Can a finite ring have a characteristic $0$?

Question

I am working on studying Abstract Algebra and doing one of the review problems for the exam. I do not think a finite ring can have characteristic zero because when I think of, for example, Integers mod $4$, this ring has a characteristic greater than $0$. I also remember learning about infinite rings and I almost naturally associate an infinite ring with having a characteristic of $0$. Could anyone help me prove the above statement? I am thinking of starting off with a contradiction: let R be a finite ring with characteristic $0$...

Answer 1

... then $z\cdot 1\neq 0$ for all $z\in \mathbb Z\setminus\{0\}$. That is, having characteristic $0$ means that nonzero integer multiples of $1$ are always nonzero.

Then $\{z\cdot 1\mid z\in\mathbb Z\setminus\{0\}\}$ must be finite, since it's a subset of your finite set.

Then $z\cdot 1 = z'\cdot 1$ for $z,z'$ two distinct nonzero integers.

Reorganizing that you get $(z-z')\cdot 1=0$. But $z-z'$ is a nonzero integer, so this contradicts what we said before that nonzero integer multiples of $1$ are always nonzero.

Answer 2

To have characteristic $0$ means that for no finite sum $1 +1 + \cdots +1$ do you ever get a repeat. So, they're all different, so the ring must be infinite.

To be clear, though, an infinite ring MAY have positive characteristic. Take $R = \mathbb{Z}_2[x]$, e.g.

Answer 3

If you consider the characteristic of a ring $R$ as being the natural number $n$ such that $$\frac{\mathbb{Z}}{n\mathbb{Z}}\cong (1),$$ where $(1)$ denotes de subring generated by the multiplicative identity, which is sometimes called the prime ring of $R$, then you can easily see that a ring of characteristic $0$ has to be infinite, for it contains an infinite subring.

Edit: this applies only if your ring has a multiplicative identity, if it doesn't, then the answer given by Slade is more suitable.

Answer 4

I'm not sure whether your ring has unity, so here is a general argument.

If a ring $R$ has characteristic zero, then, for each $n>0$, there is some $a\in R$ with additive order $>n$. This means the set $\{0,a,2a,\ldots ,na\}$ consists of distinct elements (this requires the existence of additive inverses), so that $|R|>n$.

Since $n$ was arbitrary, $R$ is infinite.

This argument is a little simpler if we can assume that $R$ has unity. In that case, $R$ must actually contain $\mathbb{Z}$ as an additive subgroup.