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$A$ is a commutative ring with identity. $I$ is a ideal of $A$.

then ideal $I$ is prime iff $A/I$ is a integral domain.

here is what I thought

$(\Rightarrow)$ We want to prove $A/I$ is a integral domain. It's equivalent to prove there are no nonzero element $a+I$ can divide $0$ such that $a\in A$. Let $a,b\in A$. Then $(a+I)(b+I)=ab+I$. Let $ab+I=I$, with $a+I\neq I$. So we need prove $b=0$ if $a\neq 0$ and $ab=0$. It's equivalent to prove $A$ is a integral domain.

$A$ is a integral domain iff the zero ideal is prime.

I stuck at here and doubt something is wrong.

($\Leftarrow$)

$A/I$ is a integral domain $\iff$ A is a integral domain

then I have no ideal how to prove.

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2 Answers 2

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You started off well and then got a little bit lost towards the end. By definition, $A/I$ is an integral domain if and only if the following statement holds.

If $a,b\in A$ are such that $(a+I)(b+I) = I$, then either $a + I = I$ or $b+I = I$. Also, we require $A/I\neq 0$, or in other words, $I\subsetneq A$.

Let's first make sure we know what the statement $(a+I)(b+I) = I$ means. You correctly point out that $(a+I)(b+I) = ab + I$. Thus it means $ab +I = I$, or equivalently, that $ab\in I$. Similarly, the statement $a+I = I$ is equivalent to $a\in I$, and $b+I = I$ is equivalent to $b\in I$. So we can rewrite the boxed statement above as

If $a,b\in A$ are such that $ab\in I$, then either $a\in I$ or $b\in I$. Also, we require $A/I\neq 0$, or in other words, $I\subsetneq A$.

Notice, this last boxed statement is exactly the definition of $I$ being a prime ideal of $A$. In this way we see that $I$ being prime is equivalent to $A/I$ being an integral domain.


EDIT: Just one additional comment. I should emphasize that the statement $A$ is an integral domain $\iff A/I$ is an integral domain is false. For instance, if $A = \mathbb{Z}$ and $I = 4\mathbb{Z}$, then $A$ is an integral domain, but $A/I = \mathbb{Z}/4\mathbb{Z}$ is not an integral domain!

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  • $\begingroup$ @Edit: Or simply $A/A=0$. $\endgroup$ Commented Feb 22, 2014 at 1:22
  • $\begingroup$ why a+I=I is equivalent to a∈I?a+I mean {a+i|i∈I},maybe exist some element a∈A-I such that a+i∈I $\endgroup$
    – amateur
    Commented Feb 22, 2014 at 2:24
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    $\begingroup$ @program666: The statement $a+I = I$ means by definition that $\{a + i : i\in I\} = I$. In particular, there is an $i\in I$ such that $a + i = 0$. But then $a = -i\in I$. Conversely, if $a\in I$, then $a + i\in I$ for all $i\in I$, so $a + I\subseteq I$. $\endgroup$
    – froggie
    Commented Feb 22, 2014 at 8:23
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Let $\mathcal{R}$ be our ring in question.

Suppose that $a,b\in\mathcal{R}$. Then $(a+I)(b+I)=0+I$ if and only if $ab\in\mathcal{R}$. Thus, if $\mathcal{R}/I$ is an integral domain, then $(a+I)(b+I)=0+I$ means that either $a+I=0+I$ or $b+I=0+I$. That is, either $a$ or $b$ lies in $I$, which shows that $I$ is prime.

The converse can then be proved by a similar argument.

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