currently I am starting my studies on Banach Algebra's and I have a question regarding the completion. Generally suppose that $A$ is an Algebra over $\mathbb C$ then one can create a unital Algebra by taking $\tilde{A} = A \times \mathbb{C}$ (We denote this by the unitisation of $A$), where multiplication is given by
$$ (a,\alpha)(b,\beta) = (ab+\alpha b + \beta a, \alpha \beta) $$
One can prove that
- $\tilde{A}$ is unital (Take $e_{\tilde{A}}=(0_A,1)$).
- $\tilde{A}$ is commutative iff A is commutative. Moreover if $A$ admits a Banach space structure then one can turn $\mathcal{A} = A \times C$ together with the above multiplication into a unital Banach Algebra by setting the norm $\|(a,\lambda)\|_{\mathcal{A}} = \|a\|_A+|\lambda|_{\mathbb C}$. Lastly the mapping $(a,\lambda) \mapsto (a,0)$ gives an isometry between both spaces.
For me the following question remains:
- Is it possible that an element in $a \in \tilde{A}$ is invertible?
What I have tried so far:
- For the first one, I believe that without more algebraic structure on $A$ the only element is invertible is the unit in $\tilde{A}=A \times \mathbb{C}$.
Take any $(a,\alpha) \in \tilde{A}$ that is invertible, we want to prove that it is actually the unit. Using the definition inverse element there exists a $(b,\beta) \in A \times \mathbb{C}$ such that $$ (ab+\alpha b + \beta a, \alpha \beta) \stackrel{!}{=} (0_A,1) $$ This directly gives that $\alpha \neq 0$ and $\beta = \frac{1}{\alpha}$, thus we are left with $$ ab + \alpha b + \alpha^{-1} a \stackrel{!}{=} 0_A $$ However here I am stuck, since $b$ does not have to be the inverse in $A$.
I don't see a way to built a contradiction to this? Any hints are welcomed.
Let $a$ be such that $a^2=0$. Then $(a,\alpha)$ has the inverse $(-\alpha^{-2}a, \alpha^{-1})$, as $$ a\cdot (-\alpha^{-2}a) + \alpha (-\alpha^{-2}a) + \alpha^{-1} a=0 $$ is satisfied. A similar construction can be done if $a^d=0$ for some $d>2$, then $b$ has to be a certain polynomial in $a$.