We work with the unitization of $C^\ast$-algebra. I.e. $C^\ast$-algebra $\mathcal{A}$ with norm $\|\cdot\|$. Let $\tilde{\mathcal{A}}=\mathcal{A}\oplus \mathbb{C}$ as a vector space. We endow it with multiplication and involution,
$$(a,\lambda)\cdot (b,\mu):=(ab+\lambda b+\mu a,\lambda \mu)$$ and $$(a,\lambda)^\ast:=(a^\ast,\bar{\lambda})$$
I managed to prove that this is a $\ast$-algebra. Now I have some questions because I want to understand this example.
We denote $\omega:\mathcal{A}\rightarrow\tilde{\mathcal{A}}$ by the map $\omega(a):=(a,0)$. I want to show that this is a $\ast$-homomorphism, i.e. it is linear and so on. The example does not show anything regarding $\ast$-homomorphism, but my attempt: Let $a,b\in\mathcal{A}$ then $\omega(a+b)=(a+b,0)=(a,0)+(b,0)=\omega(a)+\omega(b)$. Is it true?
I readed about two side ideals in my book. Is it true that $\omega(\mathcal{A})$ is a two-sided ideal in $\tilde{\mathcal{A}}$? If so, how come?
Definition: A (two-sided) ideal $J$ in a Banach algebra $\mathscr{A}$ is a subspace of $A$ with the property that $A \in \mathscr{A}$ and $S \in J$ implies $AS \in J$ and $SA \in J$.
Is this definition the right one to use? If so, then I will try to do the part "$AS \in J$" if anyone can give an example of the other part $SA \in J$.
Yes, the fact that $\omega$ is a $*$-homomorphism simply follows from the fact that $\omega(\mathcal{A}) = \mathcal{A} \times \{0\}$ inherits the $*$-algebra structure from $\mathcal{A}$. By definition, we have $$\omega(\alpha a + \beta b) = (\alpha a + \beta b, 0) = \alpha (a,0) + \beta (b,0) = \alpha \omega(a) + \beta \omega(b),$$ $$\omega(ab) = (ab,0) = (a,0)(b,0) = \omega(a)\omega(b),$$ $$\omega(a^*) = (a^*,0) = (a,0)^*.$$
$\omega(\mathcal{A}) = \mathcal{A} \times \{0\}$ is indeed a two-sided ideal in $\tilde{\mathcal{A}}$. For $(a,0),(b,0) \in \mathcal{A} \times \{0\}$ and $(c,\lambda) \in \tilde{\mathcal{A}}$ we have $$\alpha (a,0) + \beta (b,0) = (\alpha a + \beta b,0) \in \mathcal{A} \times \{0\},$$ $$(a,0)(c,\lambda) = (ac+ \lambda a,0) \in \mathcal{A} \times \{0\}, \qquad (c,\lambda)(a,0) = (ca + \lambda a, 0) \in \mathcal{A} \times \{0\}.$$