Maximal Modular Two-Sided Ideals in a Commutative Banach Algebra

Maximal Modular Two-Sided Ideals in a Commutative Banach Algebra

Recall from the Maximal Modular Two-Sided Ideals of Codimension 1 in a Banach Algebra page the following important result: If $X$ is a Banach algebra then every maximal modular two-sided ideal of codimension $1$ is the kernel of some multiplicative linear functional on $X$ and conversely, the kernel of every multiplicative linear functional on $X$ is a maximal modular two-sided ideal of codimension $1$.

When $X$ is a commutative Banach algebra then we can say more.

Lemma 1: Let $X$ be a Banach algebra. If $M$ is a maximal modular two-sided ideal of $X$ then the only two-sided ideals of $X / M$ are the trivial ideals $\{ 0 \}$ and $X / M$.
  • Suppose that $X / M$ has a proper nontrivial ideal $J$, i.e., suppose $J$ is an ideal of $X / M$ such that $J \neq \{ 0 \}$ and $J \neq X / M$. Let $K = \{ j : j' \in J \}$. We claim that $K$ is an ideal of $X$ such that $M \subset K$ and $K \neq X$.
  • To see why, let $x \in X$ and let $k \in K$. Then $k' \in J$. Since $J$ is an ideal of $X / M$ we have that $x'k' \in J$ and $k'x' \in J$. So $xk, kx \in K$, showing that $XK \subseteq K$ and $KX \subseteq K$, i.e., $K$ is a two-sided ideal of $X$. 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 $J \neq X/M$ there exists a $y' \in X/M \setminus J$. So $y \in X \setminus K$ showing that $K \neq X$.
  • Thus if we assume that $X / M$ has a proper nontrivial ideal $J$ then there exists an ideal $K$ of $X$ such that $M \subset K \neq X$, contradicting the maximality of $M$. So $X / M$ has no proper nontrivial ideal. $\blacksquare$
Theorem 2: Let $X$ be a commutative Banach algebra. Then every maximal modular two-sided ideal of $X$ if the kernel of some multiplicative linear functional on $X$. Conversely, the kernel of any multiplicative linear functional is a maximal modular two-sided ideal of $X$.
  • Proof: Let $M$ be a maximal modular two-sided ideal of the commutative Banach algebra $X$. Let $u$ be a modular unit for $M$ so that $(1 - u)X \subseteq M$ and $X(1 - u) \subseteq M$.
  • Since $X$ is a Banach algebra and $M$ is a maximal modular two-sided ideal of $X$ we have by one of the corollaries on the Basic Theorems Regarding Ideals in a Linear Space 2 page that $M$ is closed. Consider the quotient Banach algebra $X / M$. Since $M$ is a maximal ideal $X$, we have that the only ideals of $X / M$ are the trivial ones, namely $\{ 0 \}$ and $X / M$.
  • Since $u$ is a modular unit for $M$, $x - ux \in M$ and $x - xu \in M$ for all $x \in X$. So $x' = u'x'$ and $x' = x'u'$ for all $x' \in X/M$ where $x'$ and $u'$ denote the canonical images of $x$ and $u$ in $X/M$ under the quotient map $q : X \to X/M$, i.e., $u'$ is a unit for $X/M$.
  • Thus $X/M$ is a Banach algebra with unit $u'$ and such that the only ideals of $X/M$ are $\{ 0 \}$ and $X/M$. So $X/M$ is a Banach division algebra. Thus for $u' \in X/M$ we have that $X / M = \mathbb{C} u'$ and so $X/M$ has dimension $1$ so $M$ has codimension $1$.
  • So every maximal modular two-sided ideal of a commutative Banach algebra $X$ 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 $X$. $\blacksquare$
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