why $\tilde{A}/A$ is isomorphic to $\mathbb{C}$?

Question

When I am reading the answer of this question, I found the statement $\tilde{A}/A\simeq \mathbb{C}$ suspicious to me. Here $A$ is a non-unital C* algebra and $\tilde{A}$ is it unitization. I don't understand why $A$ is a maximal ideal in $\tilde{A}$, I thought $A$ is only an essential ideal. Can anyone explain why?

Answer 1

Recall how the unitization is created: we have that $\tilde{A}=A\oplus\mathbb{C}$ as vector spaces and then we set $$(a,\lambda)\cdot(b,\mu)=(ab+\mu a+\lambda b, \lambda\mu)$$ for the multiplication operation and $$(a,\lambda)^*=(a^*,\bar{\lambda})$$ for involution. It is a standard proof that $\tilde{A}$ with these operations admits a (unique) $C^*$-norm under which it it complete and also $\|(a,0)\|_{\tilde{A}}=\|a\|_A$. So we have a $*$-isomorphism $A\cong A\oplus0\subset\tilde{A}$, so we identify $A$ with the set $A\oplus0=\{(a,0):a\in A\}$. Now since $(a,0)\cdot(b,\mu)=(ab+\mu a,0)$ we have that $A$ is an ideal in $\tilde{A}$.

Also, when $A$ is non-unital, $A$ is an essential ideal in $\tilde{A}$: let $(b,\mu)\in\tilde{A}$ be such that $(b,\mu)\cdot(a,0)=0$ for all $a\in A$. We will show that $(b,\mu)=(0,0)$. We have that $ba+\mu a=0$ for all $a\in A$. For $a=b^*$ we have that $bb^*=-\mu b^*$, so $\|b\|^2=\|bb^*\|=|\mu|\cdot\|b\|$. Therefore, if $b\neq0$ we have that $|\mu|=\|b\|$. Now take an approximate unit $(e_\lambda)$ of $A$ and note that $be_\lambda+\mu e_\lambda=0$ for all $\lambda$, so $\lim_{\lambda}e_\lambda=\frac{-b}{\mu}$. But this shows that $\frac{-1}{\mu}b$ is a unit of $A$, a contradiction. So $b=0$ and $\mu=0$ follows immediately.

Also, the quotient map $\pi:\tilde{A}\to\mathbb{C}$ defined by $\pi(a,\lambda)=\lambda$ is a well-defined $*$-homomorphism. Note that this is onto and that the kernel of $\pi$ is precisely $A$. So $\tilde{A}/A\cong\mathbb{C}$ indeed.