# 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.

$$\newcommand{\pp}{\mathfrak{p}}$$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.)

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 $$\mathbb Z$$.

• This is great. How do I get counterexamples to other theorems/corollaries. I am curious about the assumptions are necessary. Any reference or any general idea about them? Oct 3 at 15:47
• Sorry that I don't have access to A&M right now. Those theorems were historically established for the sake of number theory. I guess it's not hard to find applications and counterexamples for integer rings of number fields and their localizations. Oct 3 at 15:52