# Showing that the only idempotents in $R$ are zero and one

I have the following question that I have to solve however I cannot achieve.

Let $R$ be a ring with $1$ and suppose $R$ has no zero divisors. Show that the only idempotents in $R$ are $0$ and $1$.

• Can you at least show some efforts? – Secret Math Mar 3 '14 at 21:03
• This exercice was literally just solved (albeit by mistake and misunderstanding of definitions) in your other question. – Joshua Pepper Mar 3 '14 at 21:04

Suppose $e$ is idempotent. Then $e^2=e$, so $e^2-e=0$. Trying factoring and see what happens, using the fact that there are no zero divisors.

If $a^2 = a$ (the definition of idempotence), we have $a(a - 1) = 0$. Since $R$ has no zero-divisors, either $a=0$ or $a = 1$.

Hint $\$ The idempotents $\,0\,$ and $\,1\,$ are roots of a nonzero quadratic polynomial $\,f\in R[x],\,$ which has at most two roots, since $\,R\,$ is a domain. Indeed, if $\,f\,$ has distinct roots $\,r,s\,$ then, by the Bifactor Theorem we deduce $\,f(x) = a(x-r)(x-s),\$ so if $\ 0 = f(t) = a(t-r)(t-s)\$ then $\,a\ne 0\,\color{#c00}{\Rightarrow}\,(t-r)(t-s)=0\,\color{#c00}\Rightarrow\,t-r=0\,$ or $\,t-s=0,\,$ therefore $\,t=r\,$ or $\,t = s,\,$ where the $\color{#c00}\Rightarrow$'s follow from $\,R\,$ is a domain, i.e. $\,R\,$ has no zero-divisors $\ne 0$.

Remark $\$ Since at least one reader seems to have misinterpreted the above, let me clarify that the above more general proof is intended for commutative rings (I explicitly say "$R$ is a domain" and domains are commutative by standard conventions). However, even if the OP intends $R$ to be possibly noncommutative, the latter half of the proof still yields the desired inference when applied to $\,f(x) = x(x-1).$

• I don't think the ring $R$ is supposed to be commutative. – egreg Mar 3 '14 at 21:40
• The polynomial $x^2+1$ has integral coefficient, but has at least six roots in the quaternions (actually infinitely many) and the quaternions form a division ring. – egreg Mar 4 '14 at 7:55
• Your comment was "if so, the proof still works", which is false. – egreg Mar 4 '14 at 13:33
• I didn't downvote, but I don't agree with your assumption about commutativity, particularly when idempotents are concerned. However, I remain with my opinion that referring to theorems on polynomials in this case is using a sledgehammer; moreover the easy argument works without assuming commutativity. – egreg Mar 4 '14 at 15:03
• Re "domains are commutative by standard conventions": I know Wikipedia is no authority with respect to terminological conventions (nor do I pretend to be), but at least it is somewhat democratic, and the conventions it uses tend to be employed by a significant portion of people. According to WP, in ring theory the term "domain" refers to a (for some: nontrivial) ring without left or right zero divisors, while the term "integral domain" refers to a, definitely nontrivial, commutative domain. – Marc van Leeuwen Mar 9 '14 at 11:32