Integral Extensions and Ideals

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Lying over, going up

Proposition 1 Let \(A\hookrightarrow B\) be an integral extension.

1. (Lying over) For any prime ideal \(\mathfrak{p}\) of \(A\), there exists a prime ideal \(\mathfrak{q}\) of \(B\) such that \(\mathfrak{q}\cap A=\mathfrak{p}\).
2. (Going up) Moreover, given any ideal \(\mathfrak{b}\) of \(B\) satisfying \(\mathfrak{b}\cap A\subseteq \mathfrak{p}\), the prime ideal \(\mathfrak{q}\) obtained above can be chosen so that \(\mathfrak{b}\subseteq \mathfrak{q}\).

Proof

First, for the second result, observe that if \(A\hookrightarrow B\) is an integral extension, then for any ideal \(\mathfrak{b}\) of \(B\) the induced ring homomorphism

\[\frac{A}{A\cap \mathfrak{b}}\hookrightarrow \frac{B}{\mathfrak{b}}\]

is again an integral extension. Thus we may assume without loss of generality that \(\mathfrak{b}=0\), which reduces the problem exactly to the first result.

Hence it suffices to find a prime ideal \(\mathfrak{q}\) of \(B\) satisfying \(\mathfrak{q}\cap A=\mathfrak{p}\) for the given prime ideal \(\mathfrak{p}\subseteq A\).

Now let \(S=A\setminus \mathfrak{p}\); then if \(A \hookrightarrow B\) is integral, so is \(S^{-1}A \rightarrow S^{-1}B\). Thus it suffices to treat the case where \(A\) is a local ring with maximal ideal \(\mathfrak{p}\). In this situation, the preimage of any maximal ideal of \(B\) containing \(\mathfrak{p}B\) must be \(\mathfrak{p}\); consequently, unless \(\mathfrak{p}B=B\), such a maximal ideal is the desired prime ideal of \(B\).

Suppose for contradiction that \(\mathfrak{p}B=B\). Then \(1\in B\) can be written as a \(B\)-linear combination of elements of \(\mathfrak{p}\):

\[1=\sum_{i=1}^n b_i a_i,\qquad a_i\in \mathfrak{p},\quad b_i\in B\]

Let \(B'\) be the \(A\)-subalgebra of \(B\) generated by the \(b_i\). Then every element of \(B'\) is integral, and \(B'\) is finitely generated as an \(A\)-algebra. Hence, by §Integral Extensions, ⁋Lemma 4, \(B'\) is finitely generated as an \(A\)-module. Applying §Integral Extensions, ⁋Lemma 8 (Nakayama) now yields \(B'=0\), a contradiction.

The main goal of this post is to prove Corollary 4, which roughly states that if two prime ideals \(\mathfrak{q}_1, \mathfrak{q}_2\) of \(B\) lying over a prime ideal \(\mathfrak{p}\) of \(A\) are given via Proposition 1, then neither contains the other.

Lemma 2 For two integral domains \(A\subseteq B\), if \(\Frac(A) \rightarrow \Frac(B)\) is an algebraic extension, then any nonzero ideal of \(B\) meets \(A\) nontrivially.

Proof

It suffices to consider principal ideals of \(B\). Take an arbitrary principal ideal generated by \(b\in B\). Since \(\Frac(B)\) is an algebraic extension of \(\Frac(A)\), we can arrange

\[a_nb^n+\cdots+a_1b+a_0=0,\qquad a_i\in \Frac(A)\]

Multiplying both sides by the least common multiple of the denominators of the \(a_i\), and dividing by a suitable power of \(b\) if necessary, we may assume that each \(a_i\) lies in \(A\) and that \(a_0\neq 0\). Then \(a_0\) belongs to the principal ideal generated by \(b\).

Corollary 3 Let \(A\) be an integral domain and let \(A \rightarrow B\) be an integral extension. Then a prime ideal \(\mathfrak{q}\) of \(B\) is maximal if and only if \(\mathfrak{q}\cap A\) is a maximal ideal of \(A\).

Proof

As in the proof of Proposition 1, taking quotients by \(\mathfrak{q}\cap A\) in \(A\) and by \(\mathfrak{q}\) in \(B\), it suffices to show that for two integral domains \(A,B\) with an integral extension \(A \hookrightarrow B\), \(A\) is a field if and only if \(B\) is a field. Now if \(A\) is a field, then by Lemma 2 \(B\) has no nonzero ideals; that is, \(B\) is a field.

Thus it suffices to assume that \(B\) is a field and prove that \(A\) is a field. Let \(\mathfrak{m}\) be a maximal ideal of \(A\). Then by Proposition 1 there exists a prime ideal \(\mathfrak{q}\) of \(B\) such that \(\mathfrak{q}\cap A=\mathfrak{m}\). But since \(B\) is a field, \(\mathfrak{q}=0\), and hence \(\mathfrak{m}=0\). The desired result follows.

Finally, we record the following.

Corollary 4 For an integral extension \(A\hookrightarrow B\), if two distinct prime ideals \(\mathfrak{q}_1\neq \mathfrak{q}_2\) of \(B\) satisfy \(A\cap \mathfrak{q}_1=A\cap \mathfrak{q}_2=\mathfrak{p}\), then \(\mathfrak{q}_1\not\subset \mathfrak{q}_2\) and \(\mathfrak{q}_2\not\subset \mathfrak{q}_1\).

Proof

Suppose for contradiction that \(\mathfrak{q}_1\subseteq \mathfrak{q}_2\) and \(A\cap \mathfrak{q}_1=A\cap \mathfrak{q}_2=\mathfrak{p}\). Then, taking the quotient by \(\mathfrak{p}\) in \(A\) and by \(\mathfrak{q}_1\) in \(B\), we may replace the given situation with an integral domain \(B\) in which \(\mathfrak{q}_1=0\) and \(\mathfrak{q}_2\cap A=0\). However, the integral equations satisfied by elements of \(B\) remain integral equations after quotienting by \(\mathfrak{p}\); in particular, \(\Frac(B)\) becomes an algebraic extension of \(\Frac(A)\). Thus the desired result follows from Lemma 2.

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References

[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.

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