Question about ideals and integral domains.

Suppose that A is a ring and I is an ideal of A. Prove that the quotient ring A/I is an integral domain if and only if I satisfies the following:

$\displaystyle I \neq A$ and $\displaystyle xy \in I \implies (x\in I$ or $\displaystyle y\in I)$.

I have tried it both ways using a contradictory argument but to no avail. Help much appreciated.

Re: Question about ideals and integral domains.

An ideal which satisfies the second condition is said to be prime. Use the fact that $\displaystyle x\in I$ is the same thing as the class of $\displaystyle x$ modulo $\displaystyle I$ is the class of $\displaystyle 0$, and the product of two classes is the class of the product.

Re: Question about ideals and integral domains.

this is pretty basic: suppose I is a prime ideal of A. then if (x+I)(y+I) = I in A/I, and x is not in I, then xy + I = I, so xy is in I, and since I is prime, and x is not in I, y is in I.

but this means that y+I = I, that is, A/I has no zero divisors. provided that A was a commutative ring with unity in the first place, A/I is an integral domain (some authors do not require commutativity nor an unit).

(the condition I ≠ A ensures we have some other element besides I = 0 + I in A/I, so that we have non-zero divisors at all).

the converse is proven similarly: using a direct approach, rather than contradiction, works well.