Algebra. If $R$ is a commutative associative ring with neutral element. Then if R|P is an integral domain ($ab \in R|P \implies a=0 \ or \ b=0$) , then $1 \notin $ P ? This is the only thing I can deduct from another proof, is this correct , or must I search for another reason?
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If $1\in P$, then since $P$ is an ideal every multiple of $1$ is in $P$, meaning $P=R$. $R/R$ is the ring with one element, which apparently by convention is not an integral domain.
Matt Samuel
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The zero ring is not a domain (by widespread convention). – Bill Dubuque Jun 15 '15 at 16:06
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But, by related convention $(1)$ is not considered prime, just like $1$ is not considered prime in $,\Bbb Z.\ $ This is a common oversight, e.g. see this recent answer. – Bill Dubuque Jun 15 '15 at 16:09
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A prime ideal is \emph{defined} to be a proper subset of the ring (with an additional property obviously). – lokodiz Jun 15 '15 at 16:10
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Fine. Due to peer pressure I've changed my answer. – Matt Samuel Jun 15 '15 at 16:11
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@Matt Are you aware of any author that considers units to be prime? – Bill Dubuque Jun 15 '15 at 16:14
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No. But a principal prime ideal need not be generated by a prime element. For example $(0)$ is prime in an integral domain. – Matt Samuel Jun 15 '15 at 16:15
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@Matt Yes, that's mentioned in the comments in my link. Besides units, zero is the other exceptional case. While some authors do allow $,0,$ to be prime, I don't recall any authors ever allowing a unit to be prime. – Bill Dubuque Jun 15 '15 at 16:18
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Some of these conventions are discussed in other threads, e.g. see Why doesn't 0 being a prime ideal in Z imply that 0 is a prime number?. and Why 1 is not considered to be a prime number?, and see John Conway's reason for considering $-1$ to be prime in Is -1 prime? – Bill Dubuque Jun 15 '15 at 16:29
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@MattSamuel : I agree with Bill in this case. Prime ideals are explicitly defined to be proper ideals of $R$ in every text I know (which is a lot), no exceptions. Allowing the entire ring to be a prime ideal does not "fit" into the big scheme well. However, $R$ is allowed to be a semiprime ideal in the same texts. – rschwieb Jun 15 '15 at 17:22
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I don't disagree. I haven't looked at a textbook that defines prime ideals in a long time. – Matt Samuel Jun 15 '15 at 17:26