Suppose $R$ is a finite unital and commutative ring that has exactly one maximal ideal. Prove that $\left | R \right |=p^{n}$ where $p$ is a prime number. If $R$ will be non-commutative, do we have the desired result?

Suppose $I$ is a maximal ideal of $R$, so $R/I$ is a field, because $R$ is finite, thus $| R/I|=p^{m}$, where $p$ is prime. Now I don't know what I should do next, so please help me.

  • $\begingroup$ $R$ being vector space under $R/I$?? $\endgroup$ – user87543 Dec 25 '13 at 10:21

First show that the maximal ideal of $R$, let say $M$, is nilpotent, that is, $M^n=0$ for some $n\ge 1$. If $n=1$ you are done, since then $R$ is a finite field. If $n=2$ use the following exact sequence $$0\to M/M^2\to R/M^2\to R/M\to 0.$$ Here $M/M^2$ is an $R/M$-vector space, so it has as number of elements a power of $p$, the characteristic of the finite field $R/M$. Then $R/M^2$ has also a power of $p$ as its cardinality.

Now you have all the ingredients to tackle the general case (by induction, let's say).

  • $\begingroup$ thank you very much,for your answer. $\endgroup$ – kpax Dec 25 '13 at 10:47
  • $\begingroup$ @kpax Welcome! However it remains the non-commutative part of the question which is beyond my knowledge. $\endgroup$ – user89712 Dec 25 '13 at 10:55

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