# Is there a more direct way of proving that this ring is an integral domain?

In self studying abstract algebra and I've come upon the following problem which I could not solve directly. For any $d\in \mathbb{Z}$ we are asked to show that $\mathbb{Z}[\sqrt d]=\{a+b\sqrt{d} ~\colon a,b\in \mathbb{Z} \}$ is an integral domain. This reduces to showing that the ring has no zero divisors which I tried to do through algebraic manipulation.

I have seen two solutions to this problem that use higher level techniques. One defines conjugate and norm and proves the claim using simple consequences of these definitions. The other says that since the given ring is a subring of the field $\mathbb{C}$ it can have no zero divisors.

I don't like either approach because the first uses ideas that haven't been introduced in the text and isn't a natural place to start (for me). The second, while both true and simple, is not satisfying to me. That $\mathbb{C}$ is a field has not been established and assuming it just makes the problem trivial.

My question is whether or not naive tactics like algebraic manipulation are capable of demonstrating the result. Am I lacking in cleverness or is such an approach impossible?

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But you can use a simpler form of the second approach - see my answer. – Bill Dubuque Aug 15 '14 at 1:13

If $d$ is the square of an integer, then the result follows from the fact that $\mathbb{Z}$ is an integral domain.
So suppose that $d$ is not the square of an integer. Let $(x+y\sqrt{d})(u+v\sqrt{d})=0$. Multiplying by $(x-y\sqrt{d})(u-v\sqrt{d})$, we get $(x^2-dy^2)(u^2-dv^2)=0$. Thus one of the terms is $0$, say $x^2-dy^2$. If $d$ is negative, this immediately forces $x=y=0$. If $d$ is positive, we conclude that $x=y=0$ by the usual proof that $\sqrt{d}$ is irrational.
$\Bbb Z[\sqrt{d}]\,$ is a domain since it is a subring of $\,\Bbb Q[\sqrt d],\,$ which is a field, by rationalizing denominators $\,\alpha\ne 0\,\Rightarrow \dfrac{1}\alpha = \dfrac{\bar\alpha}{\alpha\bar\alpha},\$ i.e $\ \dfrac{1}{a\!+\!b\sqrt d}\, =\, \dfrac{a\!-\!b\sqrt{d}}{a^2\!-\!b^2 d\ }^{\phantom{I^I}}\!\!\!\!\in\Bbb Q[\sqrt d].\,\$ Note $\ a^2\!-b^2d\ne 0\$ (else $\,\sqrt d\in \Bbb Q)$