In my class we define a ring to always have identity element. So then we define a Ring Homomorphisms as follows,
Let $f:R \to S$ be a mapping. Then $f$ is a ring homomorphism if
$$f(x+y) = f(x) + f(y)$$ $$f(xy)=f(x)f(y)$$
$$f(1_R)=1_S$$
I am asked to find all homomorphisms from $\mathbb{}Z_6 \to \mathbb{Z}_{15}$.
My question is, wouldn't there be no Ring Homomorphisms using the definition of a Ring homomorphism that I have?
This is because, let $f$ be some homomorphism satisfying the conditions above then,
$$0=f(0)=f(6)=6f(1)=6$$
which is a contradiciton since $f(0)=0$.
I'm aware of the other definition of a ring homomorphism when a ring isn't defined to always have an identity in which the last condition is relaxed.
Doesn't the definition that I use imply that there will never be a ring homomorphism from $\mathbb{Z}_m \to \mathbb{Z}_n$ whenever $m \neq n$?
Any help is appreciated, thanks!
There exists a ring homomorphism $\mathbf Z/m\mathbf Z\longrightarrow \mathbf Z/n\mathbf Z$ if and only if $n\mid m$, since we must have $m \cdot 1\bmod m\implies m\cdot1\bmod n=0$, in other words $n\equiv 0\mod n$.