$[K(\beta):K] \ \deg{f_i} = [K(\alpha): K] \ \deg{g_i}$

30 Views Asked by At

Let $f$,$g$ be irreducible polynomials over the field $K$, both without multiple zeros (in an algebraic closure of $K$). Let $L=K(\alpha)$ and $E=K(\beta)$ be extensions of $K$ with $f(\alpha)=0$ and $g(\beta)=0$. Show that if $f=f_1\dots f_r$ and $g=g_1 \dots g_s$ are the prime factorization of $f$ over $K(\beta)$ and of $g$ over $K(\alpha)$, we have $r=s$, and after reordering we also have $E[X]/f_i \simeq L[X]/g_i$ for all $1 \leq i \leq r$; in particular,

$$[K(\beta):K] \ \deg{f_i} = [K(\alpha): K] \ \deg{g_i}$$

Any help for solving this question would be appreciated.