If a ring has no zero-divisors and it contains a nonzero element $b$ such that $b^2 = b$, then show that $b$ is the unity for that ring

Question

Suppose that $\mathbb{R}$ is a ring with no zero-divisors and that $\mathbb{R}$ contains a nonzero element $b$ such that $b^2 = b$. Show that $b$ is the unity for $\mathbb{R}$.

I attempted to show that $ab=a$ as $b$ is the unity for that ring. Now, with that assumption, $$(ab)^2=a^2$$ $$\implies abab=a^2$$

Multiplying both sides by $b$ and using $b^2=b$ we get:

$$abab=a^2b=a(ab)$$

Not only can we not take inverse on both sides (as it is not mentioned), but this kinda seems like a circular proof. I proved what I assumed so it is definitely incorrect.

I am unable to come up with any other method. What is the correct proof here?

Answer 1

Let $a\in R$ an arbitrary, nonzero element. Then $ba = b^2a \implies (b^2-b)a =0 \implies b^2-b=0\implies b(b-1)=0 \implies b=0 \text{ or } b=1$, but since we have assumed $b\neq 0$ we have $b=1$ (unity in the ring)

Edit:

There actually isn't really any need for the arbitrary element $a$ -- we could just have easily have said $b^2=b \implies b^2-b=0$ and worked from there. Sorry for the extraneous detail.

Answer 2

One has to show that $ab=a$ for all $a$. Consider the product $x:=(ab-a)b$. By distributivity $x=ab^2-ab=ab-ab=0$. Since $b\not=0$ we get $ab-a=0$ as desired.

Note that it is not necessary to assume that the ring has a multiplicative unit. This follows from the assumptions.

Answer 3

Let $e$ be the unit element of $R$. Then $e^2=e$ and $ae=a=ea$ for all $a\in R$. Thus $b(e-b) = be-b^2= b-b=0$ and so $e-b = 0$ since $R$ has no zero divisors and $b\ne 0$ by hypothesis. Hence, $b=e$.