I've though that $[\mathbb{Q}[ \sqrt2 , \sqrt3] : \mathbb{Q}] = [\mathbb{Q}[ \sqrt2] : \mathbb{Q}].[ \mathbb{Q}[\sqrt2, \sqrt3]:\mathbb{Q}[ \sqrt2] ] $ And I know how to prove $[\mathbb{Q}[ \sqrt2] : \mathbb{Q}]$ = 2, but not sure how to calculate $[ \mathbb{Q}[\sqrt2, \sqrt3]:\mathbb{Q}[ \sqrt2] ] $.
i.e. Irreducibility over $\mathbb{Q}[\sqrt2,\sqrt3]$
Hint: since $\sqrt{3}$ is a root of $x^{2}-3$ over $\mathbb{Q}(\sqrt{2})$, you need to show that this is the minimal polynomial for $\sqrt{3}$ over $\mathbb{Q}(\sqrt{2})[x]$. This amounts to showing that $\sqrt{3}$ is not the root of any linear polynomial over $\mathbb{Q}(\sqrt{2})[x]$. Suppose, then, that the minimal polynomial was linear, i.e. $\sqrt{3} \in \mathbb{Q}(\sqrt{2})$.
Then $\sqrt{3} = a+b\sqrt{2}$ for some $a, b \in \mathbb{Q}$. Can you see the contradiction here?
The other answers are fine but there is a much stronger result that one should be aware of:
Let $F$ be a field of characteristic different than $2$. Let $D_1$ and $D_2$ be elements of $F$, neither of which is a square in $F$. Then $F(\sqrt{D_1},\sqrt{D_2})$ is of degree $4$ over $F$ if $D_1D_2$ is not a square in $F$ and is of degree $2$ over $F$ otherwise.
The case when $F(\sqrt{D_1},\sqrt{D_2})$ is of degree $4$ is referred to as a biquadratic extension of the field $F$. The result above is what we shall show.
Assume $D_1D_2$ is not a square in $F$. We need show that $F(\sqrt{D_1},\sqrt{D_2})$ is of degree $4$ over $F$. First, we show that $F(\sqrt{D_1})$ is of degree $2$ over $F$. Since $D_1$ is not a square in $F$, $\sqrt{D_1} \notin F$. Suppose that $K$ is an extension of $F$ of degree $2$, $[K:F]=2$. Let $\alpha K$ such that $\alpha \notin F$. Given that $[K:F]=2$, $\alpha$ must satisfy an equation of degree $2$ over $F$ (it cannot be of degree $1$ as $\alpha \notin F$).
Therefore, the monic minimal polynomial for $\alpha$ is $$ m_{\alpha,F}(x)=x^2+bx+c $$ for some $b,c \in F$. But as $F\subset F(\alpha) \subseteq K$ and $F(\alpha)$ has degree $2$ over $F$ and $K$ is of degree $2$ over $F$ containing $F(\alpha)$, it must be the case that $K=F(\alpha)$. Since the characteristic of $F$ is not $2$, we can use the quadratic formula so that $$ \alpha=\frac{-b\pm \sqrt{b^2-4c}}{2}=-1/2b \pm 1/2 \sqrt{b^2-4c} $$ (here $a=1$ as $m_{\alpha,F}(x)$ is monic). As $\alpha$ is not a square in $F$, $b^2-4c$ is not a square in $F$. Let $\sqrt{b^2-4c}$ denote the root of $x^2-(b^2-4c)=0$ in $K$ (we choose this positive root arbitrarily).
Now we show that $F(\alpha)=F(\sqrt{b^2-4c})$. Showing $F(\alpha) \subseteq F(\sqrt{b^2-4c})$ is trivial as $\alpha$ is given by the quadratic formula above and clearly is of the form of an element of $F(\sqrt{b^2-4c})$. The fact that $\sqrt{b^2-4c}=\mp (b+2\alpha)$ shows that $\sqrt{b^2-4c}$ is the form of an element of $F(\alpha)$. Therefore, $F(\alpha)=F(\sqrt{b^2-4c})$. Therefore, any extension $K$ of $F$ of degree $2$ is of the form $F(\sqrt{D})$, where $D$ is an element of $F$ that is not a square in $F$. Furthermore, every such extension is of degree $2$.
