This is from Judson's Abstract Algebra textbook.
We define the $\underline{characteristic}$ of a ring R to be the least positive integer $n$ s.t. $nr=0, \forall r\in R$. If there is no such n, we say that the characteristic is 0.
Lemma 16.18: Let R be a ring with identity. If 1 has order n, then the characteristic of R is n.
I am having a hard time understanding this Lemma. First, why do we suppose 1 has order n? I read in another similar post that this means that n1 = 0, but I do not see that connection and how that could be derived from Lemma 16.18.
Could someone please give me an example to help understand this.
Could an application of Lemma 16.18 be the following problem:
What is the characteristic of the field formed by the set of matrices
F= $\{ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \}$ with entries in $\mathbb{Z}_2$?
Answer: Characterisitic of this field is 2, since $1+1=2(1)\equiv0\mod2$. So, 1 has order 2.
Suppose that $n$ has order $1$. Then $n$ is the least positive integer such that $\underbrace{1 + {}\dotsm {} + 1}_{\text{$n$ times}} = 0$. It follows that, for every $r \in R$, $nr = 0$ since $$ nr = \underbrace{r + {}\dotsm{} + r}_{\text{$n$ times}} = r(\underbrace{1 + {}\dotsm {} + 1}_{\text{$n$ times}}) = r0 = 0 $$ Moreover $n$ is the least positive integer such that, for all $r \in R$, $nr = 0$ since $n$ is the order of $1$. Thus $R$ has characteristic $n$.