I read some books and articles like http://goo.gl/P4VWS1 or http://goo.gl/FFyRup, which state a theorem: Any finite ring is a direct sum of rings of prime power order. But they only state the theorem without proof. I can't find any book or article proof this theorem. (If someone know that there is a proof in some book, please tell me.) Here is my proof, is it correct?

Let $R$ be an associated finite ring, not necessarily be commutative and not necessarily has a multiplicative identity.

Since $(R,+)$ is a finite abelian additive group, by the Fundamental Theorem of Finitely Generated Abelian Group, $(R,+)$ is isomorphic to $(R_1,+)\times (R_2,+)\times \cdots \times (R_n,+)$ by a group isomorphism $\theta$. For $i=1,2,...,n$, $(R_i,+)$ is an additive group with order $p_i^{r_i}$, $p_i$ is a prime and $p_i\neq p_j$ if $i\neq j$.

For any $i\in \{1,2,...,n\}$ and $r_i, r_i' \in (R_i,+)$, consider $(0,...,0,r_i,0,...,0), (0,...,0,r_i',0,...,0)\in (R_1,+)\times \cdots \times(R_i,+)\times \cdots \times(R_n,+)$. Define the multiplication between $r_i$ and $r_i'$ by $$r_i\cdot r_i'\stackrel{\triangle}{=}\theta^{-1}(0,0,...,r_i,0,...,0)\cdot \theta^{-1}(0,0,...,r_i',0,...,0).$$ Then $(R_i,+)$ is a ring under this multiplication whose order is a prime power order.

There seem to has another proof.


Of course, the finite ring already has its own multiplication, and you would not be at liberty to redefine it to some other product...

But on the other hand, if you show your candidate is actually the only multiplication possible, you would have a solution.

The primary components of the underlying groups are surely rngs, but to show that the big ring is a product of these, we'd have to show these pieces are ideals in the big ring.

Suppose the additive order of x is a power of p, and the additive order of y is a power of q, p,q a pair of distinct primes. Can you see why xy=0? (Argue using the additive order of xy.) From this, deduce that the primary components are all ideals, and you have a decomposition into subrngs.

  • $\begingroup$ Thanks! Let me think about it. $\endgroup$ – bfhaha Jul 5 '14 at 5:10
  • $\begingroup$ Thank you for you indicate my mistake and give me the most important hint. $\endgroup$ – bfhaha Jul 6 '14 at 3:24

Since $R$ is finite, suppose $|R|=p_1^{s_1}p_2^{s_2}\cdots p_n^{s_n}$, the distinct prime divisor factorization of $|R|$.

For each $i=1,2,...,n$, let $R_i=\{r\in R\mid p_i^s r=0 \mbox{ for some }s\in \Bbb{N}^+\}$. It is easy to verify $R_i$ is an ideal in $R$ and $R_i\cap (R_1\cap \cdots \cap R_{i-1}\cap R_{i+1}\cap \cdots\cap R_n)=\emptyset$.

Consider the finite additive abelian group $(R,+)$, by the Fundamental Theorem of Finitely Generated Abelian Group, we have $(R,+)\cong (\Bbb{Z}_{p_1^{s_1}},+)\oplus (\Bbb{Z}_{p_2^{s_2}},+)\oplus \cdots\oplus (\Bbb{Z}_{p_n^{s_n}},+)$. Which implies the following two things.

(i) For every $r\in R$, $$r=r_1+r_2+\cdots +r_n,$$ where the additive order of $r_i$ is a power of $p_i$. Then $R=R_1+R_2+\cdots +R_n$.

(ii) There are exactly $p_i^{s_i}$ elements in $R$ which additive order is a power of $p_i$. Hence $|R_i|=p_i^{s_i}$.

Therefore, by the theorem in p.130, thm.2.24, Algebra, Hungerford, $R\cong R_1\oplus R_2\oplus \cdots \oplus R_n$ and $|R_i|=p_i^{s_i}$.


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