Counterexamples to Theorems/Corollary related to going down theorems We know that if $A \subset B$ is an integral extension, and $P$ is a prime ideal of $B$, then $P$ is maximal if and only if $A \cap P$ is maximal.
I was wondering why do we need prime here. I know in the proof of the theorem, we are using that but I am interested in a counter-example to justify that $P$ is prime is a necessary condition.
I have been brainstorming and I am looking for counterexample as such to verify all these hypothesis are necessary, that is being an integral extension is necessary but I am finding it very difficult. I have been unsuccessful in finding counterexamples.
Can anyone suggest me a reference where I can find examples to show that everything is neccesary in Atiyah's theorem/corollary $5.7, 5.8, 5.9, 5.10, 5.11$.
I would really appreciate that.
 A: In $\mathbb Z\subset\mathbb Z[i]$, $(2\mathbb Z[i])\cap\mathbb Z=2\mathbb Z$ is maximal, but $2\mathbb Z[i]$ is not even prime and definitely not maximal. In general, $p\mathcal O_K$ is not necessarily prime in $\mathcal O_K$ for number field $K$ and rational prime $p$, despite $p\mathbb Z$ is maximal in $\mathbb Z$, and $\mathcal O_K$ is integral over $\mathbb Z$. This is studied in ramification theory.
If integral condition is dropped, take e.g. $\mathbb Z\subset\mathbb Q$, then $\{0\}$ is maximal in $\mathbb Q$ but not in $\mathbb Z$.
A: $\newcommand{\pp}{\mathfrak{p}}$Note that the given condition is equivalent to the following (Proposition 5.7):
Let $A \subset B$ be an integral extension of integral domains. Then, $A$ is a field iff $B$ is a field.
The conclusion is not true if we don't assume that $A$ and $B$ are integral domains, to begin with.
For example, we have the extension $\Bbb C \subset \Bbb C[x]/(x^2)$. This is an integral extension since it is a finite extension. However, the latter is not a field. (It is not even an integral domain.)
This also gives a simple example where the non-prime ideal $0$ contracts to the maximal ideal $0$.
Thus, I have given counterexamples to both 5.7 and 5.8.

5.9: $A \subset B$ is integral. $\mathfrak{q} \subset \mathfrak{q}'$ are primes in $B$ such that $\mathfrak{q}^c = \mathfrak{q}'^c$. Then, $\mathfrak{q} = \mathfrak{q}'$.
Counter 1. If $A \subset B$ is not integral.
Consider $\mathbb{C} \subset \mathbb{C}[x]$. Both ideals $(0)$ and $(x)$ contract to the zero ideal.
Counter 2. If $\mathfrak{q} \not\subset \mathfrak{q}'$.
Consider $\Bbb Z \subset \Bbb Z[i]$. This is an integral extension. The distinct ideals $(1 - 2i)$ and $(1 + 2i)$ both contract to $5\Bbb Z$.

5.10: This is the lifting theorem for integral extensions.
This is false for $\Bbb Z \subset \Bbb Q$. You cannot lift $2 \Bbb Z$.

5.11: Going-up theorem.
$\Bbb Z \subset \Bbb Q$ works again. We have the chain $0 \subset 2 \Bbb Z$ inside $\Bbb Z$. We can lift $0$ but not $2 \Bbb Z$.
A more interesting example would be one where we could lift both ideals but not in a way that we have containment. (After fixing a lift for the first ideal.)
Consider $\Bbb Z \subset \Bbb Z[x]$.
As before, we have the chain $0 \subset 2 \Bbb Z$ in $\Bbb Z$.
Note that the ideal $\pp = (2x - 1) \subset \Bbb Z[x]$ contracts to $(0)$. (Look at degrees.)
Moreover, this is a prime ideal since $\Bbb Z[x]/\pp \cong \Bbb Z[\frac{1}{2}]$.
However, $(2, \pp) = \Bbb Z[x]$. Thus, there is no prime ideal of $\Bbb Z[x]$ containing $\pp$ that contracts to $2 \Bbb Z$.
Note that $\Bbb Z \subset \Bbb Z[x]$ does actually have the Lying over theorem. Simply lift any prime ideal to its extension, and it works. (In fact, it also has the Going-Down theorem since $\Bbb Z \to \Bbb Z[x]$ is flat.)

For the last one, it would be interesting to have a counterexample of the following kind: $A \subset B$ is a ring extension. $\pp \subset \pp'$ are primes in $A$. There are lifts for both $\pp$ and $\pp'$. However, there are no lifts that respect inclusion. (This would also mean that the Going-down theorem is false for this extension.)
