Let $S = \left\{\begin{bmatrix}a&b\\0&0\end{bmatrix} :a,b \in \mathbb {R} \right\}$. Prove that the set $S$ is a subring of $M_2 (R)$ with its usual addition and multiplication. Is $S$ a ring with identity? Is it a commutative ring? Describe the units of this ring.
yes this is a subring of $M_2 (R)$, $S$ has an identity but its not the identity in $M_2 (R)$ it is commutative and it has no units.
Does the fact that the identity is not the same matter?
Best Answer
$ \begin{bmatrix}1&0\\0&0\end{bmatrix} \in S$ so S is non empty.
$ \begin{bmatrix}a&b\\0&0\end{bmatrix} \begin{bmatrix}c&d\\0&0\end{bmatrix}= \begin{bmatrix}ac+0&ad+0\\0&0\end{bmatrix}$ $ac,ad \in \mathbb {R}$ closed under multiplication
$ \begin{bmatrix}a&b\\0&0\end{bmatrix} + \begin{bmatrix}c&d\\0&0\end{bmatrix} = \begin{bmatrix}a+c&b+d\\0&0\end{bmatrix}$ $(a+c),(b+d) \in \mathbb {R}$ closed under addition
$ \begin{bmatrix}a&b\\0&0\end{bmatrix} \begin{bmatrix}c&d\\0&0\end{bmatrix}= \begin{bmatrix}ac+0&ad+0\\0&0\end{bmatrix}$
$ \begin{bmatrix}c&d\\0&0\end{bmatrix} \begin{bmatrix}a&b\\0&0\end{bmatrix}= \begin{bmatrix}ca+0&cb+0\\0&0\end{bmatrix}$
so not a commutative ring.
$\begin{bmatrix}1&0\\0&1\end{bmatrix} $ is not in S so S has no unity.
Since $ \forall s \in S $ the det(s)=0 nothing is a unit.