I need help with the following:
Let $R$ be a ring that is also an integral domain, such that its characteristic is not zero. Let $n \in \mathbb{N}$ be the characteristic of $R$. Show that $n \neq 1$.
Since $R$ is NOT of characteristic $0$, then (in the notation of my Algebra class) we have $o_{R^{+}}=1$. Here $o_{R^{+}}=1$ is the order of $1$ as an element of the additive group $(R, +)$. So for $R$ a positive integer $k$ exists such that $k \cdot a = 0$ for all $ a \in R$, and in this case $n$ is the least such positive integer, so $n\cdot a=0$. By $n \cdot a$, I mean: $(a+ a+...+ a)$ $n$-times.
Now by way of contradiction, suppose $n=1$, then $1\cdot a=0$. So $1=0$ or $a=0$. But $1 \neq 0$, so $a=0$. How can I come to a contradiction? Is $a \neq 0$?
