Is $\mathbb Q_p(a_2)=\mathbb Q_p(b_2)$?

Question

Consider the $p$-adic field $\mathbb Q_p$.

Suppose $a_1,b_1$ be two algebraic numbers of $\mathbb Q_p$ such that $\mathbb Q_p(a_1)=\mathbb Q_p(b_1).$

Now let $a_2, b_2$ are algebraic numbers of $\mathbb Q_p(a_1)$ or $\mathbb Q_p(b_1)$ such that \begin{align} &\mathbb Q_p(a_2) \subset \mathbb Q_p(a_1,b_2) \\ & \mathbb Q_p(b_2) \subset \mathbb Q_p(b_1,a_2) \end{align}

Can we say $\mathbb Q_p(a_2)=\mathbb Q_p(b_2)$ ?

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I think if $a_1,b_1$ are $n^{th}$ roots of $-1$, then the answer is yes.

In the other cases, I am not sure.

Thanks