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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 $(1+x)R \cap xR = \{0\}$.

hmakholm left over Monica
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wlad
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  • Is $R$ assumed to be commutative? – MooS May 01 '15 at 13:39
  • @MooS Maybe. Does it make a difference? For simplicity, yes. – wlad May 01 '15 at 13:40
  • 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. – MooS May 01 '15 at 13:43
  • I won't be so sure that $R/xR$ has the same property. – user26857 May 01 '15 at 15:00
  • @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). – anon May 01 '15 at 15:05
  • @anon The induction hypothesis applies to the rings with the same property, that is, all non-zero elements are zero-divisors. Now suppose $a \bmod (x)$ is non-zero (in $R/(x)$). In particular $a\ne 0$, so there is $b\in R$, $b\ne 0$ such that $ab=0$. Now what if $b\in (x)$? How can we conclude that $a\bmod (x)$ is also a zero-divisor? – user26857 May 01 '15 at 15:12
  • @user26857 Ah yes, that is a good point. – anon May 01 '15 at 15:20

1 Answers1

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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 $\mathbb Z/2\mathbb Z$.

user26857
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MooS
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