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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.

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  • $\begingroup$ Does your definition of prime ideal require $P$ to be proper? $\endgroup$
    – Matthew Leingang
    Oct 1, 2017 at 20:58
  • $\begingroup$ $P\neq R$ is within the definition of prime ideal. This is more and less for the same reason that 1 is not a prime number by definition, because R is like the "identity ideal" with respect to multiplication, i.e, $RI=I$ for every ideal $I$. $\endgroup$
    – nowhere dense
    Oct 1, 2017 at 21:00
  • $\begingroup$ This might also be useful to read. $\endgroup$
    – Edward Evans
    Oct 1, 2017 at 21:02
  • $\begingroup$ The relation above between prime ideals and prime number can be a bit obscure for someone who just saw the definition of prime ideal. It get's much clear the connection when you gets to see that there are theorems of unique factorization of ideals as product of prime ideals. Try to look for some expository article about the history of ideals if you are interested in understand this relation further. $\endgroup$
    – nowhere dense
    Oct 1, 2017 at 21:12
  • $\begingroup$ Please see proof "Conversely, ~". The (=>) is OK to me. I have problem with the (<=). And I am considering that my book is wrong.(my professor made it with a transcript of his lecture.) $\endgroup$
    – T.Walker
    Oct 1, 2017 at 21:58

2 Answers 2

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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.

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  • $\begingroup$ Okay, last one question. Does $1+P\in R/P$ imply $1\notin P$? Why? $\endgroup$
    – T.Walker
    Oct 3, 2017 at 17:00
  • $\begingroup$ @T.Walker The statement that "$R/P$ has an identity not equal to $0+P$" means that $1+P\neq 0+P$. This latter thing holds iff $1\notin P$. $\endgroup$
    – rschwieb
    Oct 3, 2017 at 17:46
  • $\begingroup$ Thanks a lot. ^^ $\endgroup$
    – T.Walker
    Oct 3, 2017 at 18:01
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Let $P \subset R$ be a an ideal. Then $P$ is prime if and only if $R/P$ is an integral domain.

$\textbf{Proof:}$

($\Leftarrow$) Suppose $P$ is not prime, then there exists $a,b \in R$ such that $ab \in P$ but $a,b \not\in P$. Then if $a+P, b+P \in R/P$, then

\begin{align} (a+P)(b+P)&= ab+P && \text{definition of operation in $R/P$}\\ &=0+P && \text{as $ab \in P$}\\ & \in R/P \end{align} Thus $R/P$ has zero divisors and is not an integral domain.

($\Rightarrow$) Suppose $R/P$ is not an integral domain, thus there exists (nonzero) zero divisors $a+P,b+P \in R/P$ which implies $a,b \not\in P$. but then

\begin{align} (a+P)(b+P)&=ab+P && \text{operation in $R/P$}\\ &=0+P && \text{as $a+P,b+P$ are zero divisors} \end{align} Thus $ab \in P$ and $P$ is not prime. $\square$

Thus $P$ is prime if and only if $R/P$ is an integral domain.

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