1
$\begingroup$

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.

$\endgroup$

2 Answers 2

1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $\mathbb Z$.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$ Oct 3 at 15:47
  • $\begingroup$ 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. $\endgroup$ Oct 3 at 15:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.