Max. Modular Two-Sided Ideals of Codim. 1 in a Banach Algebra over C

# Characterization of Maximal Modular Two-Sided Ideals of Codimension 1 in a Banach Algebra over C

Maximal Modular Two-Sided Ideals of Codimension 1 in a Banach Algebra

Recall from the Multiplicative Linear Functionals on a Banach Algebra page that if $\mathfrak{A}$ is an algebra then a linear functional $f : \mathfrak{A} \to \mathbf{F}$ is said to be multiplicative if for all $x, y \in \mathfrak{A}$ we have that:

(1)
\begin{align} \quad f(xy) = f(x)f(y) \end{align}

We proved that if $\mathfrak{A}$ is a Banach algebra then every multiplicative linear functional $f$ on $\mathfrak{A}$ is bounded with $\| f \| \leq 1$. We also proved that if $\mathfrak{A}$ is a Banach algebra with unit and $f$ is a multiplicative linear functional on $\mathfrak{A}$ then $\| f \| = 1$. We now, in part, classify multiplicative linear functionals for a Banach algebra $\mathfrak{A}$ in terms of their kernels.

 Proposition 1: Let $\mathfrak{A}$ be a Banach algebra over $\mathbb{C}$. Then every maximal modular two-sided ideals of $\mathfrak{A}$ with codimension $1$ is the kernel of some multiplicative linear functional on $\mathfrak{A}$, and conversely, the kernel of every multiplicative linear functional on $\mathfrak{A}$ is a maximal modular two-sided ideal of $\mathfrak{A}$.

Here the notation "$\mathbb{C}y$" is used to denote the space of $y$, i.e., $\mathbb{C}y = \{ \alpha y : \alpha \in \mathbb{C} \}$.

