The Quotient Algebra of an Algebra A Modulo a Two-Sided Ideal J

# The Quotient Algebra of an Algebra X Modulo a Two-Sided Ideal J

Definition: Let $\mathfrak{A}$ be an algebra and let $J \subseteq \mathfrak{A}$ be a two-sided ideal of $\mathfrak{A}$. Define an equivalence relation $\sim$ on $\mathfrak{A}$ where for all $x, y \in \mathfrak{A}$ we have $x \sim y$ if and only if $x - y \in J$. For each $x \in J$ let $x'$ denote the equivalence class containing $x$, that is, $x' = \{ y \in \mathfrak{A} : x - y \in J \}$. Let $\mathfrak{A} / J = \{ x' : x \in \mathfrak{A} \}$ denote this set of equivalence classes. Then the Quotient Algebra of $\mathfrak{A}$ Modulo $J$ is the set $\mathfrak{A} / J$ with the operations of addition, scalar multiplication, and multiplication defined respectively for all $z, w \in \mathfrak{A} / J$ by:1) $z + w = (x + y)'$ where $x \in z$ and $y \in w$.2) $az = (ax)'$ where $x \in z$.3) $zw = (xy)'$ where $x \in z$ and $y \in w$. |

If $\mathfrak{A}$ is a normed algebra $J \subseteq \mathfrak{A}$ is a closed linear subspace of $\mathfrak{A}$, we can define a norm $\| \cdot \| : \mathfrak{A}/J \to [0, \infty)$ defined for all $z \in \mathfrak{A}/J$ by:

(1)\begin{align} \quad \| z \| &= \inf \{ \| x \| : x \in z \} \\ &= \inf \{ \| z - j \| : j \in J \} \end{align}

Definition: Let $\mathfrak{A}$ be a normed algebra and let $J \subseteq \mathfrak{A}$ be a closed linear subspace of $\mathfrak{A}$. The Canonical Map is the map $q : \mathfrak{A} \to \mathfrak{A} / J$ defined for all $x \in \mathfrak{A}$ by $q(x) = x'$. |

Proposition 1: Let $\mathfrak{A}$ be a normed algebra and let $J \subseteq \mathfrak{A}$ be a closed linear subspace. Let $q : \mathfrak{A} \to \mathfrak{A} / J$ be the canonical map. Then:a) If $J$ is a proper subspace of $\mathfrak{A}$ then $\| q \| = 1$.b) If $J = \mathfrak{A}$ then $\| q \| = 0$. |

**Proof of a)**Suppose that $J$ is a proper subspace of $\mathfrak{A}$. For each $x \in \mathfrak{A}$ we have that:

\begin{align} \quad \| q(x) \| = \| x' \| = \inf \{ \| y \| : y \in x' \} \leq \| x \| \end{align}

- (Where the inequality comes from the fact that $x \in x'$). So $\| q \| \leq 1$.

- On the other hand, since $J$ is a proper subspace of $\mathfrak{A}$, we can take $z \in \mathfrak{A} \setminus J$.

- Since $J$ is closed we have that $\| q(z) \| > 0$. Let $\epsilon > 0$ be such that $\epsilon < 1$. Then:

\begin{align} \quad \frac{\| q(z) \|}{1 - \epsilon} > \| q(z) \| = \inf \{ \| y \| : y \in z' \} \end{align}

- By the definition of infimum, there must exist a point $y \in z'$ such that:

\begin{align} \quad \frac{\| q(z) \|}{1 - \epsilon} &> \| y \| \\ \quad \| q(z) \| &> \| y \| (1 - \epsilon) \end{align}

- But since $y \in z'$ we have that $\| q(y) \| = \| q(z) \|$, and from above we have that:

\begin{align} \quad \| q(y) \| & > \| y \| (1 - \epsilon) \end{align}

- Since $z - y \in J$ we see that $y \neq 0$ otherwise $z$ would be in $J$ (which it cannot be since $z \in \mathfrak{A} / J$). Thus $\| y \| \neq 0$. Since for all $0 < \epsilon < 1$ there exists such a $y$ which makes the above inequality hold, we conclude that $\| q \| \geq 1$.

- Thus $\| q \| = 1$. $\blacksquare$

**Proof of b)**Suppose that $J = \mathfrak{A}$. Let $x \in \mathfrak{A}$. Then for every $y \in \mathfrak{A}$ we have that $x - y \in \mathfrak{A} = J$. So the only $J$-coset is $0'$.

- If $q : \mathfrak{A} \to \mathfrak{A} / J$ is the canonical map then $q(x) = 0$ for every $x \in \mathfrak{A}$ and so clearly $\| q \| = 0$. $\blacksquare$