-2

Please help me to solve the following question.

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

Anne Bauval
  • 34,650
gokhanakkurt
  • 145

3 Answers3

8

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.

Ben West
  • 12,366
5

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

ploosu2
  • 8,707
2

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 $\,f(x) = x(x-1).$

Community
  • 1
Bill Dubuque
  • 272,048
  • 6
    I don't think the ring $R$ is supposed to be commutative. – egreg Mar 03 '14 at 21:40
  • 4
    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 04 '14 at 07:55
  • With your reasoning, since $I-(-I)$ is cancellable in the $2\times2$ matrices over $\mathbb{C}$, you would conclude from $A^2-I=0$ that $A=I$ or $A=-I$, which is obviously false. The factor theorem you refer to in the linked answer is valid only in commutative rings. – egreg Mar 04 '14 at 11:44
  • 3
    Your comment was "if so, the proof still works", which is false. – egreg Mar 04 '14 at 13:33
  • @egreg I mean that relevant part of the proof works - see my Remark. Downvoter: if something is not clear then please feel welcome to ask questions. – Bill Dubuque Mar 04 '14 at 14:52
  • 5
    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 04 '14 at 15:03
  • It is certainly not true that the mere mention of idempotents necessarily implies that the ring is noncommutative, since idempotents do occur in the theory of commutative rings, e.g. in many treatments of the Chinese Remainder Theorem and related (coprime) factorization results. – Bill Dubuque Mar 05 '14 at 02:14
  • 2
    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 09 '14 at 11:32
  • @Marc Wikipedia is not only not an authority, but is frequently incorrect. I cannot recall any authors using said Wikipedia convention. It is questioned even on its talk page for integral domain, e.g. this remark "I am a working algebraic number theorist, and I am not so comfortable with the convention that "domain" means a ring without zero divisors and "integral" connotes commutativity...". By the way, egreg posted an associated meta question. – Bill Dubuque Mar 09 '14 at 15:17