• Proof: $\Rightarrow$ Let $M$ be a maximal modular two-sided ideal of $\mathfrak{A}$ with codimension $1$ and let $u \in \mathfrak{A} \setminus M$ be a two-sided modular unit for $\mathfrak{A}$. Then $(1 - u)\mathfrak{A} \subseteq M$ and $\mathfrak{A}(1 - u) \subseteq M$.
• Consider the quotient space $\mathfrak{A} / M$, which has dimension $1$. (Note that since $\mathfrak{A}$ is a Banach algebra and $M$ is a maximal modular two-sided ideal of $\mathfrak{A}$ we have by the theorem on the Basic Theorems Regarding Ideals in an Algebra 2 page that $M$ is closed so that $\mathfrak{A} / M$ is defined). We can write $\mathfrak{A} / M = \mathbb{C}u'$ where $u'$ is the canonical image of of $u$ under the quotient map $q : \mathfrak{A} \to \mathfrak{A}/M$ defined by $q(x) = x'$.
• Since $\mathfrak{A} / M = \mathbb{C}u'$ we have that for each $x \in \mathfrak{A}$ there exists a number $f(x) \in \mathbb{C}$ such that $x' = f(x)u'$. We claim that $f$ is a multiplicative linear functional.
• Showing that $f$ is linear is straightforward. If $x, y \in \mathfrak{A}$ then since $q : \mathfrak{A} / M \to \mathfrak{A}$ is linear we have that $f(x + y)u' = (x + y)' = x' + y' = f(x)u' + f(y)u'= [f(x) + f(y)]u'$ and so $f(x + y) = f(x) + f(y)$. If $x \in \mathfrak{A}$ and $\alpha \in \mathbb{C}$ then again since $q$ is linear we have that $f(ax)u' = (ax)' = ax' = af(x)u'$ and so $f(ax) = af(x)$.
• Lastly, observe that for $u \in \mathfrak{A}$ we have that $u' = f(u)u'$ and so $f(u) = 1$. For $x, y \in \mathfrak{A}$. Since $q$ is a homomorphism we have that:
(2)
\begin{align} \quad f(xy)u' = q(xy) = q(x)q(y) = f(x)u' f(y)u' = f(x)f(y) u'u' = f(x)f(y) (uu)' \end{align}
• Thus: $f(f(x)f(y)uu) = f(xy)$ from above. Since $f$ is linear (as proven above), $f(x)f(y)f(uu) = f(xy)$. Observe that since $J$ is a modular we have that $\mathfrak{A}(1 - u) \subseteq J$ and $(1 - u)\mathfrak{A} \subseteq J$ so that in particular, $u - uu \in J$, so $q(u) = q(uu)$, i.e., $u' = (uu)'$. So $f(u) = f(uu) = 1$. Thus $f(xy) = f(x)f(y)$ for all $x, y \in \mathfrak{A}$, showing that $f$ is multiplicative.
• $\Leftarrow$ Let $f$ be a multiplicative linear functional on $\mathfrak{A}$. Let $J = \ker (f) = \{ x \in \mathfrak{A} : f(x) = 0 \}$. Then $J$ is a subspace of $\mathfrak{A}$. Furthermore, since $f$ is multiplicative by definition we have that $f \neq 0$ and so $J$ is a proper subspace of $\mathfrak{A}$.
• We will now show that $J$ is a maximal modular two-sided ideal of $\mathfrak{A}$.
• 1. Showing that $J$ is a two-sided ideal of $\mathfrak{A}$: Let $x \in \mathfrak{A}$ and let $j \in J$. Then $f(j) = 0$. So $f(jx) = f(j)f(x) = 0f(x) = 0$ which shows us that $jx \in J$. Similarly, $f(xj) = f(x)f(j) = f(x)0 = 0$ which shows us that $xj \in J$. Thus $J\mathfrak{A} \subseteq J$ and $\mathfrak{A}J \subseteq J$, i.e., $J$ is a two-sided ideal of $\mathfrak{A}$.
• 2. Showing that $J$ has codimension $1$ ideal: Since $J$ is a proper subset of $\mathfrak{A}$, let $y \in \mathfrak{A} \setminus J$. Then $f(y) \neq 0$. We aim to show that $\mathfrak{A} = J \oplus \mathbb{C}y$. Indeed $J \cap \mathbb{C}y = \emptyset$ since $y \not \in J$, and we have that $\mathfrak{A} = J + \mathbb{C}y$ since for every $x \in \mathfrak{A}$ we have that:
(3)
\begin{align} \quad x = \underbrace{x - \frac{f(x)}{f(y)}y}_{\in J} + \underbrace{\frac{f(x)}{f(y)}y}_{\in \mathbb{C}y} \end{align}
• Thus $J$ has codimension $1$. Furthermore, $J$ is maximal for if not, say $J \subset M$ for some maximal ideal $M$ of $\mathfrak{A}$ then for $y \in M \setminus J$ we have that $\mathfrak{A} = J \oplus \mathbb{C}y \subseteq M \neq \mathfrak{A}$, a contradiction.
• 3. Showing that $J$ is modular: Since $f$ is a multiplicative linear functional, $f \neq 0$. So there exists a $x_0 \in \mathfrak{A}$ such that $f(x_0) \neq 0$. Let $u = \frac{x_0}{f(x_0)}$. Then $f(u) = 1$. We claim that $u$ is a two-sided modular unit for $J$ so we need to show that $(1 - u)\mathfrak{A} \subseteq J$ and $\mathfrak{A}(1 - u) \subseteq J$. Let $x \in \mathfrak{A}$. Then:
(4)
\begin{align} \quad f(x - ux) = f(x) - f(u)f(x) = f(x) - 1 f(x) = 0 \end{align}
• And also:
(5)
\begin{align} \quad f(x - xu) = f(x) - f(x)f(u) = f(x) - f(x)1 = 0 \end{align}
• So $x - ux \in J$ and $x - ux \in J$ for all $x \in \mathfrak{A}$. Thus $J$ is modular. $\blacksquare$