This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
Lying over, going up
Proposition 1 Let \(A\hookrightarrow B\) be an integral extension.
- (Lying over) For any prime ideal \(\mathfrak{p}\) of \(A\), there exists a prime ideal \(\mathfrak{q}\) of \(B\) such that \(\mathfrak{q}\cap A=\mathfrak{p}\).
- (Going up) Furthermore, 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, if \(A\hookrightarrow B\) is an integral extension, then for any ideal \(\mathfrak{b}\) of \(B\), the ring homomorphism
\[\frac{A}{A\cap \mathfrak{b}}\hookrightarrow \frac{B}{\mathfrak{b}}\]is also an integral extension. Thus, we may assume without loss of generality that \(\mathfrak{b}=0\), and this is exactly proving the first result.
Therefore, 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 consider only the case where \(A\) is a local ring with maximal ideal \(\mathfrak{p}\). In this situation, the preimage of a maximal ideal of \(B\) containing \(\mathfrak{p}B\) must necessarily be \(\mathfrak{p}\); hence, unless \(\mathfrak{p}B=B\), this maximal ideal is the prime ideal of \(B\) we seek.
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. Thus, by §Integral Extensions, ⁋Lemma 4, \(B'\) is finitely generated as an \(A\)-module. Now applying §Integral Extensions, ⁋Lemma 8, we obtain \(B'=0\), 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 only principal ideals of \(B\). Consider an arbitrary principal ideal generated by \(b\in B\). Since \(\Frac(B)\) is an algebraic extension of \(\Frac(A)\), we can arrange
\[a_nb^n+\cdots+a_1b+a_0=0,\qquad a_i\in \Frac(A)\]Now multiply both sides by the least common multiple of the denominators of the \(a_i\), and if necessary divide both sides by a suitable power of \(b\) so that each \(a_i\) lies in \(A\) and \(a_0\neq 0\). 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 \(\mathfrak{q}\cap A\) and \(\mathfrak{q}\) respectively, it suffices to show that when two integral domains \(A,B\) and an integral extension \(A \hookrightarrow B\) are given, \(A\) is a field if and only if \(B\) is a field. On the other hand, if \(A\) is a field, then by Lemma 2, \(B\) must have no nonzero ideals; that is, \(B\) is a field.
Thus it suffices to assume that \(B\) is a field and show 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, let us examine 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 \(B\) where \(\mathfrak{q}_1=0\) and \(\mathfrak{q}_2\cap A=0\). However, the integral equations satisfied by elements of \(B\) remain integral equations after taking the quotient by \(\mathfrak{p}\); in particular, \(\Frac(B)\) becomes an algebraic extension of \(\Frac(A)\). Thus we obtain the desired result by Lemma 2.
References
[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.
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