What if 0=1 ?
In everyday mathematics, zero and one are fundamentally different.
But abstract algebra asks: what if we relax our assumptions? What happens in a ring where the additive identity and the multiplicative identity are the same? Can such a ring even exist?
The answer reveals a simple but powerful fact: if $0 = 1$ in a ring, then the ring collapses to a single element.
The Setup
A ring $R$ possesses two binary operations, commonly denoted by $+$ and $\cdot$. Assuming we are working with a ring with unity (see ring with multiplicative identity), each operation has its own identity:
- the additive identity, denoted by $0$
- the multiplicative identity, denoted by $1$
In familiar rings like $\mathbb{Z}$, these identities are distinct: $0 \neq 1$. But what happens if we assume instead that $0 = 1$?
The Proof
Assume that $0 = 1$.
$$ a \cdot 1 = a \cdot 0 \tag{1} $$Every ring has a zero element with a special annihilation property (see why rings have zero):
$$ a \cdot 0 = 0 \tag{2} $$ $$ a \cdot 1 = 0 $$That is,
$$ a \cdot 1 = 0 \tag{3} $$But by definition of the multiplicative identity, we must also have:
$$ a \cdot 1 = a $$Comparing with (3), we conclude:
$$ a = 0 $$Since $a$ was arbitrary, every element of $R$ equals zero.
The Conclusion
We have shown that assuming $0 = 1$ forces every element of the ring to collapse into a single value.
Therefore:
A zero ring is a ring with exactly one element: $\{0\}$.
Equivalently:
A zero ring is a ring in which the additive and multiplicative identities coincide.