prime ideal, integral domain proof

**rayman** · October 21st 2012, 02:18 AM

I am going through the following proof and I need some explanation
Thm: Let A be a commutative ring, an ideal $P\subset A$ is prime iff $A/P$ is an integral domain.
Proof:
$(\Rightarrow)$ Assume that P is prime and pick $a+P, b+P\in A/P$ suppose that $(a+P)(b+P)=0+P$ is a zero in $A/P$
we need to show that either $a+P=0+P \wedge b+P=0+P$
we have now $(a+P)(b+P)=ab+P=0+P$ so $ab\in P$ (how do we know that???) $a,b\in A$ right? then by def of prime ideal $a\inP \wedge b\in P$ thus $a+P=0+P\wedge b+P=0+P$ no zero divisors and $A/P$ is int domain

$(\Leftarrow)$ suppose $ab\in P$ then $(a+P)(b+P)=ab+P=0+P$ as we have no zero divisors in $A/P$ we get $a+P=0+P \wedge b+P=0+P$ so $a\in P\wedge b\in P$

why does $ab+P=0+P$ implies that $ab\in P$ ? and
why does $ab\in P$ implies that $ab+P=0+P$ ?

thanks for the explanaition

**Bingk** · October 22nd 2012, 05:33 AM

For $ab + P = 0 + P = P \Rightarrow ab \in P$ :
Remember that $ab + P = \{ab + p: p \in P\}$ , so if $ab + P = P$ then $ab + 0 = ab \in P$ (can you see why $0 \in P$ ?)
From this, can you see why if $x \in P \Rightarrow x+P = 0+P$ ?

Yes, $a,b \in A$ , otherwise we couldn't talk about $(a+P), (b+P)$ .

**hedi** · October 22nd 2012, 06:27 AM

my answer is attached.
Hedi

**Deveno** · October 22nd 2012, 07:29 AM

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.

**rayman** · November 18th 2012, 11:34 PM

Now I get it. Thank you for all replies

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