Let $R$ be a commutative ring with $1 \neq 0$. Show that if $I$ is a prime ideal, then $R/I$ is an integral domain.

I have a proof for "Let $R$ be a commutative ring. Show that if $I$ is a prime ideal, then $R/I$ is an integral domain." Do you have to change anything when $1 \neq 0$?

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    $\begingroup$ So you proved it also for $1=0$. Then why do you care? You have shown it more generally (if you really have). $\endgroup$ – Dietrich Burde Apr 3 '20 at 10:55
  • $\begingroup$ A ring in which $1=0$ has no prime ideal, unless your definition allows $R$ to be called a prime ideal, which is usually not done. $\endgroup$ – egreg Apr 3 '20 at 13:31

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.


Suppose that $P$ is a prime ideal of a commutative ring $R$ with $1$. Then $P$$R$ implies $1+P≠0+P$. As you stated $1≠0$. 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.

  • $\begingroup$ I don't understand why $0$ is still used in this proof when $1 \neq 0$. What does $1 \neq 0$ actually mean? That $0$ isn't the identity of $R$ or something else? $\endgroup$ – MathGeek1998 Apr 3 '20 at 12:03
  • $\begingroup$ Usually we denote the additive identity by 0, so r+0=r for r in our ring and the multiplicative identity by 1, so r*1=r. This is by analogy to the kinds of rings we are used to working with, like the ring of integers for example. So 1≠0 just says that the multiplicative identity isn't equal to the additive identity. $\endgroup$ – robotsheepboy Apr 3 '20 at 12:09
  • $\begingroup$ But isn't $1 \neq 0$ true for the majority of rings then as $1$ is usually the multiplicative identity and $0$ is normallythe additive identity? $\endgroup$ – MathGeek1998 Apr 3 '20 at 12:10
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    $\begingroup$ For the usual examples of rings that we think of, like integers, then yes it is usually true, but in the more general case there's no reason why there shouldn't be a single element which is both the additive and multiplicative identity, they don't have to be distinct. $\endgroup$ – robotsheepboy Apr 3 '20 at 12:13
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    $\begingroup$ 1 = 0 in the so called zero ring, which we usually don't want to consider as it may complicate our arguments. You can see more about the zero ring by just googling it for example. $\endgroup$ – robotsheepboy Apr 3 '20 at 12:14

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