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Exercise: Show that the smallest subdomain of complex numbers containing the element $\alpha=\frac{\sqrt{5} - 1}{2}$ is ${\mathbf Z}[\alpha] = \{a + b\alpha\ |\ a,\ b \in {\mathbf Z}\}$.

I thought I could first show that $\mathbf{Z}[\alpha]$ is indeed an integral domain by verifying all axioms (that's easy) and then I would show that if there's a smaller domain $N$, $x \in \mathbf{Z}[\alpha]$ exists such that $x \notin N$. This $x = a + b\alpha$ for some $a,\ b \in \mathbf{Z}$ (as it is in $\mathbf{Z}[\alpha]$). But from the axioms it follows that for integral domain $N$, we have $\mathbf{Z} \in N$ and $\alpha \in N$, so $a + b\alpha \in N$, a contradiction.

It's easy to show that ${\mathbf Z}[\alpha]$ is a commutative ring, but I have trouble showing that $a,\ b \in {\mathbf Z}[\alpha],\ a,b \neq 0 \implies ab \neq 0$.

Have I chosen a sensible way or is there a better approach?

Thanks.

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  • $\begingroup$ "being integral domain" i.e. $a,b \neq 0 \Rightarrow ab \neq 0$ is actually the ine thing zou don't need to prove - it hold for general complex nombers, so it holds for the elements of $\mathbb{Z}[\alpha]$ especially. $\endgroup$
    – Pavel Čoupek
    Oct 24, 2014 at 21:01

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A subring of a field is automatically an integral domain: zero divisors in it would be zero divisors in the field.

The set $\{a+b\alpha:a,b\in\mathbb{Z}\}$ is surely contained in any subring of $\mathbb{C}$ containing $\alpha$. So you just have to prove it's a subring.

Closure under addition and opposite is clear; moreover $0=0+0\alpha$ and $1=1+0\alpha$. For closure under multiplication: $$ (a+b\alpha)(c+d\alpha)=ac+(bc+ad)\alpha+bd\alpha^2 $$ But $$ \alpha^2=\frac{5+1-2\sqrt{5}}{4}=\frac{3-\sqrt{5}}{2}=1-\frac{\sqrt{5}-1}{2} =1-\alpha $$ so you can conclude.

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  • $\begingroup$ Oh, sorry, that's what I meant. $\endgroup$
    – David
    Oct 24, 2014 at 21:02
  • $\begingroup$ Well, that's how far I got (except I got $\alpha^2 = \alpha + 1$). But how can I tell that for nonzero $a,b$, $ab$ is nonzero? What is the essence of "subring of field is automatically integral domain"? $\endgroup$
    – David
    Oct 24, 2014 at 21:07
  • $\begingroup$ @DavidČepelík I added a couple of words. If $a,b\ne0$ in the subring, then they are still element of the field, so $ab\ne0$. $\endgroup$
    – egreg
    Oct 24, 2014 at 21:09
  • $\begingroup$ @DavidČepelík The $2$ was wrong, but $\alpha^2=1-\alpha$, not $1+\alpha$. $\endgroup$
    – egreg
    Oct 24, 2014 at 21:11
  • $\begingroup$ Right... Too late. Will fix it now. $\endgroup$
    – David
    Oct 24, 2014 at 21:12

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