Let $f:R \rightarrow S$ be a ring homomorphism that is onto $S$. If the ring $(R, +, \cdot)$ is commutative then prove that the ring $(S, +, \cdot)$ is commutative. In continuity , give an example that the converse is not true.
Solution
We are proving that the ring $(S, +, \cdot)$ is commutative.
Let $a, b \in S$. Since $f$ is onto there exist $c, d \in R$ such that $f(c)=a, \; f(d)=b$.
$$ab= f(c)f(d)= f(cd)= f(dc)= f(d)f(c)=ba$$
That the converse does not necessarily hold we see that by picking a $2 \times 2$ real matrix that is upper triangular and diagonizable :
$$f \begin{pmatrix}
a &b \\
0& c
\end{pmatrix} = \begin{pmatrix}
a &0 \\
0& c
\end{pmatrix}$$
we easily check that the given function has all of the above properties.
Solution
We are proving that the ring $(S, +, \cdot)$ is commutative.
Let $a, b \in S$. Since $f$ is onto there exist $c, d \in R$ such that $f(c)=a, \; f(d)=b$.
$$ab= f(c)f(d)= f(cd)= f(dc)= f(d)f(c)=ba$$
That the converse does not necessarily hold we see that by picking a $2 \times 2$ real matrix that is upper triangular and diagonizable :
$$f \begin{pmatrix}
a &b \\
0& c
\end{pmatrix} = \begin{pmatrix}
a &0 \\
0& c
\end{pmatrix}$$
we easily check that the given function has all of the above properties.
Hello Tolaso J.Kos.
ReplyDeleteHere is an analytical proof that $f$ is a ring homomorphism and onto. Let $\begin{pmatrix} a & 0\\ 0 & c \end{pmatrix}\in S$ Then, $f\,\left(\begin{pmatrix} a & 1\\ 0 & c \end{pmatrix}\right)=\begin{pmatrix} a & 0\\ 0 & c \end{pmatrix}$ so the function $f$ is onto $S$. Consider $A=\begin{pmatrix} a_1 & b_1\\ 0 & c_1 \end{pmatrix} , \; B=\begin{pmatrix} a_2 & b_2\\ 0 & c_2 \end{pmatrix}\in R$. We have that : $$A+B=\begin{pmatrix} a_1+a_2 & b_1+b_2\\ 0 & c_1+c_2 \end{pmatrix}$$
and $$A\cdot B=\begin{pmatrix} a_1\,a_2& a_1\,b_2+b_1\,c_2\\ 0 & c_1\,c_2 \end{pmatrix}$$
and thus
$$f(A+B)=\begin{pmatrix} a_1+a_2 & 0\\ 0 & c_1+c_2 \end{pmatrix}=f(A)+f(B)$$
and
$$f(A\cdot B)=\begin{pmatrix} a_1\,a_2 & 0\\ 0 & c_1\,c_2 \end{pmatrix}=f(A)\cdot f(B)$$ Finally, $f(I_2)=I_2$. We conclude that the function $f$ is a ring homomorphism. Although the ring $\left(S,+,\cdot\right)$ is commutative, the ring $\left(R,+,\cdot\right)$ is not:
$$\begin{pmatrix}
0& 1\\
0 & 0
\end{pmatrix}\cdot \begin{pmatrix}
1 & 0\\
0 & 0
\end{pmatrix}=\mathbb{O}\neq \begin{pmatrix}
0 & 1\\
0 & 0
\end{pmatrix}=\begin{pmatrix}
1 & 0\\
0 & 0
\end{pmatrix}\cdot \begin{pmatrix}
0& 1\\
0 & 0
\end{pmatrix}$$