Is the function $f: \mathbf{Z}_5 \rightarrow \mathbf{Z}_{30}$ given by $f(x)=6x$ a ring homomorphism?\
Note that the definition of a ring homomorphism: a ring A to a ring B is a function $f: A \rightarrow B$ satisfying the identities $f(x_1 + x_2)= f(x_1)+f(x_2)$ and $f(x_1 \times x_2)= f(x_1) \times f(x_2)$\
if $f(x)=6x$ and we let $a,b \in \mathbf{Z}_5$
then $f(a)+f(b)= 6a+6b$
$=6(a+b)$
$f(a+b)$
Also
$f(a)f(b)=6a \times 6b$
$=36ab$
$=6ab(\mod 30)$
$=f(ab)$
Thus by definition Ring Homorphism, f is a ring homomorphism.
Any suggestions on improving this?
Is there any other way one would go about this? I think I could draw on the fact their isomorphic, but I think that would be a longer proof?