# What is the Characteristic of a local ring? [closed]

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 $$(A,m)$$

## closed as unclear what you're asking by Arnaud D., max_zorn, Javi, José Carlos Santos, RamiroApr 19 at 16:25

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• 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. – Claudius Apr 18 at 12:23
• 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? – Arnaud D. Apr 18 at 12:28
• 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? – Claudius Apr 18 at 12:29
• Like integral domain you can classify the characteristic of a local ring. – user371231 Apr 18 at 12:45
• I much as I guess $6$ cannot be characteristic of a local ring – user371231 Apr 18 at 12:49

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.

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.