Let D be an integral domain and let A be a proper ideal of D. Is D/A an integral domain? If yes, prove it; otherwise, give a counterexample.
What I did is:
Let (x+A),(y+A) be in D/A where x,y are in D.
WTS: either x+A=A or y+A=A
By def'n of quotient rings, (x+A)(y+A) = xy+A = 0+A = A
Since D is an integral domain, ab=0 implies a=0 or b=0.
Thus, (x+A)(y+A)=A implies x+A=A or y+A=A.
So, D/A has no zero divisors. Therefore, D/A is an integral domain?
Is this proof already correct or lacking to be complete?
This step doesn't work. You know that $xy+A=0+A$, but that doesn't imply $xy=0$; it just implies that $xy\in A$.
In fact, the result is not true. To find a counterexample, I suggest looking at various choices of ideals $A$ in the ring $D=\mathbb{Z}$.