Proving each nonzero element of a subfield of C has an inverse

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Homework Statement



Let S={p+qα+rα2 : p, q, r \in \mathbb{Q}}, where α=\sqrt[3]{2}. Then S is a subfield of \mathbb{C}. Prove that each nonzero element of S has a multiplicative inverse in S.

The Attempt at a Solution



Let p, q, r\in\mathbb{Q} such that not all of p, q, r are zero. If each nonzero element of S has a multiplicative inverse in S, then there are rational numbers a, b, c so that

(p+qα+rα2)(a+bα+cα2)=1

Expanding and using the fact that α3=2 and α4=2α, we get:

(pa+2qc+2rb)+α(pb+qa+2rc)+α2(pc+qb+ra)=1

Since 1, α, and α2 are linearly independent we may equate coefficients to give a system of equations:

pa+2qc+2rb=1
pb+qa+2rc=0
pc+qb+ra=0

Now to finish the problem I want to show that this system always has a solution in \mathbb{Q}. I tried using Cramer's rule (i.e. showing the determinant of the matrix whose entries are the coefficients of p, q, r is equal to zero only if p=q=r=0), but I had no luck and it just turned into a mess. Likewise, back substitution was a mess. Can anyone give me a hand on how to complete this problem?

Thanks!
 
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MissMoneypenny said:

Homework Statement



Let S={p+qα+rα2 : p, q, r \in \mathbb{Q}}, where α=\sqrt[3]{2}. Then S is a subfield of \mathbb{C}. Prove that each nonzero element of S has a multiplicative inverse in S.

The Attempt at a Solution



Let p, q, r\in\mathbb{Q} such that not all of p, q, r are zero. If each nonzero element of S has a multiplicative inverse in S, then there are rational numbers a, b, c so that

(p+qα+rα2)(a+bα+cα2)=1

Expanding and using the fact that α3=2 and α4=2α, we get:

(pa+2qc+2rb)+α(pb+qa+2rc)+α2(pc+qb+ra)=1

Since 1, α, and α2 are linearly independent we may equate coefficients to give a system of equations:

pa+2qc+2rb=1
pb+qa+2rc=0
pc+qb+ra=0

Now to finish the problem I want to show that this system always has a solution in \mathbb{Q}. I tried using Cramer's rule (i.e. showing the determinant of the matrix whose entries are the coefficients of p, q, r is equal to zero only if p=q=r=0), but I had no luck and it just turned into a mess. Likewise, back substitution was a mess. Can anyone give me a hand on how to complete this problem?

Thanks!

If ##x, y \in S## have multiplicative inverses in ##S## and if ##x + y \neq 0##, can you prove that ##x+y## has a multiplicative inverse in ##S##? Can you prove that ##1, \alpha, \alpha^2## all have multiplicative inverses in ##S##?
 
Thanks for the quick response! I think I see what you're getting at. If I can prove what you wrote, then since each element of S is a linear combination of 1, α, and α2 whose coefficients are in \mathbb{Q}, it follows that each element of S has an inverse in S. Unfortunately I really am not sure how to show that if x, y, x^{-1}, y^{-1} \in S with x \ne y, then (x+y)^{-1} \in S. I tried to solve the equation (x+y)(ax^{-1}+bx^{-1})=1 for a and b, but that didn't get me anywhere, and I'm not sure what else to try. Is it possible for you to give me a small hint to get me going in the right direction?

Thanks again.
 
Was that really the original statement of the problem? Saying "S is a subfield of mathbb{C}. Prove that each nonzero element of S has a multiplicative inverse in S" seems very strange! A subfield is, by definition, a field so each non-zero element must have a multiplicative inverse. It would make more sense to first define S then ask to show that every nonzero element has a multiplicative inverse, as part of proving that S is a subfield.
 
You're correct, what I wrote is a strange problem statement. It was my fault. The full problem was to show S is a subfield of C. However, I knew how to prove all of the other field axioms hold for S, so I changed the problem statement. As you correctly pointed out, I didn't change it very well. Would it be better if I wrote "confirm that each nonzero element of S has an identity"? Or should I just have written up the actual problem statement but mentioned I only need help proving the existence of multiplicative inverses?
 
MissMoneypenny said:
Thanks for the quick response! I think I see what you're getting at. If I can prove what you wrote, then since each element of S is a linear combination of 1, α, and α2 whose coefficients are in \mathbb{Q}, it follows that each element of S has an inverse in S. Unfortunately I really am not sure how to show that if x, y, x^{-1}, y^{-1} \in S with x \ne y, then (x+y)^{-1} \in S. I tried to solve the equation (x+y)(ax^{-1}+bx^{-1})=1 for a and b, but that didn't get me anywhere, and I'm not sure what else to try. Is it possible for you to give me a small hint to get me going in the right direction?

Thanks again.

Many of these issues are dealt with in
http://math.stackexchange.com/quest...proof-that-the-algebraic-numbers-form-a-field

If you Google 'algebraic numbers' or 'algebraic number fields' you will encounter loads of material on your topic.
 
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...

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