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What is the Characteristic of a local ring ?

We define Characteristic of a Commutative ring with $1$ say, $A$ in the following way: Define a ring homomorphism $\phi: \mathbb{Z} \to A$ by $\phi(n)=n \cdot 1.$ Since $\mathbb{Z}$ is a PID, $\text{ker}(\phi)$ is a principal ideal. If $\text{ker}(\phi)=m\mathbb{Z},$ we define the Characteristic of the ring $A$ to be $m.$ We know that Characteristic of a domain is either $0$ or a prime $p.$

How do I classify the Characteristic of a local ring $(A,m)$

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    $\begingroup$ The characteristic of a ring need not be prime: for instance, the characteristic of $\mathbb Z/(6)$ is $6$. It only needs to be a prime if the ring is an integral domain. $\endgroup$ – Claudius Apr 18 at 12:23
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    $\begingroup$ I don't really understand the question. You have the definition of the characteristic of a ring, so theoretically you now how to find it, right? $\endgroup$ – Arnaud D. Apr 18 at 12:28
  • $\begingroup$ It is not clear to me what you mean by “find the characteristic“. Are you asking about the definition of the characteristic of a local ring? $\endgroup$ – Claudius Apr 18 at 12:29
  • $\begingroup$ Like integral domain you can classify the characteristic of a local ring. $\endgroup$ – user371231 Apr 18 at 12:45
  • $\begingroup$ I much as I guess $6$ cannot be characteristic of a local ring $\endgroup$ – user371231 Apr 18 at 12:49
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The characteristic of a local ring is a power of a prime or $0$, and any of these happens in some local rings.

That they all happen is easy : you may look at fields for characteristic $0$, and $\mathbb{Z}/p^n\mathbb{Z}$ for powers of primes.

Now let $(R,m)$ be a local ring, and $n$ its characteristic, which we assume to be $>0$. Suppose $n=ab, a\land b = 1$. Then the ideals $I=\{x\in R, ax = 0\}$ and $J=\{x\in R, bx=0\}$ are comaximal : indeed $a\in J, b\in I$ and there are $u,v$ with $au+bv=1$ so $1\in I+J$.

Therefore by locality, one of them is $R$ (otherwise they would both be $\subset m$). If it is $I$, then $a = 0$ in $R$ and so $R$ has characteristic $\mid a$ so $b=1$. If it's $J$, then $a=1$. In any case, $a=1 \lor b=1$, so that $n$ is a power of a prime.

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If you already know a local ring has only trivial idempotents, then you can reason this way:

Suppose the characteristic of a ring $R$ is finite, say $n$, and is divisible by more than one prime. The ring contains a copy of $\mathbb Z/n\mathbb Z$. So to show $R$ isn't local, it suffices to show that $\mathbb Z/n\mathbb Z$ contains a nontrivial idempotent, so that $R$ will also contain a nontrivial idempotent.

By reasoning with the Chinese remainder theorem, you can quickly see that $\mathbb Z/n\mathbb Z$ is isomorphic to $\mathbb Z/p^k\mathbb Z$ for primes $p$ dividing $n$ and powers $k$ depending on $p$. Since there is more than one prime dividing $n$ (our assumption) there is at least one nontrivial idempotent splitting $\mathbb Z/n\mathbb Z$ into two pieces. This is clearly a nontrivial idempotent of $R$ too.

So by contrapositive, we have shown that a local ring $R$ must have either characteristic $0$, or else it has finite characteristic that is a power of a prime.

Since we already have examples of such rings for every such finite characteristic, we can see these are precisely the characteristics that are possible.

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