TITLE: Maximal ideal with given condition
QUESTION [5 upvotes]: I am reading the proof by B. Banaschewski of "Krull implies Zorn." I am having difficulties filling in the details in one of the steps.

We start with a partition $\mathfrak U$ of an arbitrary set $E$. A subset $X$ of $E$ is called spread if $X\cap C$ has at most one element for each $C\in\mathfrak U$. Let $\mathfrak S$ be the set of all spreads.
We consider the ring $R=\mathbb Q[E]$ of polynomials with elements of $E$ as indeterminates.
Set $T=\bigcup_{X\in\mathfrak S}\langle X\rangle_R$ and $U=R-T$, where $\langle X\rangle_R$ denotes the smallest ideal in $R$ containing $X$. One may prove that $U$ is closed under multiplication. (Since it hasn't been used elsewhere, this information may be important for the answer of my subsequent question.)
Next, let $R_0=R\left[\frac 1 u\mid u\in U\right]$, which has a maximal ideal $M$ under assumption of Krull's Theorem. Let $H=M\cap R$.
Clearly, $H$ is an ideal of $R$ which is contained in $T$. Apparently, $H$ is maximal (with respect to inclusion) as such. More precisely, if $I\supseteq H$ is an ideal of $R$ which is contained in $T$, then $I=H$.

I'm having difficulties to see why $H$ is maximal.
My idea was to pick an ideal $I\supsetneq H$ of $R$, and to prove that some $u\in U$ belongs to $I$.

REPLY [2 votes]: I'm not sure if there is any flaws with this proof but the following is my try:
In general, given a ring homomorphism $f:A\to B$ and denote $I^e$ the extension of ideal $I\subset A$ and $J^c$ the contraction of ideal $J\subset B$, we have $I\subseteq I^{ec}$.
Consider any ideal $I$ such that $H\subseteq I \subseteq T$, in particular, $U\cap I =\emptyset $ so the extension $I^e$ via the inclusion $i:R\to R_0$ is an ideal of $R_0$. Now note that we can assume that the maximal ideal $M$ contains $I^e$, otherwise, the ideal $\langle M,I^e \rangle$ contains $M$. Thus, $H\subseteq I \subseteq I^{ec} \subseteq M^c =M\cap R = H$.