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I read in an answer on MO that Nathan Jacobson had given a universal algebraic proof that a ring satisfying the equation $x^n=x$ is commutative.

The sketch given in the answer is very clear : wlog one may assume that $R$ is subdirectly irreducible, as a result of a general result in universal algebra (that is, it has a minimum nonzero ideal).

Then one proves that a subdirectly irreducible ring satisfying the equation is a finite (skew, a priori) field, and one concludes from Wedderburn's theorem that it's commutative.

But I'm having trouble with the interesting step, that is

A subdirectly irreducible ring satisfying the equation $x^n=x$ for some $n\geq 2$ is a finite division ring

I had the folliwing idea : since $R$ has no nilpotent elements it should be a subdirect product of integral domains satisfying the same equations - however I know this property for commutative rings, and it relies on the well-known fact that $\displaystyle\bigcap\{p, p\in \mathrm{Spec}R\} = \{x, x$ is nilpotent $\}$ - and I don't know whether this is true for noncommutative rings.

As a matter of fact I'm pretty much convinced that it's not true (in $M_n(K)$, $K$ a field, $n\geq 2$, the set of nilpotent elements isn't a bilateral ideal - indeed there are no nontrivial ones). So unless this idea can be saved by the specifics of the situation, I can't go any further with it.

What I also noticed (I don't know if that can help though) is that $I^2 = I$, if $I$ denotes the minimum nonzero ideal.

I can also sort of make a connection with Wedderburn's theorem by studying the case where $Z(R)$ (the center) is a field; and so $R$ is a $Z(R)$-vector space. Then, it's a finite field. I can't yet see why $Z(R)$ would be finite dimensional (this would probably help a lot).

Am I anywhere near the right direction ? Can anyone give some hints to solve this ? (If possible - I know sometimes it's not- I'd rather see some hints than a full solution; and also if someone has read the article in question and saw that the proof in question was longer than what an MS answer can suggest, I'd also like to know haha)

EDIT: Here's the MO question : https://mathoverflow.net/questions/30220/abstract-thought-vs-calculation The answer I'm mentioning should be recognizable

user26857
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Maxime Ramzi
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2 Answers2

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A subdirectly irreducible ring satisfying the equation $x^n=x$ for some $n\geq 2$ is a finite division ring

First let's consider the elementary part of this statement. Observe

Lemma. If $R$ is a subdirectly irreducible ring with no nonzero elements that square to zero, then $R$ has no nontrivial zero divisors.

Reasoning: Assume not. Then $R$ has elements $a\neq 0\neq b$ such that $ab=0$. The set $bRa$ consists of elements that square to zero, so $bRa=\{0\}$. Then $(b)(a)=RbRaR=R\{0\}R=\{0\}=(0)$. If $I$ is the least nonzero ideal of $R$, then $I\subseteq (b)$ and $I\subseteq (a)$, so $I^2\subseteq (b)(a)=(0)$. This shows that all elements of $I$ square to zero, a contradiction. \\\

Here is how the lemma applies. No nonzero element $a\in R$ which squares to zero can satisfy the equation $x^n=x$, so the lemma implies that if $R$ is subdirectly irreducible and satisfies $x^n=x$, then $R$ has no nontrivial zero divisors. Given any nonzero $a\in R$ we have $0=a^n-a = a(a^{n-1}-1)$. Since $a\neq 0$, it must be that $a^{n-1}-1=0$. Hence, any nonzero $a\in R$ satisfies $a^{n-1}=1$, showing that nonzero elements are units. Thus, $R$ is a division ring.

The non-elementary part of the statement

A subdirectly irreducible ring satisfying the equation $x^n=x$ for some $n\geq 2$ is a finite division ring

is that a division $R$ whose nonzero elements satisfy $x^{n-1}=1$ must be finite. The idea for this is to first prove that any maximal subfield of $R$ is finite (which is easy), then to prove that any finite maximal subfield of $R$ has finite index in $R$ (see Lam's A First Course in Noncommutative Rings, Theorem 15.4).

