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Let \(\displaystyle{f:R\longrightarrow S}\) be a ring homomorphism which is onto \(\displaystyle{S}\) .

1. If the ring \(\displaystyle{\left(R,+,\cdot\right)}\) is commutative, then prove that the ring \(\displaystyle{\left(S,+,\cdot\right)}\)

is commutative.

2. Give an example of a ring homomorphism \(\displaystyle{f:R\longrightarrow S}\) which is onto \(\displaystyle{S}\), where the ring

\(\displaystyle{\left(S,+,\cdot\right)}\) is commutative but the ring \(\displaystyle{\left(R,+,\cdot\right)}\) is not commutative.

Hello!

Here is an answer to the second part.Let \(R\) be the ring of the 2X2 upper triangular real matrices (which is a subring of \(M_2(\mathbb R))\).

Consider a mapping \( \phi : R\rightarrow\mathbb R \) such that

\( \phi\left( \begin{array}{cc}a & b \\0 & c \end{array} \right)=a\)

Obviously, \( \phi \) is onto \( \mathbb R\) .

Also, since :
\( \left( \begin{array}{cc}a & b \\0 & c \end{array} \right)\left( \begin{array}{cc}d & e \\0 & f \end{array} \right)=\left( \begin{array}{cc}ad & ae+bf \\0 & cf \end{array} \right) \)

and \( \left( \begin{array}{cc}a & b \\0 & c \end{array} \right) + \left( \begin{array}{cc}d & e \\0 & f \end{array} \right)= \left( \begin{array}{cc}a+d & b+e \\0 & c+f \end{array} \right)\)

we can see that \( \phi \) is a ring homomorphism onto \(\mathbb R\) with \(R\) being non-commutative.

To conclude the exercise, let's prove the first part:

Consider arbitrary elements $ s_{1},s_{2} \in S $. Since $f$ is onto $S$, there exist $ r_{1},r_{2} \in R $ such that $f(r_{1}) = s_{1}$ and $f(r_{2}) = s_{2} $. Now, by using the facts that $f$ is a ring homomorphism and $R$ is a commutative ring, we have that \[ s_{1} s_{2} = f(r_{1}) f(r_{2}) = f(r_{1} r_{2}) = f(r_{2} r_{1}) = f(r_{2}) f(r_{1}) = s_{2} s_{1} \]Therefore, $S$ is commutative.