0
$\begingroup$

Let $A \subseteq B$ be two integral domains over a field $k$ of characteristic zero. Let $p$ be a prime ideal of $A$.

A comment to this question says: "Extensions of prime ideals pretty much never remain prime. If $A \subseteq B$ are rings, and $p$ is a prime of $A$, then $pB$ is prime in $B$ if and only if $B/pB = B \otimes_A A/p$ is an integral domain, which it seldom is, even if $A$ and $B$ are integral domains."

Further assume that $p$ is a maximal ideal of $A$. Then $B/pB = B \otimes_A k$; What can be said about that tensor product? Probably not much? When is it an integral domain?

What if we further assume that $p=A \cap P$, for some prime ideal $P$ of $B$; should this help somehow?

Thank you very much.

$\endgroup$
4
  • $\begingroup$ What do you mean by "over a field $k$ of characteristic zero?" What is $k$ in relation to $A$ and $B$? $\endgroup$
    – D_S
    Mar 10 at 23:42
  • $\begingroup$ Thank you. I mentioned $k$ so that I can replace $A/p$ for $k$, when $p$ is a maximal ideal of $A$. $\endgroup$
    – user237522
    Mar 10 at 23:46
  • 1
    $\begingroup$ In the shaded section you seem to be assuming that $A/p \cong k$. This is true if, for instance, $k$ is algebraically closed and $A$ is a finitely generated $k$ algebra. For instance, if $A= \mathbb Q[x]$ and $p=(x^2+1)$ then $A/p \cong \mathbb Q(i)$, not $\mathbb Q$. Additionally, I suggest that you look into Dedekind domains and ramification of primes to investigate a particular case of this question. Any algebraic number theory book should mention this topic. $\endgroup$ Mar 10 at 23:51
  • $\begingroup$ @paulblartmathcop, thank you very much, interesting. Actually, it is good enough for me to deal with $k=\mathbb{C}$ and $A,B$ finitely generated $\mathbb{C}$-algebras. Your suggestion to first deal with Dedekind domains is good (though I prefer not to restrict to one-dimensional rings only). $\endgroup$
    – user237522
    Mar 11 at 0:00
4
$\begingroup$

The situation where $A = \mathbb Z$, and $B$ is the integral closure of $A$ in a finite field extension $K$ of $\mathbb Q$ is particularly enlightening on this matter.

For example, if $K = \mathbb Q(i) = \{ a+bi : a, b \in \mathbb Q\}$, then $B = \mathbb Z[i] = \{a+bi : a,b \in \mathbb Z\}$.

For general $K$, the following results may be interesting to you:

  • Every nonzero prime ideal of $B$ is a maximal ideal.

  • Every prime ideal $\mathfrak p$ of $A$ is equal to $P \cap A$ for a (not necessarily unique) prime ideal $P$ of $B$). In fact, $P$ is unique if and only if $\mathfrak p B$ is a power of a prime ideal of $B$ (in which case, we have $P^e = \mathfrak p B$ for some $e \geq 1$).

Let us specialize to the case $B = \mathbb Z[i]$. If $\mathfrak p$ is a prime ideal of $\mathbb Z$, generated by a prime number $p$, then sometimes $\mathfrak p$ remains prime in $B$, and sometimes not. Specifically:

  • If $p = 2$, then $\mathfrak p B$ is not a prime ideal of $\mathbb Z[i]$, but it is equal to $P^2$, where $P$ is the ideal in $\mathbb Z[i]$ generated by $1+i$.

  • If $p \equiv 1 \pmod{4}$, then $\mathfrak pB$ is not a prime ideal of $B$, but is equal to the intersection of two distinct prime ideals of $B$.

  • If $p \equiv 3 \pmod{4}$, then $\mathfrak pB$ is a prime ideal of $B$.

This exhausts all the possibilities.

For the general case of a finite field extension $K$ of $\mathbb Q$, the problem of determining for which prime ideals $\mathfrak p$ of $A$ have the property that $\mathfrak p B$ remains a prime ideal in $B$ is unsolved. When $K$ is a Galois extension of $\mathbb Q$ such that $\operatorname{Gal}(K/\mathbb Q)$ is an abelian group, the problem is essentially solved by the Kronecker-Weber theorem.

$\endgroup$
1
  • $\begingroup$ Nice answer, thank you! (It would be nice to have further special cases, for example in polynomial rings). $\endgroup$
    – user237522
    Mar 11 at 0:08

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.