Is there a non-unital ring such that one of its quotient ring is unital?

Question

Suppose $R$ is a non unital ring, but it has an ideal $I$ such that $R/I$ is unital, does this kind of animal exist?

This problem arouse when I try to show “$R$ is unital, $R/I$ is division ring, then $I$ is maximal. “ I am wondering whether the condition $R$ is unital can be dropped or not.

Thank you in advance.

By the way, is this kind of problem meaningful? I am pretty addicted to this kind of problems but it seems that nobody cares about it.

Answer 1

$R = 2 \mathbb Z$, $I = ( 6 )$

Answer 2

Take a look at the non-unital commutative ring

$R = \Bbb Z \oplus 2\Bbb Z, \tag 1$

with component-wise addition and multiplication. Let $I \subset R$ be the ideal

$I = \{(0, b) \mid b \in 2\Bbb Z \}; \tag 2$

then $R$ has no unit, but

$R/I \simeq \Bbb Z. \tag 3$

Answer 3

Yes, it can happen. Consider e.g. $R:=\{p\in\Bbb Z[x]:p(0)\in 2\Bbb Z\} $ and $I:=(2x-2,x^2-x)$.
Then $2ax+I=2a+I$ for $a\in\Bbb Z$ and by induction, $x^n+I=x+I$, so $x+I$ will act as a multiplicative identity on $R/I$, and by the way, $R/I\cong\Bbb Z$.