now, we note that $b = d = \pm 1$ and that neither cases yield integer solutions of above system and thus arrive at contradiction that $p$ is reducible.
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