Ring Without Unity may have Quotient Ring with Unity

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Theorem

Let $\struct {R, +, \circ}$ be a ring.

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

Let $\struct {R / I, +, \circ}$ be the associated quotient ring.


Then $\struct {R / I, +, \circ}$ may have a unity even if $\struct {R, +, \circ}$ 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 $\set 0 \times 2 \Z$ of $\Z \oplus 2 \Z$.

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

$\tuple {0, b} - \tuple {0, c} = \tuple {0, b - c}$
$\tuple {a, b} \cdot \tuple {0, c} = \tuple {0, b \cdot c}$

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


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

$\paren {\Z \oplus 2 \Z} / \paren {\set 0 \times 2 \Z} \cong \paren {Z / \set 0} \oplus \paren {2 \Z / 2 \Z}$

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


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

$\blacksquare$


Also see