Every ring must have a zero
Nathan Kamgang
When studying abstract algebra, we might wonder: why do we always use “0” for the additive identity in rings? Is this just convention, or is there something deeper? Why don’t we choose a less opinionated notation like $i$ to distinguish the additive identity of a ring? This article proves there’s a mathematical inevitability to calling it zero.
To see why, let’s use the notation $i$ to denote the additive identity of a ring as opposed to the usual $0$, as suggested above. Where does it get us?
By the definition of the identity element of a group, we have that:
$$ \forall a \in G \text{, } a + i = a $$We choose $a$ to denote any element in $G$, but remember that $ i \in G $, so by replacing $a$ with $i$, the above statement can be used to assert that:
$$ i + i = i $$Using the previous statement again, we can assert that:
$$ a \cdot i = a \cdot (i+i) $$But by the distributive law in the ring, $ a \cdot (i+i) = a \cdot i + a \cdot i $, so:
$$ a \cdot i = a \cdot i + a \cdot i $$But since $ a \cdot i = i + a \cdot i $, we have:
$$ i + a \cdot i = a \cdot i + a \cdot i $$Now by applying the cancellation law on both sides, we arrive at the realization that:
$$ i = a \cdot i $$In simpler terms, the conclusion we obtained says that if you multiply any element in $G$ by the additive identity $i$, then you get $i$ again.
Wait a minute. On one hand, any number multiplied by zero equals zero.
Any ring element multiplied by $i$ equals $i$. The additive identity in a ring behaves exactly like our familiar $0$! We might as well use the symbol $0$ to denotes the additive identity of the ring since they behave exactly the same.
Take a moment to appreciate the conclusion we just reached. From the definition of a ring and the distributive property, we proved that the additive identity of a ring behaves like zero. If you want, every ring has a zero.