Quotients of Banach algebras by ideals

Question

I am currently working through Banach Algebra Techniques in Operator Theory and am hung up on a detail on 2.32. When trying to show that the quotient of a Banach space $\mathcal{B}$ by a closed ideal $\mathcal{M}$ is again a Banach algebra, Douglas argues that it is a Banach space and an algebra (I'm okay with these steps). What I am stuck on is showing that $\lVert [1]\rVert = 1$. He says that $\lVert [1] \rVert = \inf_{g\in\mathcal{M}}\lVert 1-g\rVert = 1$. Clearly $\lVert [1]\rVert \le 1$ since $0\in\mathcal{M}$ but to show that it cannot be less than $1$, he argues that if $\lVert 1-g\rVert < 1$, $g$ would be invertible (which is true) and somehow concludes this cannot happen. Why can't $g\in\mathcal{M}$ be invertible? Clearly if $\mathcal{M}$ were to be taken to be a proper ideal, then it could not contain $1$ and thus no elements could be invertible. However $\mathcal{B}$ is a closed ideal over itself so these two seem at odds. Does Douglas mean for $\mathcal{M}$ to be a proper ideal (without saying so)?

Answer 1

Yes, the ideal must be proper. Otherwise $\mathcal{B}/\mathcal{M} = \{0\}$, and there is no element of norm $1$ in the quotient at all.