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 Suppose we are given an integral extension \(A\hookrightarrow B\).

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) The \(\mathfrak{q}\) obtained above may be chosen so that \(\mathfrak{b}\subseteq \mathfrak{q}\) whenever we are given an ideal \(\mathfrak{b}\) of \(B\) with \(\mathfrak{b}\cap A\subseteq \mathfrak{p}\).

Proof

First, for the second result, we observe that 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 \(\mathfrak{b}=0\), which amounts to proving the first result.

So, given a prime ideal \(\mathfrak{p}\subseteq A\), it suffices to find a prime ideal \(\mathfrak{q}\) of \(B\) with \(\mathfrak{q}\cap A=\mathfrak{p}\).

Now set \(S=A\setminus \mathfrak{p}\); whenever \(A \hookrightarrow B\) is integral, so is \(S^{-1}A \rightarrow S^{-1}B\). Hence it suffices to consider only the case in which \(A\) 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 \(\mathfrak{p}\), so as long as \(\mathfrak{p}B\neq B\), this maximal ideal is the prime ideal of \(B\) we seek.

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

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

Now 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. Therefore, by §Integral Extensions, ⁋Lemma 4, \(B'\) is finitely generated as an \(A\)-module. Applying §Integral Extensions, ⁋Lemma 8 (Nakayama), we obtain \(B'=0\), a contradiction.

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

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

Proof

For this it suffices to consider only the 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 may write

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

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

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

Proof

Again, as in the proof of Proposition 1, taking quotients by \(\mathfrak{q}\cap A\) and \(\mathfrak{q}\) respectively, it suffices to show that, given two integral domains \(A,B\) and 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 \(B\) cannot have a nonzero ideal; 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 \(\mathfrak{m}=0\). This yields the desired result.

Finally, we consider the following.

Corollary 4 For an integral extension \(A\hookrightarrow B\), if two 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 \(\mathfrak{q}_1\not\subseteq \mathfrak{q}_2\) and \(\mathfrak{q}_2\not\subseteq \mathfrak{q}_1\).

Proof

Assume for contradiction that \(\mathfrak{q}_1\subseteq \mathfrak{q}_2\), and set \(A\cap \mathfrak{q}_1=A\cap \mathfrak{q}_2=\mathfrak{p}\). Taking the quotient of \(A\) by \(\mathfrak{p}\) and of \(B\) by \(\mathfrak{q}_1\), we may reduce to the situation in which \(B\) is an integral domain with \(\mathfrak{q}_1=0\) and \(\mathfrak{q}_2\cap A=0\). Now, the integral equations satisfied by the elements of \(B\) remain integral equations after taking the quotient by \(\mathfrak{p}\); in particular, \(\Frac(B)\) becomes an algebraic extension of \(\Frac(A)\). Therefore, by Lemma 2, we obtain the desired result.

Whereas Proposition 1 and Corollary 4 concern lifting chains of prime ideals upward, the reverse direction, going-down, does not hold for a general integral extension. It does hold, however, when the base is integrally closed, and its proof rests on the following lemma.

Lemma 5 In an integral extension \(A\hookrightarrow B\), suppose that \(A\) is an integrally closed domain and that \(B\) is also a domain. Let \(f_b\in K[\x]\) be the minimal polynomial over \(K=\Frac(A)\) of an arbitrary \(b\in B\).

1. All coefficients of \(f_b\) lie in \(A\).
2. Given a prime ideal \(\mathfrak{p}\) of \(A\) with \(b\in\mathfrak{p}B\), all coefficients of \(f_b\) other than the leading coefficient lie in \(\mathfrak{p}\).

Proof

Since \(b\) is integral, there exists a monic polynomial \(g\in A[\x]\) having \(b\) as a root. In \(K[\x]\) we have \(f_b\mid g\), and the roots of \(f_b\) are all roots of \(g\), hence integral over \(A\). Therefore the coefficients of \(f_b\), being symmetric functions of these roots, are also integral over \(A\), and since \(A\) is integrally closed they lie in \(A\).

