For Closed Left Ideals J, A / J is a Normed Left A-Module

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

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

The quotient algebra $\mathfrak{A} / J$ then becomes an $\mathfrak{A}$-module with module multiplication $f : \mathfrak{A} \times \mathfrak{A} / J \to \mathfrak{A} / J$ defined for all $x \in \mathfrak{A}$ and all $z' \in \mathfrak{A} / J$ by:

(1)
\begin{align} \quad f(x, z') = (xy)' \end{align}

Where $y \in z'$. We now show that $\mathfrak{A}/J$ is a normed left $\mathfrak{A}$-module.

• 1. For each fixed $x \in \mathfrak{A}$ the map $f_x : \mathfrak{A} / J \to \mathfrak{A} / J$ defined by $f_x(z') = (xy)'$, $y \in z'$ is linear since for all $z_1', z_2' \in \mathfrak{A} / J$ 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.
• 2. Similarly, for each fixed $z' \in \mathfrak{A} / J$ the map $f_z : \mathfrak{A} \to \mathfrak{A} / J$ defined by $f_{z'}(x) = (xy)'$, $y \in z'$ is linear since for all $x_1, x_2 \in \mathfrak{A} / J$ 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.
• 3. For all $x, y \in \mathfrak{A}$ and all $z' \in \mathfrak{A} / J$ we have that:
(2)
\begin{align} \quad x(yz') = x(ys)' \: (s \in z') = (xt) \: t \in ys \end{align}
• By choosing the representative $t = ys$ we have that $x(yz') = xyz' = (xy)z'$. So axiom $LM3$ is satisfied. So indeed, $\mathfrak{A} / J$ is a left $\mathfrak{A}$-module.
• 4. Lastly, for all $x \in \mathfrak{A}$ and all $z' \in \mathfrak{A} / J$ we have that:
(3)
\begin{align} \quad \| xz' \|_{\mathfrak{A}/J} = \inf \{ \| y \|_{\mathfrak{A}} : y \in xz \} \leq \| xz \|_{\mathfrak{A}} \leq \| x \|_{\mathfrak{A}} \| z \|_{\mathfrak{A}} \overset{(*)} = \| x \|_{\mathfrak{A}} \| q(z) \|_{\mathfrak{A}/J} = \| x \|_{\mathfrak{A}} \| z' \|_{\mathfrak{A}/J} \end{align}
• (The equality at $(*)$ comes from the fact that the canonical map $q : \mathfrak{A} \to \mathfrak{A} / J$ is an isometry.)

So $\mathfrak{A}/J$ is a normed left $\mathfrak{A}$-module.

It can similarly be shown that if $J$ is a closed right ideal then $\mathfrak{A} / J$ is a normed right $\mathfrak{A}$-module.