proof of spectrum of an element in any unital complex banach algebra is not empty

This is related to Lang Real and Functional Analysis Chpt XVI, Sec 1.

Let $$A$$ be a banach algebra. Let $$v\in A$$. Spectrum of $$v$$ is defined as $$\{z\in C\vert$$ $$v-ze$$ is not invertible$$\}$$.

Thm 1.2 Let $$A$$ be an unital commutative normed algebra over real. Assume there is $$j\in A$$ s.t. $$j^2=-e$$. Let complex number $$C=R+Rj$$. Given $$v\in A,v\neq 0$$, there exists $$c\in C$$ s.t. $$v-ce$$ is not invertible in $$A$$.

Cor 1.4 The spectrum of an element in any complex Banach algebra(commutative or not) with unit element is not empty.

I am having trouble to follow the proof.

"If $$A$$ is a banach algebra with unit and $$v\in A$$, then closure of algebra generated by $$v$$ and unit is a commutative banach algebra. Hence, it follows from Thm 1.2 above."

$$\textbf{Q:}$$ Why it follows from Thm 1.2 above? Consider $$v\in A$$. Denote spectrum of $$v$$ in $$A$$ as $$Sp_A(v)$$. Denote the closure of algebra generated by $$v$$ and unit by $$B$$. It follows from Thm 1.2 that $$Sp_B(v)\neq\emptyset$$. Why is $$Sp_A(v)\neq\emptyset$$? This is more or less like the following statement. Given a ring inclusion map, $$B\subset A$$, non-units of $$B$$ are non-units of $$A$$. Take $$B=k[x]$$ and $$A=k(x)$$. Certainly non-units of $$B$$ other than $$0$$ are invertible in $$A$$.

The proof is indeed wrong for the reason that you say. To fix it, you have to let $$B$$ be the closure of the algebra generated by $$v$$ and the elements $$(v-ce)^{-1}$$ for all $$c\in\mathbb{C}$$. Then clearly $$Sp_B(v)=\emptyset$$, and $$B$$ is still commutative (here we use the fact that if $$a$$ and $$b$$ commute and $$b$$ is invertible then $$a$$ and $$b^{-1}$$ commute, since $$ab^{-1}=b^{-1}bab^{-1}=b^{-1}abb^{-1}=b^{-1}a$$).