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Let $R$ be a commutative ring with unity, and $A$ be an ideal of $R$, and $R/A$ be an integral domain.

In my book, I read the following:

$R/A$ is a commutative ring with unity for any proper ideal $A$.

I have a hard time understanding what this phrase means. More precisely, what does it mean that $R/A$ is commutative ring with unity for any proper ideal $A$? Does this mean that $R/A$ is not a commutative ring with unity for an improper ideal $A$? Doesn't make sense to me. Would appreciate a clarification.

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A proper ideal is an ideal $A$ such that $A\neq R$. If you have a "non-proper" ideal, i.e. $A=R$, then $R/A$ is the zero ring, which doesn't have a unity $1\neq 0$.

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  • $\begingroup$ But why is $R/R$ a zero ring if $r+R \in R/R$ for any $r\in R$? Say, if $R=\mathbb{Z}$, then $\mathbb{Z}/\mathbb{Z}$ must be equal to $\mathbb{Z}$, since $5+\mathbb{Z} = \mathbb{Z}$. $\endgroup$
    – sequence
    Feb 6 '17 at 19:31
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    $\begingroup$ @sequence Okay I'll try to explain. So how do we define the quotient? It is the set of cosets, i.e. $\{r+A:r\in R\}$, where $r+A=s+A$ if and only if $r-s\in A$. But for any $r\in R$, this exactly means that $r+R=0+R$, which is the zero element of $R/R$, so all elements of $R/R$ are zero. $\endgroup$ Feb 6 '17 at 20:25
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    $\begingroup$ $0$ is unity in the zero ring. It's just also $0$. (Wanted to make the distinction. You haven't said anything wrong, I just feel like the sentence "which doesn't have a unity $1\neq 0$." is potentially misleading.) $\endgroup$
    – Stahl
    Feb 8 '17 at 3:28

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