# Proving that a ring has an identity and $R$ only has $1$

Let $$R$$=ring with identity $$1$$. If $$d^2=a$$ for all $$d \in R$$,The only unit in $$R$$ is $$1$$...show this.

I was thinking $$1\cdot d^2=d^2\cdot 1$$ which equals $$1\cdot d=d\cdot 1$$.

I found out a unit means that $$uv=vu=1$$.

• Something's up here. If a ring has $1$, then by definition it also has $0$ and $-1$. May 31, 2021 at 6:29
• @innerproduct note you can have $1=-1$ May 31, 2021 at 6:42
• You need to work with the definition of a unit. May 31, 2021 at 6:45
• @innerproduct We need $0$ but it’s not a unit (except in the zero ring), so no claim about it is made in the question. The equation $1=-1$ holds in some rings, in particular in all rings with the additional property that $a^2=a$ for all $a\in R$. May 31, 2021 at 7:11
• @innerproduct The definition of a ring gives you $0$ of course, but the proof here does not require it. With the idempotent property it is easy to see that $1=(1)^2=(-1)^2=-1$. Think about the ring (even field) with two elements. I said "you can have" not "every ring must have". The rings in this question are Boolean Rings and all have $1=-1$ May 31, 2021 at 7:13

If $$a$$ is invertible, then $$1=aa^{-1}=a^2a^{-1}=a$$.
Suppose $$a \in R$$ is a unit. Then there exists $$b \in R - \{0\}$$ such that $$ab = 1.$$ By the definition of $$R,$$ we have $$a^2 = a,$$ and so $$a(a-1) = 0.$$ But then $$a - 1 = ab(a-1) = ba(a-1) = 0,$$ hence $$a = 1.$$
(Note that $$R$$ must be commutative, so the order of multiplication does not matter.)
• @Amanda if $a^2 = a,$ then $0 = a^2 - a = a(a-1).$ May 31, 2021 at 6:57