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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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