An affine scheme is integral if and only if it is connected and all stalks are integral domains.

Question

Let X=Spec A. So far I have shown the easy side, i.e, if the affine scheme is integral, then it's irreducible and so connected. Since all stalks are localization of A at prime ideals, we have that all stalks are integral domains since A is an integral domain. I am having trouble proving the other way. I'm trying to prove that A is an integral domain by looking at A as the ring of global sections of X. So let f,g be two global sections of X such that fg=0. Then f_x.g_x=0(product of stalks) for all x. Then the points where the stalks of f are zero form an open set. I'm not able to show that it's also closed to get a disconnection.