Problem: Let $\mathcal{A}$ be a unital Abelian Banach algebra, and $I$ a closed ideal in $\mathcal{A}$. Then the quotient $\mathcal{A}/I$ has no nonzero singular elements $\iff$ $\mathcal{A}/I$ has no nonzero proper ideals. (Please note that I only have a slight acquaintance with abstract algebra, so I might be confused about something very basic in that subject here)

Attempt: $\Rightarrow)$ Suppose the quotient $\mathcal{A}/I$ has no nonzero singular elements, that is, $\forall I_x \in \mathcal{A}/I,$ $\exists I_{x^{-1}} \in \mathcal{A}/I$. Then for any ideal $J' \subset \mathcal{A}/I$ and taking some $I_x \in J'$, we have $I_{y x^{-1}} J' \subset J' \Rightarrow I_y \in J', \forall y \in \mathcal{A}$, so $J' = \mathcal{A}/I$.

$\Leftarrow )$ I am having trouble with this direction. Suppose $\mathcal{A}/I$ has no nonzero proper ideals, how does this imply all $I_x \in \mathcal{A}/I$ nonzero has an inverse? Any help is appreciated!

  • $\begingroup$ What is a "singular element"? One without an inverse? $\endgroup$ – Eric Wofsey Apr 11 '18 at 19:50
  • $\begingroup$ Yes, singular means without inverse here. $\endgroup$ – Longti Apr 11 '18 at 19:51

Follows an attempt to handle the ($\Longleftarrow$) direction:

($\Longleftarrow$)Suppose $\mathcal B$ is a unital, commutative algebra with no non-zero proper ideals. Suppose $0 \ne b \in \mathcal B$, and consider the principal ideal

$\langle b \rangle = b \mathcal B = \{ bs \mid s \in \mathcal B \}; \tag 1$


$\langle b \rangle \ne \{ 0 \}, \tag 2$


$0 \ne b = b 1_{\mathcal B} \in b \mathcal B = \langle b \rangle; \tag 3$

thus the ideal $b \mathcal B$ does not vanish, so we must have $b\mathcal B$ improper, i.e.,

$\langle b \rangle = b \mathcal B = \mathcal B \tag 4$

which implies there must be some $c \in \mathcal B$ with

$cb = bc = 1_{\mathcal B}; \tag 5$

thus $c$ is an inverse for $b$. Thus in fact $b$ is non-singular for all $b \in \mathcal B$.

Now we simply take

$\mathcal B = \mathcal A / I, \tag6$

and we attain a similar result for $\mathcal A / I$.

Note : We have not used the assumption that $\mathcal A$ or $\mathcal B$ are Banach; this part of the proof is purely algebraic.


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