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 $\alpha \in K$.

up vote 3 down vote accepted

$\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 $g$.

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