If A has characteristic 3, and 5a = 0, then a = 0.

Question

[A. 3, P. 202, A Book of Abstract Algebr by Pinter] Let A be a finite integral domain. Prove that if A has characteristic 3, and 5a = 0, then a = 0.

Now, this is how I try to prove it. since characteristic with 3, it must be a group with three elements, i.e. $Z_3$. Since $Z_3$ is a an integral domain, then it has no zero divisors. Thus, we can use cancellation property. I.e., $5a=0$ then either 5=0 or a=0 and since $5 \neq 0$, then it must be that $a=0$. Done.

Is there something wrong or is my proof isn't enough since this is first exercise I did in characteristic of a ring and I don't really fully understand.

Thanks in advance.

Answer 1

Hint $ $ the kernel of $\,n\to n\cdot a\,$ is an ideal in $\Bbb Z$ containing $\,3,5\,$ so it contains $\,\gcd (3,5)=1$

More simply: such $n$ are closed under subtraction, and contains $3$ and $5$ so contains $\,\gcd(3,5).\,$

More conceptually: the additive order of $a$ divides the coprimes $3$ and $5$ so it must be $1$.

More generally: see order ideals and denominator ideals.