Here is the theorem and proof in my textbook.
Thm. Let $P$ be an ideal of a commutative ring $R$ with identity $1$. Then $P$ is a prime ideal of $R$ if and only if $R/P$ is an integral domain.
Proof.
Suppose that $P$ is a prime ideal of a commutative ring $R$ with $1$.
Then $P$≠$R$ implies $1+P≠0+P$.
Hence $R/P$ is a commutative ring $R$ with identity.
Assume that $(a+P)(b+P)=0+P$.
Then $ab+P=0+P$ and $ab∈P$.
By the definition of a prime ideal P we get $a∈P$ or $b∈P$.
That is, $a+P=0+P$ or $b+P=0+P$.
Thus $R/P$ is an integral domain.
Conversely, if $R/P$ is an integral domain, then $1+P≠0+P$ and $R/P$ is a commutative ring $R$ which has no zero divisors.
Hence $P≠R$.
Assume $ab∈P$.
Then $ab+P=0+P$ and $(a+P)(b+P)=0+P$.
Since $R/P$ is an integral domain, we get $a+P=0+P$ or $b+P=0+P$.
So $a∈P$ or $b∈P$.
Thus $P$ is a prime ideal.
I am having problems understanding proof with "Conversely, if $R/P$ is an integral domain, then $1+P≠0+P$.".
I know that $R/P$ has an identity. (cause it is ID)
But, why doesn't "$P$" have an identity?
Isn't there a case $P=R$?
Please help me understanding.
I think $P≠R$ should be with condition of Thm..
Thank you in advance.
Best Answer
Who says it can't? If $F$ is a field, then $P=\{0\}\times F$ is a prime ideal of $F\times F$, and $P$ has an identity element $(0,1)$.
If you are assuming the condition on the left that $P$ is prime, then no: prime ideals are defined to be proper ideals of their rings.
If you are assuming the condition on the right ($R/P$ an integral domain) then again no: integral domains are defined to have a nonzero multiplicative identity, so this quotient ring must have at least two elements. Saying that $R=P$ would imply that the quotient has only one element.
I say both these things from the position of standard definitions of "prime ideal" and "integral domain." Of course you can always make nonstandard definitions and change the answer here, but I think the point is to know what the mainstream reasoning is.