Prove that a ring $R$ has a unique 1

Question

Problem: Prove that a ring $R$ has a unique 1.

My attempt: Suppose we have another unit denoted by $1'$. Then $\forall a \in R, 1.a = a = 1'.a \Rightarrow 1.a = 1'.a$. Apply cancellation law for both side we have $1=1'$. Is this proof completely? Thank all!

Answer 1

The cancellation law is in full generality only applicable to nonzero divisiors $a$.

If $1,1'$ are two ones, then $1\cdot 1'=1$ since $1'$ is a one, and $1\cdot 1' = 1'$ since $1$ is a one. So $1=1'$.

Answer 2

Your proof doesn't work because you don't necessarily have a cancellation law in a ring. (For example, $4 \cdot 5 \equiv 2 \cdot 5 \operatorname{mod} 10$ but $2 \neq 4 \operatorname{mod} 10$.) But if you consider the product $1 \cdot 1'$, I think you'll see why you have uniqueness.