Now assume \(b\in\mathfrak{p}B\). Write \(b=\sum_i p_i b_i\) (\(p_i\in\mathfrak{p}, b_i\in B\)), and let \(B'\) be the \(A\)-subalgebra of \(B\) generated by the \(b_i\); then \(B'\) is a finitely generated \(A\)-module. Since \(b\in\mathfrak{p}B'\), by §Integral Extensions, ⁋Lemma 8 (Nakayama) we obtain a monic polynomial \(g\in A[\x]\) having \(b\) as a root and whose non-leading coefficients all lie in \(\mathfrak{p}\). By what we showed above, \(h=g/f_b\in A[\x]\), and reducing \(g=f_b h\) modulo \(\mathfrak{p}\) gives \(\x^{\deg g}=\overline{f_b}\overline{h}\) in \((A/\mathfrak{p})[\x]\). Since \(A/\mathfrak{p}\) is a domain, \(\overline{f_b}\) must be a power of \(\x\), which means that the coefficients of \(f_b\) other than the leading coefficient lie in \(\mathfrak{p}\).

Theorem 6 (Going down) In an integral extension \(A\hookrightarrow B\), suppose that \(A\) is an integrally closed domain and that \(B\) is a domain. Let \(\mathfrak{p}_1\subseteq\mathfrak{p}_2\) be prime ideals of \(A\), and let \(\mathfrak{q}_2\) be a prime ideal of \(B\) satisfying \(\mathfrak{q}_2\cap A=\mathfrak{p}_2\). Then there exists a prime ideal \(\mathfrak{q}_1\) of \(B\) with \(\mathfrak{q}_1\subseteq\mathfrak{q}_2\) and \(\mathfrak{q}_1\cap A=\mathfrak{p}_1\).

Proof

Since \(B\) is a domain, we have \(B\hookrightarrow B_{\mathfrak{q}_2}\), and composing this with \(A\hookrightarrow B\) yields the ring extension \(A\hookrightarrow B_{\mathfrak{q}_2}\). By the same argument used in the proof of Proposition 1, if we show that \(\mathfrak{p}_1 B_{\mathfrak{q}_2}\cap A\subseteq\mathfrak{p}_1\), then there exists a prime ideal of \(B_{\mathfrak{q}_2}\) lying over \(\mathfrak{p}_1\). This corresponds to a prime ideal \(\mathfrak{q}_1\) of \(B\) contained in \(\mathfrak{q}_2\), giving the desired result.

Let \(a\in\mathfrak{p}_1 B_{\mathfrak{q}_2}\cap A\). Write \(a=p/s\) (\(p\in\mathfrak{p}_1 B\), \(s\in B\setminus\mathfrak{q}_2\)) and assume \(a\neq 0\). By Lemma 5, the minimal polynomial \(f=\x^n+c_{n-1}\x^{n-1}+\cdots+c_0\) of \(p\) satisfies \(c_i\in\mathfrak{p}_1\) for all \(i\). On the other hand, since \(a\in A\subseteq K\) and \(p=as\), the polynomial obtained from \(f\) by the substitution \(\x\mapsto a\x\) followed by normalization,

\[\frac{1}{a^n}f(a\x)=\x^n+\frac{c_{n-1}}{a}\x^{n-1}+\cdots+\frac{c_0}{a^n}\]

is irreducible over \(K\) and has \(s\) as a root. That is, this is the minimal polynomial of \(s\), and by Lemma 5 its coefficients lie in \(A\). Writing it as \(\x^n+c'_{n-1}\x^{n-1}+\cdots+c'_0\), we have \(c'_{n-i}a^i=c_{n-i}\in\mathfrak{p}_1\). If \(a\notin\mathfrak{p}_1\), then since \(\mathfrak{p}_1\) is prime, all \(c'_{n-i}\in\mathfrak{p}_1\); hence \(s^n=-c'_{n-1}s^{n-1}-\cdots-c'_0\in\mathfrak{p}_1 B\subseteq\mathfrak{q}_2\), contradicting the fact that \(\mathfrak{q}_2\) is prime. Therefore \(a\in\mathfrak{p}_1\).

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References

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

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