-1
$\begingroup$

I know that if $F$ is a field, and $x,y$ are roots of an irreducible polynomial over $F$ lying in an algebraic extension of $F$, then $F(x) \cong F(y)$.

But it seems it is not true (in general) that if $x,y,z$ are roots of an irreducible polynomial in an extension of $F$ then $F(x,y) \cong F(x,z)$.

What might a counterexample look like? I assume I should not be looking at algebraic extensions? Or is there such an example of an algebraic extension?

  • 8
    Use F = Q (the rational numbers) and the polynomial T^4-2. Let x and y be the two real fourth roots of 2 and z be a non-real fourth root of 2. y = -x and z = ix (say), so Q(x,y) = Q(x) has degree 4 over Q and Q(x,z) = Q(x,i) has degree 8 over Q.2011-02-13
  • 20
    Commenting so that this will be visible at the top: This question is Problem 2.3 on the homework for Math 620, at the University of Buffalo. As user "Student" points out, every question user6560 has asked is a homework question from that course. http://www.math.buffalo.edu/~badzioch/MTH620/Homework_files/hw2.pdf2011-02-13
  • 0
    @David: The professor for the course should be contacted then.2011-02-13

2 Answers 2