# 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 $\alpha \in K$.

$\newcommand{\Q}{\mathbb{Q}}$$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}In the notation of the linked post, assume the degrees$$ \Size{K:E} = m, \Size{E:F} = n $$are finite, and you have bases$$ \beta_{1}, \dots, \beta_{m}, \gamma_{1}, \dots, \gamma_{n} $$of the two extensions, so that the \beta_{i} \gamma_{j} form a basis for K/F. If A is the matrix of the F-linear map x \mapsto x \alpha on K, then the minimal polynomial of A will be the same as the characteristic polyomial of A will be the same as the minimal polynomial of \alpha over F. As an example, if F = \Q, E = \Q(\sqrt{2}), K = E(\sqrt{3}), then the matrix A for \alpha = \sqrt{2} + \sqrt{3} with respect to the basis 1, \sqrt{2}, \sqrt{3}, \sqrt{6} will be (my vectors are row vectors)$$ \begin{bmatrix} 0 & 1 & 1 & 0\\ 2 & 0 & 0 & 1\\ 3 & 0 & 0 & 1\\ 0 & 3 & 2 & 0\\ \end{bmatrix} $$with characteristic polynomial x^{4} - 10 x^{2} + 1. • Do you have an idea for proving that A = P a P^{-1} for some invertible matrix P, where A is the matrix you obtain and a is mine ? – reuns Jan 19 '17 at 11:18 If f is an irreducible polynomial of F[x], then let A be its companion matrix. Note A is a root of f, and the matrix field F(A) is isomorphic to F(\alpha) \simeq F[X]/(f(X)) where \alpha is any root of f. Hence by induction, any algebraic extension E/F is isomorphic to a matrix field : E \simeq F(A_1,\ldots A_m). and if K/E/F is a tower of algebraic extensions, then K \simeq F(A_1,\ldots A_l) for some (invertible) matrices A_1,\ldots,A_l \in F^{n\times n} where n = [K:F]. Ok so now we think to K as a matrix field over F, and guess what, for any a \in F(A_1,\ldots A_l) we have the field norm :$$N_{K/F}(a) = \det(a) \in F$$which is clearly an injective morphism of multiplicative groups K^\times \to F^\times, and as usual with the polynomial$$g(x) = \det(a-x I) \in F[x]$$then$a$is a root of$g$(see the Caley-Hamilton theorem). Hence$a$is algebraic over$F$, its minimal polynomial being one of the irreducible factors of$g\$.