I came across this problem: Let $R$ be a integral domain. Show that the map $b \mapsto ba$, with $a \neq 0$, is injective.
My proof went as follows:
Since $R$ is an integral domain, we have $ab = 0 \Leftrightarrow a = 0$ or $b = 0$. But $a$ is nonzero, therefore $b$ must be $0$ and thus the kernel of the map is $\{0\}$. Thus the map is injective.
First of all, is this correct?
Secondly, my friend pointed out that we're dealing with multiplication here, therefore I should use the multiplicative identity element $1$ for the definition of the kernel, and not the additive identity element $0$. I looking this up and turns out the kernel of a ring homomorphism is defined as follows:
$$ker(f) = \{a \in R : f(a) = 0\}$$
So my friend was wrong(?) But I'm curious, why is the definition like this? Usually, the kernel is just defined through the identity element. But in a ring with $1$ we have two identity elements to choose from.