Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
7  
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
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. 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

3 Answers 3

Please note that every question USER6560 has asked has come from the HW assigned in MATH 620 at university at buffalo. The link is http://www.math.buffalo.edu/~badzioch/MTH620/Homework.html

USER6560: stop asking questions from the HW.

share|improve this answer
    
Contact the professor teaching the class to alert him/her. –  Arturo Magidin Feb 13 '11 at 19:50
    
i have done so, thank you. –  student Feb 13 '11 at 19:54
    
I was not looking for direct answers. I have received hints in the past, and I have not copied anything. I used whatever hints or in one case an answer to learn about the question and nothing was plagiarized. –  user6560 Feb 13 '11 at 21:07
4  
@user6560: And if you used the hints or answers but did not explicitly acknowledge them, then that would be considered plagiarism even if you did not copy it word for word: Buffalo defines plagiarsm as "Copying or receiving material from any source and submitting that material as one’s own, without acknowledging and citing the particular debts to the source (quotations, paraphrases, basic ideas), or in any other manner representing the work of another as one’s own." (Undergraduate Catalog 2008-2010, page 14), undergrad-catalog.buffalo.edu/archive/0910/pdf/… –  Arturo Magidin Feb 13 '11 at 21:32
7  
@user6560: The policies of the course you are taking are something you are to take up with your professor, not with us. When I see a problem labeled "homework", then I treat my answer as I would treat one of my students coming to see me during office hours with questions for that problem, giving what I consider to be appropriate help; this is different from the kind of answer I give a colleague, or a student who is just wondering about something that is not part of an assignment. When I was a student, if a friend helped me, or if I got an answer from a book, I cited it or indicated this. –  Arturo Magidin Feb 13 '11 at 22:46

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 $\mathbb{Q}(x,-x) \not \cong \mathbb{Q}(x,ix)$.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.