Let $R\subseteq A$ be an integral extension of integral domains, and $\mathfrak m$ be a maximal ideal of $R$. The proof that there is a maximal ideal $\mathfrak M$ of $A$ lying over $\mathfrak m$ (that is $\mathfrak m=R\cap\mathfrak M$) consists of two steps:
- $A\mathfrak m\subsetneq A$.
- $A$ has a maximal ideal containing $\mathfrak m$.
Step 1 is an easy argument using the usual trick with a characteristic polynomial. Step 2 is just an immediate consequence of Zorn's Lemma.
While teaching a class in Galois theory, I was wondering if Step 2 can be done without resorting to Zorn's Lemma if the field extension $\text{Quot}(R)\subseteq\text{Quot}(A)$ of the field of fractions is finite. If it helps, I'm fine if we additionally assume that $R=A\cap\text{Quot}(R)$.
Question: Does anyone know a reference or slick proof for Step 2 in this situation without using Zorn's Lemma (or axiom of choice)?
One attempt which I didn't complete is as follows: $A/I$ is an $R/\mathfrak m$-vector space for every ideal $I$ of $A$ containing $\mathfrak m$. So the existence would follow for dimension reasons if $A/\mathfrak mA$ has finite dimension as an $R/\mathfrak m$-space. I haven't checked if that is actually true.
(Remark: Here integral domain is a commutative ring with a multiplicative identity and no zero divisors.)
Not a complete answer but I think this might provide some insight--see my previous comment. You can at least reduce to the case where the dimension of $\mathrm{Frac}(A)$ over $\mathrm{Frac}(R)$ is 1. First, if we can prove this case then we can always replace $R$ with its integral extension $R'=A \cap \mathrm{Frac}(R)$ which itself satisfies $R'=A \cap \mathrm{Frac}(R')$. Second, if $R=A \cap \mathrm{Frac}(R)$ and $A \neq R$ then we can find some $x \in A \backslash R$ and therefore $x \notin \mathrm{Frac}(R)$, so $R'=R[x]$ is such that $[\mathrm{Frac}(R'):\mathrm{Frac}(R)] > 1$, and we can replace $R$ with its integral extension $R'$. Alternating these two steps, we eventually terminate, and we can easily reduce further to the case where $A$ is the integral closure of $R$ (in its field of fractions.)
I am not aware of any constructivity results for this case of the lying-over theorem, but I would keep in mind that according to Banaschewski (edit: Hodges) the existence of maximal ideals in every ring is equivalent to the axiom of choice while the existence of maximal ideals in Noetherian rings follows from the axiom of dependent choice, so without the axiom of choice it's possible that all Noetherian rings have a maximal ideal and some non-Noetherian rings do not, but (with AC) the integral closure of a Noetherian domain is not necessarily Noetherian. This isn't a proof of a consistency result that you might want as a negative answer, but if the problem is really about the integral closure, then it seems hard to believe the answer is yes, since integral closures can be highly pathological and even if maximal ideals are easy to construct in the base domain they may be hard to construct in its integral closure.