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Let $(B, +, \cdot)$ be a non-trivial ring with the property that every $x \in B$ satisfies $x \cdot x = x$. How does one prove that such a ring $(B, +, \cdot)$ must have a unit element $1_B$? (Or, in case this is not true in general, what is a counterexample?)

BTW, I'm looking for an elementary proof, not requiring anything more than the definition of a ring, the definition of $(B, +, \cdot)$, and, if necessary, the easily shown facts that $x + x = 0$ and $x\cdot y = y\cdot x,\,\forall\, x,y \in B$.

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    $\begingroup$ My definition of ring includes a $1$, and my definition of boolean algebra also includes a $1$. $\endgroup$
    – Zhen Lin
    May 14, 2013 at 8:35
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    $\begingroup$ @ZhenLin I learned it as "ring" and then "ring with unity". But whatever the definition, I think it's clear that this post in particular is referring to rings without a 1. $\endgroup$
    – Caleb Stanford
    May 14, 2013 at 8:44
  • $\begingroup$ @ZhenLin: What do you call something that is like $2\mathbb{Z}$, which is similar to what you call a "ring", except that it does not include a unit? I will gladly rephrase my question to use your terminology. $\endgroup$
    – kjo
    May 14, 2013 at 8:49
  • $\begingroup$ I wonder, can anyone produce an example of a Boolean rng as a proper ideal of a Boolean ring? This is the "obvious" thing to try until one runs into the problem that these ideals often have identities (which is not the identity of the original ring). The larger Boolean ring must necessarily be infinite. $\endgroup$
    – Ragib Zaman
    May 14, 2013 at 9:12
  • $\begingroup$ @RagibZaman Unless I did something silly I think that's easy :) Please review my solution. $\endgroup$
    – rschwieb
    May 15, 2013 at 13:28

3 Answers 3

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An example: the family of all finite subsets of a given infinite set.

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    $\begingroup$ @kjo: Let your ground set be $\mathbb{N}$ and the elements be binary sequences, where a $1$ in the $n$ coordinate means that $n$ is in your set, and a zero means that is not. Then define addition and multiplication coordinate wise, and work modulo $2$. Then the unit would have to be $(1,1,1,1...)$ which will fail to be in your ring since you are consider finite families. $\endgroup$
    – Daniel Montealegre
    May 14, 2013 at 8:53
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    $\begingroup$ I really think you should be more explicit than this (although the questioner just accepted this answer...) $\endgroup$
    – Rhys
    May 14, 2013 at 8:57
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    $\begingroup$ Yes, the ring operations should be specified. Or just take $\oplus_n \mathbb{F}_2$, as in Daniel's comment. $\endgroup$
    – Martin Brandenburg
    May 14, 2013 at 9:13
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    $\begingroup$ I think the operations are evident and natural. $\endgroup$
    – Boris Novikov
    May 14, 2013 at 9:19
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    $\begingroup$ @BorisNovikov: I don't think so. I happen to know some operations that work (namely, symmetric difference for $+$ and intersection for $\cdot$), and that's why I decided to accept your answer (reluctantly), but I consider them neither "evident" nor "natural". $\endgroup$
    – kjo
    May 14, 2013 at 9:30
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In the ring $\prod_{i=1}^\infty\Bbb Z_2$, consider the ideal $\oplus_{i=1}^\infty\Bbb Z_2$. It is a subrng without identity.

(Ultimately I think this is isomorphic to Boris' example, but here the operations are clear.)

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Here is a positive result: Let $R$ be a commutative ring such that every element can be written as a sum of products of elements of $R$ (i.e. the multiplication map $R \otimes_{\mathbb{Z}}R \to R$ is surjective). If $R$ is finite, then $R$ is unital. In particular, every finite boolean ring is unital (which also can be proven directly, of course).

Proof: Consider the unitalization $R^+$ as an $R$-module. Then we have $R R = R$, hence by Nakayama there is some $e \in R$ with $(1-e) R=0$. But this means that $e$ is a unit of $R$.

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  • $\begingroup$ Thanks for your answer. It flies so many miles above my head that I can't reasonably ask you to explain it further (my question requested elementary proofs for a reason). Needless to say, I can't even begin to evaluate it, but others may find it useful. $\endgroup$
    – kjo
    May 14, 2013 at 9:38
  • $\begingroup$ Yes, and one day, perhaps you will find it useful too. ;) $\endgroup$
    – Martin Brandenburg
    May 14, 2013 at 9:48
  • $\begingroup$ @YACP He did use it: it would not be possible to write the $1$ in $(1-e)R$ otherwise. $\endgroup$
    – rschwieb
    May 16, 2013 at 10:38
  • $\begingroup$ @MartinBrandenburg Looks like you meant rather "consider $R$ as an $R^+$ module," rather than the other way around, and then $M=R\lhd R^+$ puts you on the road to a nice application of Nakayama's lemma :) $\endgroup$
    – rschwieb
    May 16, 2013 at 11:31
  • $\begingroup$ Using "by Nakayama" is not helping, but showing off. $\endgroup$
    – nilo de roock
    Oct 10, 2020 at 8:32

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