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Let $(R,+,.)$ be a commutative ring without a multiplicative identity and with no zero divisors (i.e. if $r.s = 0$ then $r=0$ or $s=0$). Is it possible that $R$ has a proper ideal $I$ for which the quotient ring $R/I$ is a zero ring?

Note 1: Here by a zero ring I mean a ring for which $r.s = 0$ for any two elements $r$ and $s$ (I am not sure if I should use "trivial ring" instead of "zero ring").

Note 2: I worked on constructing examples like $R=n \mathbb{Z} / m \mathbb{Z}$, but it appears that $R/I$ can never be a zero ring for any proper ideal $I$.

user114285
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3 Answers3

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Sure. For instance, let $R=2\mathbb{Z}$ and $I=4\mathbb{Z}$.

More generally, to say that $R/I$ is a zero ring just means that $R^2\subseteq I$. So, you're just asking for an example of $R$ without zero divisors such that $R^2$ is a proper ideal.

Eric Wofsey
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No, $x$ is zero in $R/I$ if and only if $x \in I$.

If your quotient is trivial, then $1 \in R$ would go to the zero class, i.e. $1 \in I$, meaning that $r = r \cdot 1 \in I$ for all $r \in R$, that means that $I = R$.

EDIT: If the ring is not unital, repeat the same argument with a fixed element $r_0 \in R$ instead of $1 \in R$.

EDIT: this proves that the quotient can't be trivial, not a 'zero' ring.

k76u4vkweek547v7
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For nonunital rings, consider $2\Bbb Z /4\Bbb Z$, this is a zero ring with two elements.

Berci
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