Let $R$ be the ring of all $2\times 2$ matrices with integer coefficients:
\[R=\left\{\, \begin{bmatrix}
a & b\\
c& d
\end{bmatrix} \quad \middle| \quad a, b, c, d\in \Z \,\right\}.\]

Let $S$ be the subset of $R$ given by
\[S=\left\{\, \begin{bmatrix}
s & 0\\
0& s
\end{bmatrix} \quad \middle | \quad s\in \Z \,\right\}.\]

In fact, let
\[A= \begin{bmatrix}
t & 0\\
0& t
\end{bmatrix}\text{ and } B=\begin{bmatrix}
s & 0\\
0& s
\end{bmatrix}\]
be arbitrary elements in $S$ with $t, s\in \Z$.

Then we have
\begin{align*}
A+B=\begin{bmatrix}
t+s & 0\\
0& t+s
\end{bmatrix} \in S
\end{align*}
and
\begin{align*}
AB=\begin{bmatrix}
ts & 0\\
0& ts
\end{bmatrix} \in S.
\end{align*}
Hence $S$ is closed under addition and multiplication.

Note that the $2\times 2$ identity matrix is the unity element of $R$ as well as the unity element of $S$.

Thus, the subset $S$ is a subring of $R$.

(b) True or False: $S$ is an ideal of $R$.

False.

To see that $S$ is not an ideal of $R$, consider the element
\[\begin{bmatrix}
1 & 1\\
1& 1
\end{bmatrix} \in R\]
and the element
\[\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix} \in S.\]
Then we have
\begin{align*}
\begin{bmatrix}
1 & 1\\
1& 1
\end{bmatrix}\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix}=\begin{bmatrix}
1 & 1\\
1& 1
\end{bmatrix},
\end{align*}
which is not in $S$.

This implies that $S$ is not an ideal of $R$.
(If $S$ were an ideal of $R$, then an element of $S$ multiplied by an element of $R$ would stay in $S$.)

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