Ideal $I$ contained in a non-trivial ideal and quotient algebra

Question

I read, in Tikhomirov's appendix to Kolmogorov-Fomin's Элементы теории функций и функционального анализа, that in order that the ideal $I$ be contained in a non-trivial ideal $I'\subset X$, it is necessary and sufficient that the algebra $X/I$ has a non-trivial ideal.

The theorem is stated in the context of commutative Banach (unitary) algebras, but the proof (p. 521 here) seems to show that it is valid for any commutative algebra defined as a linear space where a commutative, associative and distributive (with respect to the addition) multiplication is defined such that $\forall\alpha\in\mathbb{K}\quad \alpha(xy)=(\alpha x)y=x(\alpha y)$.

In any case, whether it concerns only commutative Banach unitary algebras or commutative algebras as defined above, I think we must intend contained as properly contained. Am I right? Thank you so much!!!

Answer 1

in Lemma 2 (p 521) only necessity is demonstrated, and the demonstration does no more than point to the general structure theorem for arbitrary commutative rings - that for a ring $A$ and an ideal $I$ the ideals over $I$ in $A$ are in 1-1 correspondence with the ideals of $A/I$.

the answer to your question is, therefore, yes.