Let $I$ be an ideal in ring $R$. Prove that every element in $R/I$ is a solution of $x^2=x$ if and only if for every $a$ in $R$, $a^2-a$ is in $I$.

Let $a \in R/I$. Suppose $a$ is a solution to $x^2 = x$. Thus $a*a|a$, which implies $a*a*b = a : \exists b \in R$. Because $a^2 = a$, then $a*b = a$. But clearly $b = a$ is a solution. Setting $b = 1$ works. So the set $Z = R/I = \{0, 1\}$ is non-empty.

But also, $b*a*a = b*a = a = a*b = a*a*b$, so the absorption law and commutativity holds. So $Z$ is an ideal.

Thus $b*a*a - a*a*b = 0 \Longrightarrow b(a^2 - a^2) = b(a^2 - a) = 0$. Consider $b \neq 0$. Then $a^2 - a = 0 \Longrightarrow a^2 = a$, and so we have proven every $a \in Z$ is a solution to $x^2 = x$.

I have a dilemma: Can I show that $0$ and $1$ are the only elements in the ideal, rigorously instead of just picking out elements?

I wasn't sure how to phrase my question, but I wanted to emphasize the latter half of my proof.

  • $\begingroup$ @Mike: Fixed... $\endgroup$
    – Don Larynx
    Dec 9, 2013 at 3:05

2 Answers 2


Some comments and rhetorical questions.

  • Going from $a^2=a$ to $a^2\mid a$ actually loses information; no point.
  • That $b=1,a$ are solutions to $ab=a$ doesn't say anything about $R/I$.
  • A quotient $R/I$ is always nonempty; at worst $I=R$ and so $R/I$ is the trivial ring containing its zero element $I$. I see no need to show $R/I$ is empty, in this problem or in general.
  • You cannot say $R/I=\{0,1\}$, even if you know $R/I$ has distinct $0$ and $1$ elements.
  • What does $a^2b=ba^2$ for particular $a$s and $b$s have to do with absorption and commutativity?
  • $R/I$ is an ideal of what exactly? Itself? Every ring is an ideal of itself. Hardly needs proving, and I don't see how that's relevant. $R/I$ is not a subset of $R$, certainly not an ideal of $R$.
  • Why are you working with two letters here, $a$ and $b$? What are they exactly?
  • An ideal always contains $0$, but never contains $1$ unless $I=R$ is the whole ring.
  • You cannot show that $0$ and $1$ are the only elements in $R/I$ (don't call it an ideal), since there may be more elements. As plattnum points out, if $R={\Bbb F}_2[t]$ and $I=(t^2-t)$ then the quotient ring $R/I$ has elements besides $0$ and $1$. (Namely, $t$ and $1+t$.)

If $x^2=x$ for all $x\in R/I$, then $x^2-x=0$ for all $x\in R/I$. Every $x\in R/I$ is $\bar{x}:=x+I$ for some element $x\in R$. The equation $\bar{x}^2-\bar{x}=0$ in $R/I$ is equivalent to $x^2-x\in I$ by definition (since something is $0$ in $R/I$ precisely when its representatives are elements of $I$).


You will not be able to prove that 0 and 1 are the only elements of $I$. Consider $R=\mathbb F_2[t]$, $I=(t^2-t)$, where $\mathbb F_2$ is the field with two elements. Then $R/I$ has three elements $0,1,\overline{t}$ each of which satisfies $x^2=x$, $I$ does not contain 1 and does contain $t^2-t$.


Thanks to @anon, I neglected 1+$\overline{t}$. But we still have $$(1+\overline{t})^2=1+2\overline{t}+\overline{t}~^2=1+\overline{t}$$ where the last equality comes from the fact that 2=0 in our ring and $\overline{t}~^2=\overline{t}$ by construction.

  • $\begingroup$ $I$ does contain $1$, it is the field with two elements $0, 1$. $\endgroup$
    – Don Larynx
    Dec 9, 2013 at 5:05
  • 1
    $\begingroup$ No. It absolutely does not. Any ideal that contains 1 is the entire ring. Also you should notice that $I$ contains infinitely many elements namely 0 and $(t^2-t)^k$ for $k>0$. $\endgroup$ Dec 9, 2013 at 5:49
  • $\begingroup$ @anon please help me, I have my algebra final in 6 hours and I don't know what to do. I've tried!! I am lost. $\endgroup$
    – Don Larynx
    Dec 9, 2013 at 6:01
  • $\begingroup$ @Don Larynx the solution to the problem follows from the fact that the natural map $\pi: R\to R/I$ is a ring homomorphism with the property $\pi(r)=0 \iff r\in I$. Hopefully that helps. $\endgroup$ Dec 9, 2013 at 6:17

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