prime ideal, integral domain proof

• Oct 21st 2012, 02:18 AM
rayman
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 $\displaystyle P\subset A$ is prime iff $\displaystyle A/P$ is an integral domain.
Proof:
$\displaystyle (\Rightarrow)$ Assume that P is prime and pick $\displaystyle a+P, b+P\in A/P$ suppose that $\displaystyle (a+P)(b+P)=0+P$ is a zero in $\displaystyle A/P$
we need to show that either $\displaystyle a+P=0+P \wedge b+P=0+P$
we have now $\displaystyle (a+P)(b+P)=ab+P=0+P$ so $\displaystyle ab\in P$(how do we know that???) $\displaystyle a,b\in A$ right? then by def of prime ideal $\displaystyle a\inP \wedge b\in P$ thus $\displaystyle a+P=0+P\wedge b+P=0+P$ no zero divisors and $\displaystyle A/P$ is int domain

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

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

thanks for the explanaition
• Oct 22nd 2012, 05:33 AM
Bingk
Re: prime ideal, integral domain proof
For $\displaystyle ab + P = 0 + P = P \Rightarrow ab \in P$:
Remember that $\displaystyle ab + P = \{ab + p: p \in P\}$, so if $\displaystyle ab + P = P$ then $\displaystyle ab + 0 = ab \in P$ (can you see why $\displaystyle 0 \in P$?)
From this, can you see why if $\displaystyle x \in P \Rightarrow x+P = 0+P$?

Yes, $\displaystyle a,b \in A$, otherwise we couldn't talk about $\displaystyle (a+P), (b+P)$.
• Oct 22nd 2012, 06:27 AM
hedi
Re: prime ideal, integral domain proof
Hedi
• Oct 22nd 2012, 07:29 AM
Deveno
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.
• Nov 18th 2012, 11:34 PM
rayman
Re: prime ideal, integral domain proof
Now I get it. Thank you for all replies:)