For what $c \in \mathbb{Q}$ is $\sqrt{2}+c\sqrt{3}$ a primitive element of $\mathbb{Q}(\sqrt{2},\sqrt{3})$? I am trying to find the $c\in\mathbb{Q}$ so that $\sqrt{2}+c\sqrt{3}$ is a primitive element of $\mathbb{Q}(\sqrt{2},\sqrt{3})$,  i.e. $\mathbb{Q}(\sqrt{2}+c\sqrt{3})=\mathbb{Q}(\sqrt{2},\sqrt{3})$.
In class we have already shown that $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$, so I know that c=1 is possible.
If $c=0\in\mathbb{Q}$, then we would have $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2})$, which is not possible since $\sqrt{3}\notin\mathbb{Q}(\sqrt{2})$, I am not sure if I have to prove this further.
So know considering $c\in\mathbb{Q}\setminus\{0\}$, I followed the prove for $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$:
$"\supseteq"$: Since $\sqrt{2},\sqrt{3}\in\mathbb{Q}(\sqrt{2},\sqrt{3})$, then $\sqrt{2}+c\sqrt{3}\in\mathbb{Q}(\sqrt{2},\sqrt{3})$, by the closure of $\mathbb{Q}(\sqrt{2},\sqrt{3})$.
$"\subseteq"$: Since $(\sqrt{2}+c\sqrt{3})^{-1}=\frac{1}{\sqrt{2}+c\sqrt{3}}\in\mathbb{Q}(\sqrt{2}+c\sqrt{3})$ and $\frac{\sqrt{2}-c\sqrt{3}}{\sqrt{2}-c\sqrt{3}}=1\in\mathbb{Q}$, then $\frac{1}{\sqrt{2}+c\sqrt{3}}\frac{\sqrt{2}-c\sqrt{3}}{\sqrt{2}-c\sqrt{3}}=c\sqrt{3}-\sqrt{2}\in\mathbb{Q}(\sqrt{2}+c\sqrt{3})$. Therefore, $\sqrt{2}+c\sqrt{3}+(c\sqrt{3}-\sqrt{2})=2c\sqrt{3}\in\mathbb{Q}(\sqrt{2}+c\sqrt{3})$. And since $\frac{1}{2c}\in\mathbb{Q}\subseteq\mathbb{Q}(\sqrt{2}+c\sqrt{3})$, $\sqrt{3}\in\mathbb{Q}(\sqrt{2}+c\sqrt{3})$. Using the same procedure we can show that $\sqrt{2}\in\mathbb{Q}(\sqrt{2}+c\sqrt{3})$.
Is this correct?

We need to decide whether $\theta=\sqrt{2}+c\sqrt{3}$ has degree $4$ over $\mathbb Q$.
This reduces to deciding whether $1,\theta,\theta^2,\theta^3$ are linearly independent over $\mathbb Q$.
We have
$$
\pmatrix{1 \\ \theta \\ \theta^2 \\ \theta^3}
=
\pmatrix{
1 & 0 & 0 & 0 \\
0 & 1 & c & 0 \\
2+3c^2 & 0 & 0 & 2c \\
0 & 2+9c^2 & 3+6c & 0 \\
}
\pmatrix{1 \\ \sqrt2 \\ \sqrt3 \\ \sqrt6}
$$
The determinant of the matrix is $2 c (9 c^3 - 4 c - 3)$ and so is zero for rational $c$ iff $c=0$.
This argument relies on knowing that $\mathbb{Q}(\sqrt{2},\sqrt{3})$ has degree $4$ over $\mathbb Q$ and that $1,\sqrt2,\sqrt3,\sqrt6$ is a basis.
