I have been confusing myself a lot with the following and I am sure I must be missing something obvious, so sorry for this probably stupid question.
Given $\alpha = \sqrt{2} - \sqrt{3}$ and $\beta = \sqrt{3} - \sqrt{5}$, we extend the rational numbers to $\mathbb{Q}(\alpha,\beta) = \mathbb{Q}(\alpha)(\beta)$. We determine the minimal polynomial of $\alpha$ over $\mathbb{Q}$ to be $x^4 - 10x^2 + 1$ and the minimal polynomial of $\beta$ over $\mathbb{Q}(\alpha)$ to be $x^4-16x^2+4$. Hence $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)] = 4$ and $[\mathbb{Q}(\alpha):\mathbb{Q}] = 4$ and so $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}] = 16$. We can now construct the following basis for $\mathbb{Q}(\alpha,\beta)$ over $\mathbb{Q}$:
$\mathcal{B} = \{1, \alpha, \alpha^2, \alpha^3, \beta, \alpha\beta, \alpha^2\beta, \alpha^3\beta, \beta^2, \alpha\beta^2, \alpha^2\beta^2, \alpha^3\beta^2, \beta^3, \alpha\beta^3, \alpha^2\beta^3, \alpha^3\beta^3\}$
(I hope this is correct so far.)
The roots of the minimal polynomial of $\alpha$ are
$\alpha = \alpha_1 = \sqrt{2} - \sqrt{3}$
$\alpha_2 = -\sqrt{2} - \sqrt{3}$
$\alpha_3 = -\sqrt{2} + \sqrt{3}$
$\alpha_4 = \sqrt{2} + \sqrt{3}$
The roots of the minimal polynomial of $\beta$ are
$\beta = \beta_1 = \sqrt{3} - \sqrt{5}$
$\beta_2 = -\sqrt{3} - \sqrt{5}$
$\beta_3 = -\sqrt{3} + \sqrt{5}$
$\beta_4 = \sqrt{3} + \sqrt{5}$
Now we can choose some primitive element $\gamma = \alpha + c\beta$ , where $c \neq \frac{\alpha_i - \alpha_1}{\beta_j + \beta_1}$ ($i = 1, 2, 3, 4$ and $j = 1, 2, 3, 4$). We choose $c = -1$ and get $\gamma = \sqrt{2}-2\sqrt{3}+\sqrt{5}$ and $\mathbb{Q}(\gamma) = \mathbb{Q}(\alpha, \beta)$.
(Now my confusion comes in.)
We can now determine the minimal polynomial of $\gamma$ over $\mathbb{Q}$ to be $x^8 - 76x^6 + 1414x^4 - 6540x^2 + 225$. This means that $[\mathbb{Q}(\gamma):\mathbb{Q}] = 8$. (Contradicting the previous result that $[\mathbb{Q}(\alpha, \beta):\mathbb{Q}] = 16$.)
There must be something really obvious I am missing. Could anybody please point out my mistake?
