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.

Good day all, I am trying to show that, for $d \neq 0,1$ a square-free integer $(d=-1$ is allowed$)$ the space $A$ defined as below, is an integral domain: \begin{equation*} A = \{a + b\sqrt{d} : a,b \in \mathbb{Q} \} \cap \overline{\mathbb{Z}} \end{equation*}

where $\overline{\mathbb{Z}}$ is the set of all roots of monic polynomials with coefficients in $\mathbb{Z}$.

I can prove the majority of properties associated with an integral domain, except for closure under addition and non-existence of zero divisors.

Let $x_1 = a_1 + b_1\sqrt{d_1}$ and $x_2=a_2 + b_2\sqrt{d_2}$ be elements of $A$. Then: \begin{eqnarray*} x_1 + x_2 & = & a_1 + b_1\sqrt{d_1} + a_2 + b_2\sqrt{d_2} \\ & = & a_1 + a_2 + b_1\sqrt{\frac{d_1d_2}{d_2}}+b_2\sqrt{\frac{d_1d_2}{d_1}} \\ & = & a_1 + a_2 + \left(\frac{b_1}{\sqrt{d_2}} + \frac{b_2}{\sqrt{d_2}}\right)\sqrt{d_1d_2} \end{eqnarray*}

But $d_1d_2$ is not necessarily squarefree and $\left(\frac{b_1}{\sqrt{d_2}} + \frac{b_2}{\sqrt{d_2}}\right)$ is definitely not always in $\mathbb{Q}$. How do I remedy this!?

Lastly, would it be alright to state that the non-existence of zero divisors follows because $A$ is a subset of $\mathbb{C}$?

share|improve this question
1  
I have the impression that $d$ is fixed. As you are not working in e.g. a quotient of a space embedded in $\Bbb C$, it is indeed clear that $A$ has no nontrivial zero divisors. –  Lord_Farin Oct 13 '12 at 7:28
    
Thanks Andre and Farin for catching my mistake. It is clear now. –  Jack Rousseau Oct 13 '12 at 7:53

1 Answer 1

up vote 2 down vote accepted

1) The parameter $d$ is fixed, else we do not have closure under addition.

2) So there is no serious issue about sum, product being of the shape $x+y\sqrt{d}$. However, the $\overline{\mathbb{Z}}$ part needs to be dealt with.

2) Yes, the fact that we are dealing with a subset of the complex numbers is enough to ensure there are no zero-divisors.

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.