Is i in the Field Generated by α Where α^3 + α + 1 = 0?

  • Thread starter Thread starter antiemptyv
  • Start date Start date
  • Tags Tags
    Field
antiemptyv
Messages
31
Reaction score
0

Homework Statement



Is i \in \mathbb{Q}(\alpha), where \alpha^3 + \alpha + 1 = 0?

Homework Equations



The Attempt at a Solution



Suppose i \in \mathbb{Q}(\alpha). Then the field \mathbb{Q}(i) generated by the elements of \mathbb{Q} and i is an intermediate field, i.e.

\mathbb{Q} \subset \mathbb{Q}(i) \subset \mathbb{Q}(\alpha).

But the degree [\mathbb{Q}(i):\mathbb{Q}] = 2 does not divide the degree [\mathbb{Q}(\alpha):\mathbb{Q}] = 3, so i \notin \mathbb{Q}(\alpha).

Is that right?
 
Last edited:
Physics news on Phys.org
Sounds good. (You did verify that your polynomial is irreducible, right?)
 
Great. Ahh, yes, that would certainly need to be shown. Thanks.

I just wanted to make sure I'm getting these basic ideas down correctly, and not missing something completely obvious. We're just beginning Galois theory, and I'm using a couple supplementary texts because the one we use in class (Algebra, Michael Artin) is a bit tough for a first exposure to this stuff. It's great, though, after you've got a good handle on things. Thanks again!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top