Keith Kearnes
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  • For the last bit, can't just one do the same proof as for fields ? I.e. a polynomial has a finite number of roots ? – Maxime Ramzi May 11 '18 at 07:52
  • No. Polynomials need not have finitely many roots in a division ring. For example, $x^2+1=0$ has infinitely many roots in the quaternion algebra. – Keith Kearnes May 11 '18 at 07:55
  • Damn me and my commutative mind ! Thank you for your help, I'll try to see why maximal subfields have finite index – Maxime Ramzi May 11 '18 at 12:41
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I had the following idea: since $R$ has no nilpotent elements it should be a subdirect product of integral domains satisfying the same equations - however I know this property for commutative rings, and it relies on the well-known fact that $\displaystyle\bigcap\{p, p\in \mathrm{Spec}R\} = \{x, x \text{ is nilpotent}\}$ - and I don't know whether this is true for noncommutative rings.

That much is true for noncommutative rings.

The condition that $x^n = x$ for some $n$ implies $R$ is von Neumann regular, and it is well-known that a reduced VNR ring (which is called a strongly regular ring) is a subdirect product of division rings. Therefore a subdirectly irreducible ring with these properties is a division ring.

The thing working for us here is that in a strongly regular ring, quotients by prime ideals are division rings, so the prime ideals actually satisfy the commutative definition of "prime." As such, their intersection contains all nilpotent elements, and their intersection is zero. So, you can rely on a similar argument.

ViktorStein
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rschwieb
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  • Isn't $M_n(K)$ a counterexample ? Indeed, the set of nilpotent elements can't be a bilateral ideal (there are no bilateral ideal except for $0$ and $M_n(K)$)... – Maxime Ramzi May 09 '18 at 17:57
  • Ok, I see one inclusion : the forward one, $\subset$, but the $\supset$ one doesn't seem true... except of course in a setting where prime $\implies$ completely prime. I'll try that out, thanks ! – Maxime Ramzi May 09 '18 at 18:10
  • @max I thought we were working under the assumption $R$ is reduced, and $M_n(K)$ is not reduced. The condition that $x^n=x$ trivially implies the ring is reduced. If $x$ is nilpotent, just take any $k>n$ such that $x^k=0$ and $k$ is a power of $n$. Then $0=x^k=x$. – rschwieb May 09 '18 at 18:51
  • Ah alright ! Then yes I agree (the other inclusion is then trivial). I thought you meant the equality was true regardless of $R$ – Maxime Ramzi May 09 '18 at 19:02
  • @Max I guess that is the point of confusion. I'm saying that any reduced ring satisfying $x^n=x$ for some $n$ is a subdirect product of division rings. I wasn't speaking to the proof about the intersection of primes, although I'm pretty sure it is true in this special case. – rschwieb May 09 '18 at 19:04
  • @Max Yeah, it's definitely not true in general. Rings for which the nilpotent elements form an ideal are called $2$-primal, and a nontrivial matrix ring is an example of a ring that isn't $2$-primal. – rschwieb May 09 '18 at 19:05
  • Anyway, thanks to you I've reduced the problem to showing the property you mention, i.e. if $P$ is prime then $R/P$ is a division ring; which must be a classical result (though I don't yet see how to do it ! I'm so used to the commutative setting where it's obvious) – Maxime Ramzi May 09 '18 at 19:35
  • @Max To summarize, in our case, $P$ prime (noncommutative definition) happens to mean $P$ is prime with the commutative definition. That means $R/P$ is a von Neumann regular domain, which is a division ring. That is, if $axa=a$ for nonzero $a$, you cancel $a$ to get $xa=ax=1$, i.e. $a$ is invertible. – rschwieb May 09 '18 at 19:58
  • Yes, yes, I simply need to see why $R/P$ is a domain (i.e. why the two definitions agree here) – Maxime Ramzi May 09 '18 at 22:12
  • @Max I'm trying to think of a slicker proof, but right now I only see a plodding way. $R/P$ is strongly regular, meaning that for every element $a$ there exists $x$ such that $a^2x=a=axa$. You can show that idempotents are central, and $ax$ and $xa$ are idempotents. Then suppose $a,b$ are nonzero, and $axa=a$ so that $xa$ is central. We have that $xaRb=Rxab=0$, and since ${0}$ is prime in $R$ it implies $b=0$ or $ax=0$. This is a contradiction. – rschwieb May 10 '18 at 00:31
  • Sorry, last ax should be xa – rschwieb May 10 '18 at 00:38
  • Mhm I must not have the same definition of prime as you do ! (By the way there's no need to assume $a,b\neq 0$ in your proof, so it can be direct and without proof by contradiction) – Maxime Ramzi May 10 '18 at 09:15
  • Ok I get how you get this property of prime ideals ! Ok pfew I'm almost there, only need to prove that idempotents are central (I'll get it at some point) – Maxime Ramzi May 10 '18 at 09:35