What possible characteristics may a local commutative unital ring have?

Question

What possible characteristics may a local commutative unital ring have?

We define the characteristic of a commutative unital ring $R$ in the following way: there is a unique ring homomorphism $\phi: \mathbb{Z} \to R$ defined as $\phi(n)=n \cdot 1$. Since $\mathbb{Z}$ is a PID, $\mathrm{ker}(\phi)$ is a principal ideal. If $\text{ker}(\phi)=c\mathbb{Z}$, then we'll say that the characteristic of $R$ is $c$.

We can prove that the characteristic of a domain is either $0$ or a prime $p$. But how do we classify the characteristic of a local ring $R$ with max ideal $\mathfrak{m}$?

Answer 1

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.

Answer 2

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.

Answer 3

Using the definition of the characteristic being the unique map out of $\mathbb{Z}$, here's a different approach: $R/\mathfrak{m}$ will be a field, and so must have characteristic either $0$ or $p$. and so the kernel of the unique homomorphism $\iota\colon\mathbb{Z} \to R/\mathfrak{m}$ will be either $(0)$ or $(p)$ respectively. Now $\iota$ will pull back to a ring homomorphism $\tilde{\iota}\colon \mathbb{Z} \to R$ such that, letting $\pi \colon R \to R/\mathfrak{m}$ be the quotient map, we have $\tilde{\iota}\pi = \iota$. But then $\mathrm{ker}(\tilde{\iota}) \subset \mathrm{ker}(\iota)$ as a subgroup, and so $\mathrm{ker}(\tilde{\iota})$ must be of the form either $(0)$ or $(p^n)$ depending on the characteristic of $R/\mathfrak{m}$.