Remember that , so if then (can you see why ?)
From this, can you see why if ?
Yes, , otherwise we couldn't talk about .
I am going through the following proof and I need some explanation
Thm: Let A be a commutative ring, an ideal is prime iff is an integral domain.
Assume that P is prime and pick suppose that is a zero in
we need to show that either
we have now so (how do we know that???) right? then by def of prime ideal thus no zero divisors and is int domain
suppose then as we have no zero divisors in we get so
why does implies that ? and
why does implies that ?
thanks for the explanaition
basically, we when take an ideal P of A, and create A/P, we're effective saying:
"everything in P goes to 0".
that is: P is the additive identity of A/P (the "0" of the ring A/P).
now consider what happens to any zero-divisor x in A in A/P. we have (since x is a zero-divisor in A):
xy = 0 (in A). thus:
(x + P)(y + P) = P (in A/P).
since in ANY quotient ring: (x + P)(y + P) = xy + P, we have xy + P = P = 0 + P, so xy - 0 = xy is in P.
all of the above is true for ANY ring A, and ANY ideal P.
now when P is a PRIME ideal, we know that xy in P means x in P, or y in P, which shows that A/P is an integral domain (all the zero-divisors lie in P, so A/P no longer has any).
running the argument backwards:
suppose A/P is an integral domain.
then if (x + P)(y + P) = P, we must have: x + P = P, or y + P = P (suppose we write [x] instead of x + P for the cosets, then we have in A/P:
[x]*[y] =  implies [x] =  or [y] = , by the definition of an integral domain).
either way, we conclude that xy in P implies x in P, or y in P, so P is a prime ideal.
suppose P is NOT a prime ideal. then we must have SOME xy in P, with x not in P, and y not in P.
so in A/P, [xy] = , but [x] ≠ , and [y] ≠ 0, so A/P is not an integral domain (it has the zero-divisors [x] and [y]).
Z/(5) is an integral domain: suppose ab = 5n. clearly 5 must divide ab, and since 5 is prime, 5 divides a or b, that is: one of a or b is in (5). so if:
ab = 0 (mod 5), either a = 0 (mod 5), or b = 0 (mod 5).
Z/(4) is not an integral domain: we have 2*2 = 4, but 2 is not in (4), and in Z/(4) we have:
2*2 = 0 (mod 4), so 2 (mod 4) is a zero-divisor.