# Conjoining elements to fields which are the roots of irreducible polynomials

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?

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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. – KCd Feb 13 '11 at 9:15
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. math.buffalo.edu/~badzioch/MTH620/Homework_files/hw2.pdf – David Speyer Feb 13 '11 at 19:26
@David: The professor for the course should be contacted then. – Arturo Magidin Feb 13 '11 at 19:50

Let $P$ be an irreducible polynomial in $F[X]$ and pick $x,y,z$ three roots of $P$ in an extension.

You do have $F(x) \cong F[X]/P \cong F(y)$.

If you want to go further, you have to look at the factorization of $P$ in $F(x)[X]$ : $P(X) = (X-x)Q(x)$ where $Q$ has coefficients in $F(x)$.

If $Q$ is irreducible, then $F(x,y) = F(x)(y) \cong F(x)[X]/Q \cong F(x)(z) = F(x,z)$.

If $Q$ factorizes further in $F(x)[X]$, the minimal polynomials of $y$ and $z$ on the field $F(x)$ can be different, and then $F(x,y) \not \cong F(x,z)$.

In KCd's example, $P(X) = X^4-2 = (X-x)(X^3+xX^2+x^2X+x^3) = (X-x)(X+x)(X^2+x^2)$. You can see that if you know one root $x$, then the others are $-x, ix, -ix$, and $\mathbb{Q}(x,-x) \not \cong \mathbb{Q}(x,ix)$.

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The problem is, although p(x) is irreducible over F, it might not be so over F(x), and hence F(x,y) and F(x,z) might not be isomorphic.
I hope you will accept this intuitive explanation as a supplement of the comment above by@KCd.

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