# Galois Theory: Morphisms

by wattsup03
Tags: galois, morphisms, theory
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,348 $(2^{1/4}^2= 2^{1/2}$ and $(2^{1/4}^3=2^{3/4}$ are irrational but $(2^{1/4})^4= 2$ is rational so any number in [/itex]Q(2^{1/4})[/itex] is of the form $a+ b2^{1/4}+ c2^{1/2}+ d2^{3/4}$ for rational numbers a, b, c, d. For any automorphism, f, $f(a+ b2^{1/4}+ c2^{1/2}+ d2^{3/4})= a + bf(2^{1/4})+ cf(2^{1/2})+ df(2^{3/4})$ so the possible values of f depend entirely upon the possible values of $f(2^{1/4})$, $f(2^{1/2})$ and $f(2^{3/4})$. Further, $(f(2^{1/4})^4= f(2)$ is rational so $f(2^{1/4}$ must be a fourth root of 2. Also, $f(2^{1/2})^2= f(2)$ so $f(2^{1/2}) must be [itex]2^{1/2} (or [itex]-2^{1/2}$ which is just a rational number times $2^{1/2}. Each f permutes the fourth roots of 1 while fixing [itex]2^{1/2}$.