Is it true that for every finite (for simplicity, commutative) ring $R$ in which every element not equal to $1$ is a zero divisor, is isomorphic to the zero ring or $\mathbb{Z}/2\mathbb{Z}$, $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ etc. The characterizing property of this sequence of rings is every element is a root of $x^2 - x$.

My work: First, consider the characteristic of $R$. If $R$ has characteristic not $1$ or $2$ then $1 \not= -1$ and $-1$ would then be a unit, not a zero divisor. If $R$ has characteristic $1$ then it's the zero ring. So take $R$ to have characteristic $2$. $R$ is a finite-dimensional algebra over $\mathbb{Z}/2\mathbb{Z}$; this implies its cardinality is $2^n$ for some $n$.

Then induction over $n$. Base case: $n=1$ so the cardinality of $R$ is $2^1$ which means $R \cong \mathbb{Z}/2\mathbb{Z}$. Inductive case: Assume it's true for all such rings with cardinality less than $2^n$; now we'll show it's true for $R$ with cardinality $2^n$. Take any element $x$ and forms its ideal $xR$. We see that $R/xR$ is a ring in the same form and so we apply the induction hypothesis to conclude that for any $y \in R$, $y^2 - y \in xR$. At this point I'm thinking of showing that $(1+x)R \cap xR = \{0\}$.

  • $\begingroup$ Is $R$ assumed to be commutative? $\endgroup$ – MooS May 1 '15 at 13:39
  • $\begingroup$ @MooS Maybe. Does it make a difference? For simplicity, yes. $\endgroup$ – man and laptop May 1 '15 at 13:40
  • $\begingroup$ The commutative case seems a little bit easier to me. Honestly I do not really know if there is a counterexample in the non-commutative case. $\endgroup$ – MooS May 1 '15 at 13:43
  • $\begingroup$ I won't be so sure that $R/xR$ has the same property. $\endgroup$ – user26857 May 1 '15 at 15:00
  • $\begingroup$ @user26857 If $x$ is nonzero then $R/(x)$ has strictly smaller cardinality, which the induction hypothesis covers. Although it still remains to be seen if we can induct. Also I don't think we've proven it's commutative, and so $R/xR$ isn't automatically a ring (maybe that's what you're getting at). $\endgroup$ – anon May 1 '15 at 15:05

The commutative case is very easy:

$R$ is reduced, since $1+a$ is a unit, if $a$ is nilpotent. Hence the product of all maximal ideals is zero. The Chinese Remainder Theorem then implies $$R = R/\mathfrak m_1 \times \dotsc \times R/\mathfrak m_s,$$

so $R$ is a product of fields.

We consider the unit groups and deduce

$$\{1\} = R^* = (R/\mathfrak m_1)^* \times \dotsc \times (R/\mathfrak m_s)^*,$$

which shows that each field occuring is $\mathbb Z/2\mathbb Z$.

  • 1
    $\begingroup$ I get that there are no nilpotents apart from $0$. I don't know what you mean by reduced and maximal ideal. Sorry. Maybe I asked a question beyond my level, or I didn't clearly state my level of knowledge. $\endgroup$ – man and laptop May 1 '15 at 14:04
  • $\begingroup$ I'll think about your answer for a bit. $\endgroup$ – man and laptop May 1 '15 at 14:10
  • $\begingroup$ Definitions for maximal ideal and reduced ring. $\endgroup$ – André 3000 May 1 '15 at 14:14
  • $\begingroup$ @user3491648 Also proved along the way in this answer. $\endgroup$ – Bill Dubuque May 1 '15 at 14:52

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