Ring Without Unity may have Quotient Ring with Unity

Theorem
Let $\left({R, +, \circ}\right)$ be a ring.

Let $I$ be an ideal of $R$.

Let $\left({R / I, +, \circ}\right)$ be the associated quotient ring.

Then $\left({R / I, +, \circ}\right)$ may have a unity even if $\left({R, +, \circ}\right)$ has not.

Proof
Consider the external direct product of rings $\Z \oplus 2 \Z$.

From Integer Multiples form Commutative Ring, $2 \Z$ does not admit a unity.

By Unity of External Direct Sum of Rings, neither does $\Z \oplus 2 \Z$.

Now consider the ideal $\left\{{0}\right\} \times 2 \Z$ of $\Z \oplus 2 \Z$.

We have for all $a \in \Z$ and $b, c \in 2 \Z$ that:


 * $\left({0, b}\right) - \left({0, c}\right) = \left({0, b - c}\right)$
 * $\left({a, b}\right) \cdot \left({0, c}\right) = \left({0, b \cdot c}\right)$

so by the Test for Ideal, indeed $\left\{{0}\right\} \times 2 \Z$ is an ideal in $\Z \oplus 2 \Z$.

By Quotient Ring of External Direct Sum of Rings, we have:


 * $\left({\Z \oplus 2 \Z}\right) / \left({\left\{{0}\right\} \times 2 \Z}\right) \cong \left({Z / \left\{{0}\right\}}\right) \oplus \left({2 \Z / 2 \Z}\right)$

By Quotient Ring by Null Ideal and Quotient Ring Defined by Ring Itself is Null Ideal, this last ring is isomorphic to $\Z \times \left\{{0}\right\} \cong \Z$.

Since $\Z$ has a unity, this construction provides an example of the required kind.

Also see

 * Quotient Ring of Ring with Unity is Ring with Unity - this result demonstrates that its converse does not hold.