Like structure theorem for finite abelian groups or modules over PID, is there any structure theorem for finite rings? Thanks.

  • $\begingroup$ Finite commutative rings or finite rings in general? $\endgroup$ – user26857 Apr 21 '13 at 16:38
  • $\begingroup$ finite rings in general. thanks $\endgroup$ – GA316 Apr 21 '13 at 16:40
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    $\begingroup$ For finite commutative rings look here. In general, the question is similar to find a structure theorem for finite groups. $\endgroup$ – user26857 Apr 21 '13 at 16:41
  • $\begingroup$ Related: math.stackexchange.com/questions/1825661 $\endgroup$ – Watson Dec 11 '16 at 18:43

There is no known complete classification of finite rings.

In the following, I'm assuming a ring is unitary, but not necessarily commutative.

The first step is to see that the decomposition of the additive group into groups of pairwise coprime prime power orders is also a ring-theoretic decomposition. In this way, we get a unique decomposition of any finite ring into rings of prime power order. Hence it suffices to classify rings of order $p^n$ with $p$ prime.

For $n=1$, there is only the ring $\mathbb Z/p\mathbb Z$.

For $n=2$, it's a nice exercise to show that up to isomorphism, there are the 4 rings $\mathbb Z/p^2\mathbb Z$, $\mathbb F_{p^2}$, $\mathbb Z/p\mathbb Z \times \mathbb Z/p\mathbb Z$ and $\mathbb Z/p\mathbb Z[X]/(X^2)$. See here and here.

The case $n = 3$ is already quite tedious. It can be found in Theorem 14 of the article R. Raghavendran, Finite Associative Rings, Composito Mathematica 21 (1969), 195—229. The result is that there are 11 rings of order 8 and 12 rings of order $p^3$ for $p$ an odd prime. The only non-commutative ring among them is the ring of upper $2\times 2$ triangular matrices over $\mathbb F_p$.

If I remember correctly, today the classification is known up to $p^5$ or $p^6$. It gets increasingly nasty, of course.

For some further classification-related properties of finite rings, see this discussion.

  • $\begingroup$ What do you mean by "primary decomposition" here? $\endgroup$ – Martin Brandenburg Sep 12 '13 at 14:27
  • $\begingroup$ And why does a decomposition of the additive group yield such a one for the ring? $\endgroup$ – Martin Brandenburg Sep 12 '13 at 14:34
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    $\begingroup$ @MartinBrandenburg: Sorry, that was not correct, of course. What I thought of is: We decompose the additive group uniquely into groups of prime power order (one for each prime). This decomposition induces a decomposition as rings. $\endgroup$ – azimut Sep 12 '13 at 14:37

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