Looking at our case, $D_1,D_2 \in F$ but $\sqrt{D_1},\sqrt{D_2} \notin F$. By above, $F(\sqrt{D_1})$ and $F(\sqrt{D_2})$ are of degree $2$ over $F$. Now observe that $$ [ F(\sqrt{D_1},\sqrt{D_2}):F]=[F(\sqrt{D_1},\sqrt{D_2}):F(\sqrt{D_2})][F(\sqrt{D_2}),F] $$ The term on the right is clearly $2$ from above. Now $[F(\sqrt{D_1},\sqrt{D_2}):F(\sqrt{D_2})]$ is $2$ if and only if $x^2-D_1$ is irreducible over $F(\sqrt{D_2})$. But since $x^2-D_1$ is a polynomial of degree $2$, this implies that it is reducible if and only if it has a linear term, meaning that $x^2-D_1$ has a root in $F(\sqrt{D_2})$. If this is the case, then $\sqrt{D_1} \in F(\sqrt{D_2})$. Assume this is the case and $\sqrt{D_1}=a+b \sqrt{D_2}$, where $a,b \in F$. Then we have $$ \begin{align} \sqrt{D_1}&=a+b \sqrt{D_2} \\ D_1 &= (a+b \sqrt{D_2})62 \\ D_1&= (a^2+b^2D_2)+2ab\sqrt{D_2} \end{align} $$ If $ab\neq 0$, then we would have $$ \sqrt{D_2}=\frac{D_1-a^2-b^2D_2}{2ab} $$ and as $D_1,D_2,a,b \in F$, $\sqrt{D_2} \in F$ contradicting the fact that $D_2$ is not a square in $F$. Because $F$ is a field (hence without zero divisors), it must be that $ab=0$ forcing $a=0$ or $b=0$. If $a=0$ then we have $$ \begin{align} D_1&= (a^2+b^2D_2)+2ab\sqrt{D_2} \\ D_1&=b^2D_2 \\ \sqrt{D_1} &= \pm b\sqrt{D_2} \end{align} $$ But since $\sqrt{D_2} \neq 0$, we have $$ \begin{align} \sqrt{D_1} &= b\sqrt{D_2} \\ \sqrt{D_1} \sqrt{D_2} &= b \sqrt{D_2} \sqrt{D_2} \\ \sqrt{D_1D_2} &= b D_2 \end{align} $$ But as $b,D_2 \in F$, this implies that $\sqrt{D_1D_2} \in F$, contradicting the fact that $D_1D_2$ is not a square in $F$. So it must be that $b=0$. Then we have $$ \begin{align} D_1&= (a^2+b^2D_2)+2ab\sqrt{D_2} \\ D_1 &= a^2 \\ \end{align} $$ But as $a,a^2 \in F$, this contradicts the fact that $D_1$ is not a square in $F$. Therefore, we have arrived at a contradiction. So $\sqrt{D_1} \notin F(\sqrt{D_2})$. This shows that $[F(\sqrt{D_1},\sqrt{D_2}):F(\sqrt{D_2})]=2$. This finally shows that $$ [ F(\sqrt{D_1},\sqrt{D_2}):F]=[F(\sqrt{D_1},\sqrt{D_2}):F(\sqrt{D_2})][F(\sqrt{D_2}),F]=2 \cdot 2 =4 $$ Now suppose that $D_1D_2$ were a square in $F$. Then by the work above, we know that $\sqrt{D_1} \in F(\sqrt{D_2})$ so that $F(\sqrt{D_1},\sqrt{D_2})$ has degree $1$ over $F(\sqrt{D_2})$. Then we have $$ [ F(\sqrt{D_1},\sqrt{D_2}):F]=[F(\sqrt{D_1},\sqrt{D_2}):F(\sqrt{D_2})][F(\sqrt{D_2}),F]=1 \cdot 2 =2 $$
This probably needs looking at for the details since I wrote this so long ago and haven't looked back at this verbiage in some time but this gives the idea of it at the very least.
