Let $A \subseteq B \subseteq C$ be fields and let $\alpha$, $\beta$, $\gamma$ be such that $A(\alpha) = B$, $B(\beta) = C$, $A(\gamma) = C$. Assume $B$ and $C$ have finite degree over $A$.
Let $m(\alpha,A)$ be the minimal polynomial of $\alpha$ over $A$, let $m(\beta,B)$ be the minimal polynomial of $\beta$ over $B$, let $m(\beta,A)$ be the minimal polynomial of $\beta$ over $A$, and let $m(\gamma,A)$ be the minimal polynomial of $\gamma$ over $A$.
Let $c \in C$. We have, on the one hand, $ c=\sum_{i = 1}^{[C:B]} b_i \beta^{i-1}, \quad b_i \in B $ $ b_i=\sum_{j = 1}^{[B:A]} a_{ij} \alpha^{j-1}, \quad a_{ij} \in A $ and on the other hand $ c=\sum_{k = 1}^{[C:A]} c_k \gamma^{k-1}, \quad c_k \in A $ We also have $ \gamma = \sum_{i = 1}^{[C:B]} d_i \beta^{i-1}, \quad d_i \in B $ $ d_i=\sum_{j = 1}^{[B:A]} e_{ij} \alpha^{j-1}, \quad e_{ij} \in A $ If the polynomials $m(\alpha,A)$, $m(\beta,B)$, $m(\beta,A)$, $m(\gamma,A)$ and the numbers $a_{ij}$, $e_{ij}$ are known explicitly, how can I calculate the $c_i$?
ADDED: Now that Joriki and Jyriki have helped me to formulate the problem correctly, I see the solution is not so hard. Since $ \gamma = \sum_{i = 1}^{[C:B]} \sum_{j = 1}^{[B:A]} e_{ij} \alpha^{j-1}\beta^{i-1} $ we can find numbers $f_{ijk}$ such that $ \gamma^{k-1} = \sum_{i = 1}^{[C:B]} \sum_{j = 1}^{[B:A]} f_{ijk} \alpha^{j-1}\beta^{i-1} $ by using the polynomials $m(\alpha,A)$, $m(\beta,B)$ to express large powers of $\alpha$ and $\beta$ in terms of smaller powers. Then $ c = \sum_{i = 1}^{[C:B]} \sum_{j = 1}^{[B:A]} a_{ij} \alpha^{j-1}\beta^{i-1} $ and also $ c = \sum_{k=1}^{[C:A]} c_k \left( \sum_{i = 1}^{[C:B]} \sum_{j = 1}^{[B:A]} f_{ijk} \alpha^{j-1}\beta^{i-1} \right) = \sum_{i = 1}^{[C:B]} \sum_{j = 1}^{[B:A]} \left( \sum_{k=1}^{[C:A]} c_k f_{ijk} \right) \alpha^{j-1}\beta^{i-1} $ Therefore $ a_{ij} = \sum_{k=1}^{[C:A]} c_k f_{ijk} $ So we are reduced to solving this system of system $[C:A] = [C:B] \cdot [B:A]$ linear equations for the $c_k$.