# Examples of Modules - For Closed Left Ideals J, X / J is a Left X-Module

Let $X$ be an algebra and let $J$ be a closed left ideal. Recall that if $\sim$ is the equivalence relation defined for all $x, y \in X$ by $x \sim y$ if and only if $x - y \in J$ then $X / J$ is the set of all equivalence classes and $X / J$ with the operation of addition defined for all $z', w' \in X / J$ by $z' + w' = (x + y)'$, $az' = (az)'$, and $z'w' = (xy)'$ where $x \in z'$, $y \in w'$, and $a \in \mathbf{F}$ forms an algebra called the quotient algebra of $X$ modulo $J$.

The quotient algebra $X / J$ is an $X$-algebra with module multiplication $f : X \times X / J \to X / J$ defined for all $x \in X$ and all $z' \in X / J$ by:

(1)Where $y \in z'$

For each fixed $x \in X$ the map $f_x : X / J \to X / J$ defined by $f_x(z') = (xy)'$, $y \in z'$ is linear since for all $z_1', z_2' \in X / J$ we have that $(x[y_1 + y_2])' = (xy_1 + xy_2)' = (xy_1)' + (xy_2)'$, where $y_1 \in z_1'$ and $y_2 \in z_2'$. So axiom $LM1$ is satisfied.

Similarly, for each fixed $z' \in X / J$ the map $f_z : X \to X / J$ defined by $f_{z'}(x) = (xy)'$, $y \in z'$ is linear since for all $x_1, x_2 \in X / J$ we have that $([x_1 + x_2]y)' = (x_1y + x_2y)' = (x_1y)' + (x_2y)'$, where $y \in z'$. So axiom [[$ LM2 $] is satisfied.

Lastly, for all $x, y \in X$ and all $z' \in X / J$ we have that:

(2)By choosing the representative $t = ys$ we have that $x(yz') = xyz' = (xy)z'$. So axiom $LM3$ is satisfied. So indeed, $X / J$ is a left $X$-module.

Now since $J$ is closed, we are able to define a norm on $X / J$ for all $z \in X / J$ by:

(3)For all $x \in X$ and all $z' \in X / J$ we have that:

(4)(The equality at $(*)$ comes from the fact that the canonical map $q : X \to X / J$ is an isometry.)

It can similarly be shown that if $J$ is a closed right ideal then $X / J$ is a right $X$-module.