Let $A$ and $R$ be rings and $f : A → R$ a ring homomorphism.
Give a concrete example of $A, R$ and a ring homomorphism $f : A → R$ such that $A$ is commutative and $R$ is not commutative.
Prove that if $A$ is commutative and $f$ is surjective, then $R$ is commutative.
Here is my random thought:
Not sure(even though matrices count)
Rough sketch: Since, $A$ is commutative, $ab = ba$, for some $a,b \in A$. And, since $f$ is a ring homomorphism, $f(ab) = f(a)f(b)$ and $f(a) = x$ and $f(b) = y$, for some $x,y \in R$, since $f$ is surjective. Then, $f(ab) = f(a)f(b) = f(b)f(a) = f(ba)$...Since A is commutative. Hence, $f(a)f(b)= xy = f(b)f(a) = yx.$ In particular, $xy = yx$ for some $x,y \in R$ Does this make sense? Please correct me if I am wrong.