# Let $A$ and $R$ be rings and $f : A → R$ a ring homomorphism.

Let $A$ and $R$ be rings and $f : A → R$ a ring homomorphism.

1. Give a concrete example of $A, R$ and a ring homomorphism $f : A → R$ such that $A$ is commutative and $R$ is not commutative.

2. Prove that if $A$ is commutative and $f$ is surjective, then $R$ is commutative.

Here is my random thought:

1. Not sure(even though matrices count)

2. 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.

• Assertion 2 is even true for semigroups. In your argument you have “for some $a,b\in A$” which should be “for all $a,b\in A$”. Feb 21, 2015 at 21:19
• Thanks for pointing out my mistake. Feb 21, 2015 at 21:50

For the first item, you said it yourself: matrices. Try $$f: (\Bbb R \setminus \{0\}, \cdot) \to ({\rm Mat}(2 \times 2, \Bbb R),\cdot) \quad \text{given by}\quad f(a) = \begin{pmatrix} a & 0 \\ 0 & a\end{pmatrix}.$$ For the second item, your idea is in the right track, let's just rephrase it: take $x,y \in R$. We want to prove that $xy=yx$. Since $f$ is surjective, we have $a,b \in A$ such that $f(a)=x$ and $f(b) = y$. This way:$$xy = f(a)f(b) = f(ab) = f(ba)=f(b)f(a)=yx.$$
• Are you writing down $f$ as a group homomorphism? For ring homomorphisms we can include $0$.