## Central multiples of a module are submodules

Let $R$ be a ring with 1, let $M$ be a left $R$-module, and let $z$ be an element in the center of $R$. Prove that $z \cdot M$ is a submodule of $M$. If $F$ is a field, $R = \mathsf{Mat}_2(F)$, $M = R$ a left $R$-module under multiplication, and $e = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$, then $eR$ is not a left $R$-submodule of $R$.

Let $z \in Z(R)$. We use the submodule criterion to show that $z \cdot M$ is a submodule of $M$. Note first that $z \cdot 0 \in z \cdot M$, so that $z \cdot M$ is not empty. Now suppose $z \cdot x, z \cdot y \in z \cdot M$ and $r \in R$. Then $(z \cdot x) + r \cdot (z \cdot y) = z \cdot x + rz \cdot y$ $= z \cdot x + zr \cdot y = z \cdot (x + r \cdot y) \in z \cdot M$. Thus $z \cdot M$ is a submodule of $M$.

Now for the example, note that $eR = \left\{ \begin{bmatrix} a & b \\ 0 & 0 \end{bmatrix} \ |\ a,b \in F \right\}$. Letting $a = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $r = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$, we have $r \cdot (ea) = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \notin eR$. Thus $eR$ is not a submodule of $R$.