$P$ is a prime ideal of $R$ iff $R/P$ is an integral domain. : $P≠R$

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.

But, why doesn't "$P$" have an identity?

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 $(0,1)$.

Isn't there a case $P=R$?

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.