# Contraction of certain ideals

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.

• What do you mean by "over a field $k$ of characteristic zero?" What is $k$ in relation to $A$ and $B$?
– D_S
Mar 10 at 23:42
• Thank you. I mentioned $k$ so that I can replace $A/p$ for $k$, when $p$ is a maximal ideal of $A$. Mar 10 at 23:46
• 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. Mar 10 at 23:51
• @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). 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.

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