# Proving idempotent ideals are solutions to $x^2 = x$

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.

• @Mike: Fixed... Dec 9, 2013 at 3:05

• 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 $$t^2-t$$.

# Edit

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.

• $I$ does contain $1$, it is the field with two elements $0, 1$. Dec 9, 2013 at 5:05
• 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 $k>0$. Dec 9, 2013 at 5:49
• @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. Dec 9, 2013 at 6:01
• @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. Dec 9, 2013 at 6:17