Transitivity of Algebraic Field Extensions (explicit) Transitivity of Algebraic Field Extensions Here we get a proof of Transitivity of Algebraic Field Extensions. But can we find explicitly a polynomial $p(x)$ with coefficients from $F$ such that $p(\alpha) = 0$, where $\alpha \in K$.

$\
ewcommand{\Q}{\mathbb{Q}}$$\
ewcommand{\Size}[1]{\left\lvert #1 \right\rvert}$In the notation of the linked post, assume the degrees $$ \Size{K:E} = m, \Size{E:F} = n $$ are finite, and you have bases $$ \beta_{1}, \dots, \beta_{m}, \gamma_{1}, \dots, \gamma_{n} $$ of the two extensions, so that the $\beta_{i} \gamma_{j}$ form a basis for $K/F$.

If $A$ is the matrix of the $F$-linear map $x \mapsto x \alpha$ on $K$, then the minimal polynomial of $A$ will be the same as the characteristic polyomial of $A$ will be the same as the minimal polynomial of $\alpha$ over $F$.

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As an example, if $F = \Q$, $E = \Q(\sqrt{2}), K = E(\sqrt{3})$, then the matrix $A$ for $\alpha = \sqrt{2} + \sqrt{3}$ with respect to the basis $1, \sqrt{2}, \sqrt{3}, \sqrt{6}$ will be (my vectors are row vectors) $$ \begin{bmatrix} 0 & 1 & 1 & 0\\\ 2 & 0 & 0 & 1\\\ 3 & 0 & 0 & 1\\\ 0 & 3 & 2 & 0\\\ \end{bmatrix} $$ with characteristic polynomial $x^{4} - 10 x^{2} + 1$.