prime ideal, integral domain proof

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.

Proof:

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

Re: prime ideal, integral domain proof

For :

Remember that , so if then (can you see why ?)

From this, can you see why if ?

Yes, , otherwise we couldn't talk about .

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Re: prime ideal, integral domain proof

my answer is attached.

Hedi

Re: prime ideal, integral domain proof

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] = [0] implies [x] = [0] or [y] = [0], 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.

OR:

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] = [0], but [x] ≠ [0], and [y] ≠ 0, so A/P is not an integral domain (it has the zero-divisors [x] and [y]).

************

an example:

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.

Re: prime ideal, integral domain proof

Now I get it. Thank you for all replies:)