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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 $J = K$. (Also, I made a choice here regarding this definition. We could have insisted that $I$ be immediately at least one maximal ideal extending it instead of being immediately under all of them. I don't have a principled reason for choosing one definition or the other.)

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.

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Yes, it's possible. Let $R = \mathbb{Q}[[x^{\mathbb{R}_{\ge 0}}]]$ be the ring of formal power series $\sum c_r x^r$ with non-negative real exponents and such that $\{ r : c_r \neq 0 \}$ is well-ordered (this guarantees that multiplication is well-defined); these are known as Hahn series. For $f \in R$ write $\nu(f)$ for the smallest exponent of a nonzero term in $f$. You can see this thread for a proof that the ideals of $R$ have the form

$$I_r = \{ f : \nu(f) \ge r \}$$ $$J_r = \{ f : \nu(f) > r \}$$

and that the only maximal ideal is $J_0$. Any other ideal strictly contains $J_0$ and there are uncountably many ideals strictly in between, e.g. $I_{r/2}$ for $I_r$ and $J_{r/2}$ for $J_r$.

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