I'm attempting to solve exercise 5 from chapter 3 in Atiyah-McDonald's text on commutative algebra; in particular, it asks if 'being an integral domain' is a local property. The counter-example I found is the following, and was interested in asking whether it made any sense.
Consider the ring $A=\mathbb{Q}[x,y]/(xy)$; evidently, $A$ is not an integral domain, as it is a quotient of $\mathbb{Q}[x,y]$ by an ideal which is not prime. Because prime ideals in $A$ are such that their contraction in $\mathbb{Q}[x,y]$ is a prime ideal that contains $(xy)$, the only possible prime ideals in $A$ are $p=(x)/(xy)$ and $q=(y)/(xy)$; one can verify directly that these are both in fact prime.
My intuition behind this exercise is that $A_p$ should be an integral domain, because the expression $\frac{x+(xy)}{1}\cdot\frac{y+(xy)}{1}=0$ fails to be a product of two non zero elements in $A_p$, seeing as $\frac{x+(xy)}{1}=0$ in the localization, becuase $y+(xy)\in (A-p)$ and $(x+(xy))(y+(xy))=0$.
I'm not quite sure if this observation is enough to assert that $A_p$ (and hence, by symmetry, $A_q$ as well) is an integral domain. I would greately appreciate any advice or suggestion which may help, thanks a lot!
Your intuition that $A_p$ and $A_q$ are domains is correct - a little more pushing will show that $A_p\cong k[x]_{(x)}$. Unfortunately, your claim that $p$ and $q$ are the only primes is false: $(x,y)$ is a prime ideal in $A$ which is not of the form you claim (there are many more).