> [!definition] > > Let $\alg$ be an [[Algebra over Ring|algebra over]] a [[Field|field]] $F$, then it is also a [[Ring|ring]]. A **left ideal** of $\alg$ is a *subalgebra* that is closed under left multiplication by $\alg$. An ideal $\mathcal I$ is *proper* if $\mathcal I \ne \alg$. > [!theorem] > > Let $\alg$ be a commutative unital [[Banach Algebra|Banach algebra]], and $\mathcal I \subset \alg$ be a proper ideal. > 1. $\mathcal I$ contains no invertible elements. > 2. The [[Topological Closure|closure]] $\ol{\mathcal I}$ is also a proper ideal. > 3. $\mathcal I$ is contained in a maximal ideal. > 4. Every maximal ideal is closed. > > *Proof*. If an ideal is proper, then it is contained in the set of non-invertible elements of $\alg$, which is closed, so $e$ cannot be in it.