Remark 1

Let $(A,1)$ be a unital algebra. If $x\in A$, one has

Sp$'_A(x) =$ Sp$_A(x)\cup \{0\}$. Indeed, one verifies that $(e-1)\cdot A = A\cdot (e-1) = 0$, hence that $\widetilde{A}$ is the unital algebra product of A and $K(e-1)$. Our assertion therefore follows from Remark 8 of I, p. 3.