I need to find three fields $K_1 \leq K_2 \leq K_3$ and $\alpha, \beta \in K_3$ such that the minimal polynomial $m_{\beta}(x)$ of $\beta$ has the same degree over $K_1$ as over $K_2$ and $\deg(m_{\beta})>1$. The minimal polynomial $m_{\alpha}(x)$ of $\alpha$ must have instead a different degree over $K_1$ than over $K_2$.
I can't figure it out, especially with the condition $\deg(m_{\beta})>1$. Would appreciate some help.
$K_1=\mathbb{Q}, K_2=\mathbb{Q}(\sqrt 2), K_3=\mathbb{Q}(\sqrt 2, \sqrt 5)$. $\alpha=\sqrt{2}$ and $\beta=\sqrt{5}$. Clearly $m_{\beta}=t^2-5$ for $K_1$ and $K_2$. But $m_{\alpha}=t^2-2$ in $K_1$ and $t-\sqrt{2}$ in $K_2$