Max. Mod. Two-Sided Ideals in a Commutative Banach Algebra over C

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

Recall from the Characterization of Maximal Modular Two-Sided Ideals of Codimension 1 in a Banach Algebra over C page the following important result: If $\mathfrak{A}$ is a Banach algebra over $\mathbb{C}$ then every maximal modular two-sided ideal of 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 codimension $1$.

When $\mathfrak{A}$ is a commutative Banach algebra then we can say more.

 Lemma 1: Let $\mathfrak{A}$ be a Banach algebra. If $M$ is a maximal modular two-sided ideal of $\mathfrak{A}$ then the only two-sided ideals of $\mathfrak{A} / M$ are the trivial ideals $\{ 0 \}$ and $\mathfrak{A} / M$.
• Proof: 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 and so the quotient algebra $\mathfrak{A} / M$ is defined.
• Suppose that $\mathfrak{A} / M$ has a proper nontrivial ideal $J$, i.e., suppose $J$ is an ideal of $\mathfrak{A} / M$ such that $J \neq \{ 0 \}$ and $J \neq \mathfrak{A} / M$. Let $K = \{ j : j' \in J \}$. We claim that $K$ is an ideal of $\mathfrak{A}$ such that $M \subset K$ and $K \neq \mathfrak{A}$.
• To see why, let $x \in \mathfrak{A}$ and let $k \in K$. Then $k' \in J$. Since $J$ is an ideal of $\mathfrak{A} / M$ we have that $x'k' \in J$ and $k'x' \in J$. So $xk, kx \in K$, showing that $\mathfrak{A}K \subseteq K$ and $K\mathfrak{A} \subseteq K$, i.e., $K$ is a two-sided ideal of $\mathfrak{A}$. Furthermore, note that if $m \in M$ then $m' = 0' \in J$ so $m \in K$, i.e., $M \subset K$ (where the inclusion is not strict since $J \neq \{ 0 \}$). Lastly, since $J \neq \mathfrak{A}/M$ there exists a $y' \in \mathfrak{A}/M \setminus J$. So $y \in \mathfrak{A} \setminus K$ showing that $K \neq \mathfrak{A}$.
• Thus if we assume that $\mathfrak{A} / M$ has a proper nontrivial ideal $J$ then there exists an ideal $K$ of $\mathfrak{A}$ such that $M \subset K \neq \mathfrak{A}$, contradicting the maximality of $M$. So $\mathfrak{A} / M$ has no proper nontrivial ideal. $\blacksquare$
 Theorem 2: Let $\mathfrak{A}$ be a commutative Banach algebra over $\mathbb{C}$. Then every maximal modular two-sided ideal of $\mathfrak{A}$ if the kernel of some multiplicative linear functional on $\mathfrak{A}$ and conversely, the kernel of any multiplicative linear functional is a maximal modular two-sided ideal of $\mathfrak{A}$.
• Proof: Let $M$ be a maximal modular two-sided ideal of the commutative Banach algebra $\mathfrak{A}$. Let $u$ be a modular unit for $M$ so that $(1 - u)\mathfrak{A} \subseteq M$ and $\mathfrak{A}(1 - u) \subseteq M$.
• Since $\mathfrak{A}$ is a Banach algebra and $M$ is a maximal modular two-sided ideal of $\mathfrak{A}$ we have by one of the corollaries on the Basic Theorems Regarding Ideals in an Algebra 2 page that $M$ is closed. Consider the quotient Banach algebra $\mathfrak{A} / M$. Since $M$ is a maximal ideal of $\mathfrak{A}$, we have by Lemma 1 that the only ideals of $\mathfrak{A} / M$ are the trivial ones, namely $\{ 0 \}$ and $\mathfrak{A} / M$.
• Since $u$ is a modular unit for $M$, $x - ux \in M$ and $x - xu \in M$ for all $x \in \mathfrak{A}$. So $x' = u'x'$ and $x' = x'u'$ for all $x' \in \mathfrak{A}/M$ where $x'$ and $u'$ denote the canonical images of $x$ and $u$ in $\mathfrak{A}/M$ under the quotient map $q : \mathfrak{A} \to \mathfrak{A}/M$, i.e., $u'$ is a unit for $\mathfrak{A}/M$.
• Thus $\mathfrak{A}/M$ is a Banach algebra with unit $u'$ and such that the only ideals of $\mathfrak{A}/M$ are $\{ 0 \}$ and $\mathfrak{A}/M$. So $\mathfrak{A}/M$ is a Banach division algebra. By The Gelfand-Mazur Theorem we have that for $u' \in \mathfrak{A}/M$ that $\mathfrak{A} / M = \mathbb{C} u'$ and so $\mathfrak{A}/M$ has dimension $1$ so $M$ has codimension $1$.
• So every maximal modular two-sided ideal of a commutative Banach algebra $\mathfrak{A}$ has co-dimension $1$, we see by the Theorem mentioned at the top of this page that every maximal modular two-sided ideal is the kernel of some multiplicative linear functional and conversely, the kernel of any multiplicative linear functional is a maximal modular two-sided ideal of $\mathfrak{A}$. $\blacksquare$