Remark 7

Let A be a unital algebra, R its radical (A, VIII, p. 150, Def. 2), and let $\varphi$ be the canonical morphism of A onto $B = A/R$. If $x\in A$, one has Sp$_B(\varphi (x)) =$ Sp$_A(x)$. In fact, it is enough to prove that if $\varphi (x)$ is invertible in B, then $x$ is invertible in A. Now, if $y\in A$ is such that $\varphi (x)\varphi (y) =\varphi (y)\varphi (x) =\varphi (e)$, one has $xy\in e+ R,yx\in e+ R$, hence $xy$ and $yx$ are invertible (A, VIII, p. 151, Th. 1) and consequently $x$ is invertible. In particular, if $x\in R$, one has Sp$_A(x) =\{0\}$ if $A\not=\{0\}$.