Let rings be commutative and unital.
Let an immediately-submaximal ideal be a non-maximal ideal $I$ such that, for all maximal ideals $K$ such that $I \subset K$, for every ideal $J$ such that $I \subset J \subset K$, it holds that $I = J$ or $J = K$.
The intuition is that this is an ideal that is immediately below the maximal ideals that contain it in the lattice of ideals for my ring $R$.
When constructing a maximal ideal in an arbitrary ring using transfinite induction (or, equivalently, Zorn's lemma) we rely on the fact that an ideal contains $1$ if and only if it is a unit ideal. And, therefore, intuitively, if $1$ is present in our putative maximal ideal, it must have been introduced at some point and not "approached" in the sense of appendix A.
This argument doesn't work for producing immediately-submaximal ideals for the same reason it doesn't go through in non-unital rings, there's no $1$ to tell us when we've failed.
The mere failure of one particular argument, though, doesn't demonstrate that there are rings with no immediately-submaximal ideals though.
This leads me to my question: are there any rings with no immediately-submaximal ideals?
Appendix A
Union of an ascending chain of proper ideals is a proper ideal.
Let $C$ be an ascending chain of ideals. Suppose for contrapositive that $\cup C$ is a unit ideal. $\cup C$ contains $1$. If $1$ is present in $\cup C$ then it must be present in some ideal in $C$, that ideal is a unit ideal, therefore $C$ is not an ascending chain of proper ideals, completing the contrapositive argument.
Every nonzero ring has a maximal ideal
Fix a ring $R$. Well-order its elements by $\kappa$ and denote the nth element by $r_\alpha$ for some ordinal $\alpha \in \kappa$. I will now define a function $f : \kappa^+ \to 2^R$, it sends ordinals less than or equal to $\kappa$ to subsets of $R$.
- Let $f(0)$ be $(r_0)$ if it isn't a unit ideal else $(0)$.
- Let $f(\alpha^+)$ be $(r_{\alpha^+}, f(\alpha))$ if it isn't a unit ideal else $f(\alpha)$.
- Let $f(\lambda)$ where $\lambda$ is a limit ordinal be $(r_\lambda, f(\alpha) : \alpha \in \lambda)$ if it isn't unit ideal else $(f(\alpha) : \alpha \in \lambda)$.
$(f(\alpha) : \alpha \in \lambda)$ is never a unit ideal by our ascending chain of proper ideals lemma above.