If $R$ is a domain and $a \in R$ satisfies $a^2=a$, prove that either $a=0$ or $a=1$

Question

Problem: If $R$ is a domain and $a \in R$ satisfies $a^2=a$, prove that either $a=0$ or $a=1$

My attempt: $R$ is a domain, so the cancellation law holds for it. $a^2=a \Rightarrow a.a = 1.a \Rightarrow a=1$. How can I prove that $a=0$? Thank all!

Answer 1

HINT : $a^2 = a \implies a^2 - a = 0 \implies a(a-1) = 0$ in the integral domain $R$. Can you conclude using the property of an integral domain?

Also, note that more generally, it can be proven that any polynomial of degree $d$ in $R[x]$ where $R$ is an integral domain has at most $n$ roots in $R$. If we consider the polynomial $x^2 - x$, then it has the roots $0$ and $1$ in $ R$, so we have already hit the maximum, namely the degree of $x^2 - x$ which is $2$. Therefore, no other roots in $R$ can exist i.e. if $a^2 = a$ then necessarily $a = 0$ or $a = 1$.

Furthermore, the contrapositive is an interesting statement : if there is an $a \neq 0,1$ such that $a^2 = a$, then $R$ is not an integral domain. This can be used to prove that some domain is not integral. An explicit example is the group $\mathbb Z_{10}$ of the $10$-adic integers. Indeed, in recreational mathematics, the "automorphic numbers" (numbers whose square end with themselves e.g. $25, 76$ etc.) extend to non-trivial roots of $x^2-x$ in $\mathbb Z_{10}$, showing that it is not integral.

Answer 2

$$\begin{align} a^2 & = a. \\ \implies a^2 -a & = 0. \\ \implies a(a-1) & = 0. \\ \implies \text {either}\ a & = 0\ \text {or}\ a = 1. \end{align}$$ (The last implication follows from the fact that $R$ is a domain).

Answer 3

Maybe $a^2=a$,then $aa-a=0$ and $a(a-1)=0$?