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$R$ is a finite ring and for every $a\in\,R$ there exist natural number $n(a)>1$ that $a^{n(a)}=a$ . Is $R$ a ring with identity?

If this question is correct then, for every $a \in R\,\,,a^{n(a)-1}$ is identity of $R$. I prove that $a^{n(a)-1}\in Z(R)$. But how I can prove that for every $a \in R\,\,,a^{n(a)-1}$ is identity of $R$?

I find a proof for my quastion as follows :
Let $R=\{a_1,a_2,...,a_n\}$.
We know for any $a_i$ there exit $n_i$ s.t $a_i^{n_i}=a_i$ and $a_i^{n_i-1}\in Z(R)$.
It is enough to show that there exit a some element $e$ s.t $ea_i=a_ie=a_i$.
I'm going to construct an identity's element.
Consider:
$e_1=a_1^{n_1-1}$ $\Rightarrow \, e_1a_1=a_1e_1=a_1$
$e_2=e_1+a_2^{n_2-1}-e_1a_2^{n_2-1}\Rightarrow \, e_2a_1=a_1e_2=a_1 \,\,, \,\,e_2a_2=a_2e_2=a_2$
$e_3=e_2+a_3^{n_3-1}-e_2a_3^{n_3-1}\Rightarrow \, e_3a_1=a_1e_3=a_1 \,\,, \,\,e_3a_2=a_2e_3=a_2\,\,, \,\,e_3a_3=a_3e_3=a_3$
$$. \\ . \\ . $$ $e_n=e_{n-1}+a_n^{n_n-1}-e_{n-1}a_n^{n_n-1}\Rightarrow$ for every $0\leq i\leq n$ we have $e_na_i=a_ie_n=a_i \Rightarrow \, 1_R=e_n$.

Mojtaba
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1 Answers1

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I think the famous theorem of Jacobson about such a ring being commutative holds, even if there is no identity.

The hypothesis implies the ring is von Neumann regular. Therefore every ideal is idempotent, including the entire ring.

Since the ring is finite, by Nakayama's lemma, we have that R has an identity.

It would follow that $R$ is a finite product of fields.

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rschwieb
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    Does the "famous theorem of Jacobson" hold for nonunital rings? It's quite some trouble to figure this out from the sources... – darij grinberg May 19 '15 at 02:58
  • @darij found a better way now, I think. – rschwieb May 19 '15 at 03:19
  • why for each x in R,x=ey for some y? – Mojtaba May 19 '15 at 05:50
  • @Mojtaba-CH That's what I intended with "generates the ideal as a right ideal," but looking at this after a night of sleep, I realize this may not be one of the equivalent characterizations of von Neumann regularity for rings without identity. I'll keep looking at the problem. – rschwieb May 19 '15 at 12:31
  • @rschwieb:thanks for looking. – Mojtaba May 19 '15 at 14:57
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    @darijgrinberg The original paper (Structure Theory for Algebraic Algebras of Bounded Degree, Theorem 11) does deal with rings without identity such as nil rings and radical rings. Since he seems to be specifying when identity is necessary, it seems like the theorem also applies to rings without identity. However I haven't traced the proof yet to be certain... – rschwieb May 19 '15 at 15:30
  • Another way to see this, I think: the condition implies that the ring has no nilpotent elements, so its Jacobson radical is zero because the ring is artinian. It is then Jacobson-semisimple and a semisimple artinian ring has a unit (this is proved in Herstein's Noncommutative rings, for example) Then you can use Jacobson's theorem with unit. – Mariano Suárez-Álvarez May 19 '15 at 17:39
  • @rschwieb:thank you but I am looking for to solve this question by an other way. – Mojtaba May 19 '15 at 17:54
  • @MarianoSuárez-Alvarez So it can be proved that an Artinian rng with Jacobson radical zero has an identity? That would be pretty useful... – rschwieb May 19 '15 at 19:12
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    This is Corollary 2 to Theorem 1.4.2: A semisimple Artinian ring has a two sided unit element. – Mariano Suárez-Álvarez May 19 '15 at 20:23