I want to find $[L=\mathbb{Q}(\sqrt{2}, \sqrt[3]{3}):K=\mathbb{Q}]_s$.
In lecture I learned that for every algebraic closure $\iota_a:K\rightarrow K^a$ it's true that...
\begin{align*} [L:K]_s = \#\{\text{different roots of } \iota^*_a (m_x)\} \end{align*} where $\iota^*_a:K[X]\rightarrow K^a[X]$ and $m_x$ is the minimal polynomial for $x$. I took $K^a = \mathbb{A}$ (field of all algebraic numbers)
I have for \begin{align*} &m_{\sqrt{2}} = X^2-2 =(X-\sqrt{2})(X+\sqrt{2}) \quad \text{ and } \\ &m_{\sqrt[3]{3}} = X^3-3 = (X-\sqrt[3]{3})(X-\sqrt[3]{3}\cdot e^{\frac{2\pi i}{3}})(X-\sqrt[3]{3}\cdot e^{\frac{4\pi i}{3}}) \end{align*}
These roots are different from each other and are all in $\mathbb{A}$. Therefore,
\begin{align*} [L:K]_s =[L:\mathbb{Q}(\sqrt{2})]_s \cdot [\mathbb{Q}(\sqrt{2}):K]_s = 3\cdot 2 = 6 \end{align*}
Am I correct?