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):

$$ a \cdot 0 = 0 \tag{2} $$

Using (2) in (1), we obtain:

$$ a \cdot 1 = 0 $$

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.