Integral Extensions and Ideals

Lying over, going up

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

1. (Lying over) For any prime ideal of \(A\), \(\mathfrak{p}\), there exists a prime ideal of \(B\), \(\mathfrak{q}\), such that \(\mathfrak{q}\cap A=\mathfrak{p}\).
2. (Going up) The prime ideal \(\mathfrak{q}\) obtained in the above proposition can be chosen so that whenever \(\mathfrak{b}\cap A\subseteq \mathfrak{p}\) holds for an ideal of \(B\), \(\mathfrak{b}\), we have \(\mathfrak{b}\subseteq \mathfrak{q}\).

Proof

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

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

is also an integral extension. Hence, without loss of generality, it suffices to assume \(\mathfrak{b}=0\), which is precisely showing the first result.

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

Meanwhile, if we 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, since the preimage of any maximal ideal containing \(\mathfrak{p}B\) in \(B\) must be \(\mathfrak{p}\), 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 linear combination of elements of \(\mathfrak{p}\) with coefficients in \(B\):

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

Now, let the subalgebra generated by the \(b_i\) in \(B\) over \(A\) be \(B'\). 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 gives \(B'=0\), a contradiction.

An important part of this post is proving Corollary 4, which roughly states that if, over a prime ideal of \(A\), \(\mathfrak{p}\), two prime ideals of \(B\), \(\mathfrak{q}_1, \mathfrak{q}_2\), obtained through Proposition 1 are given, they do not contain each 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

It suffices to consider only principal ideals of \(B\). Consider any principal ideal generated by \(b\in B\). Then, since \(\Frac(B)\) is an algebraic extension of \(\Frac(A)\), we can arrange that

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

holds. Now multiplying both sides by the product of the denominators of the \(a_i\) and, if necessary, dividing both sides appropriately by a power of \(b\), we can arrange that each \(a_i\) is an element of \(A\) and \(a_0\neq 0\). Then \(a_0\) belongs to the principal ideal generated by \(b\).

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

Proof

As in the proof of Proposition 1, here too we pass to the quotients by \(\mathfrak{q}\cap A\) and \(\mathfrak{q}\), respectively. Since \(\mathfrak{q}\) is prime, \(B/\mathfrak{q}\) is a domain, and \(A/(\mathfrak{q}\cap A)\), being its subring, is also a domain. Hence 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. On the other hand, if \(A\) is a field, then by Lemma 2, \(B\) cannot have any nonzero proper ideal; that is, \(B\) is a field.

Therefore, it suffices to assume that \(B\) is a field and show that \(A\) is a field. In \(A\), let a maximal ideal \(\mathfrak{m}\) be given. Then by Proposition 1, we can find in \(B\) a prime ideal \(\mathfrak{q}\) such that \(\mathfrak{q}\cap A= \mathfrak{m}\). But since \(B\) is a field, \(\mathfrak{q}=0\), and hence \(\mathfrak{m}=0\). From this we obtain the desired result.

Finally, consider the following.

Corollary 4 For an integral extension \(A\hookrightarrow B\), if in \(B\) two prime ideals \(\mathfrak{q}_1\neq \mathfrak{q}_2\) 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

Suppose, contrary to the conclusion, that \(\mathfrak{q}_1\subseteq \mathfrak{q}_2\), and let \(A\cap \mathfrak{q}_1=A\cap \mathfrak{q}_2=\mathfrak{p}\). Then, taking the quotient of \(A\) by \(\mathfrak{p}\) and of \(B\) by \(\mathfrak{q}_1\), we can reduce the situation to the case where \(B\) is an integral domain, \(\mathfrak{q}_1=0\), and \(\mathfrak{q}_2\cap A=0\). However, the integral equations satisfied by elements of \(B\) remain integral equations even after passing to the quotient by \(\mathfrak{p}\); in particular, \(\Frac(B)\) is an algebraic extension of \(\Frac(A)\). Therefore, by Lemma 2, we obtain the desired result.

While Proposition 1 and Corollary 4 are in the direction of lifting chains of prime ideals upward, going-down in the opposite direction does not hold for general integral extensions. It does hold, however, when the base is an integrally closed domain, that is, a normal domain (§Integral Extensions, ⁋Definition 3), and its proof rests on the following lemma.

Lemma 5 Let \(A\hookrightarrow B\) be an integral extension, where \(A\) is an integrally closed domain and \(B\) is also a domain. For any \(b\in B\), let its minimal polynomial over \(K=\Frac(A)\) be \(f_b\in K[\x]\).

1. Every coefficient of \(f_b\) belongs to \(A\).
2. If \(b\in\mathfrak{p}B\) for a prime ideal of \(A\), \(\mathfrak{p}\), then all coefficients of \(f_b\) other than the leading coefficient belong to \(\mathfrak{p}\).

Proof

Since \(b\) is integral, there exists, with \(b\) as a root, a monic polynomial \(g\in A[\x]\). 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 polynomials in these roots, are also integral over \(A\), and since \(A\) is integrally closed, they belong to \(A\).

Now suppose \(b\in\mathfrak{p}B\). Write \(b=\sum_i p_i b_i\) (\(p_i\in\mathfrak{p}, b_i\in B\)), and if we let the subalgebra generated by the \(b_i\) in \(B\) over \(A\) be \(B'\), then \(B'\) is a finitely generated \(A\)-module. Since \(b\in\mathfrak{p}B'\), by §Integral Extensions, ⁋Theorem 1, we obtain, having \(b\) as a root and all coefficients other than the leading coefficient belonging to \(\mathfrak{p}\), a monic polynomial \(g\in A[\x]\). By what was shown above, \(h=g/f_b\in A[\x]\), and reducing \(g=f_b h\) modulo \(\mathfrak{p}\) yields, in \((A/\mathfrak{p})[\x]\), \(\x^{\deg g}=\overline{f_b}\overline{h}\). Since \(A/\mathfrak{p}\) is a domain, \(\overline{f_b}\) must be a power of \(\x\), which means that all coefficients of \(f_b\) other than the leading coefficient belong to \(\mathfrak{p}\).

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

Proof

Since \(B\) is a domain, \(B\hookrightarrow B_{\mathfrak{q}_2}\), and composing this with \(A\hookrightarrow B\) gives the ring extension \(A\hookrightarrow B_{\mathfrak{q}_2}\). By the same argument as in the proof of Proposition 1, if we show that \(\mathfrak{p}_1 B_{\mathfrak{q}_2}\cap A\subseteq\mathfrak{p}_1\), there exists, lying over \(\mathfrak{p}_1\), a prime ideal of \(B_{\mathfrak{q}_2}\). This corresponds to a prime ideal of \(B\) contained in \(\mathfrak{q}_2\), \(\mathfrak{q}_1\), which gives 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 of \(p\), \(f=\x^n+c_{n-1}\x^{n-1}+\cdots+c_0\), 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. In other words, this is the minimal polynomial of \(s\), and by Lemma 5 its coefficients belong to \(A\). Writing this 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 \(c'_{n-i}\in\mathfrak{p}_1\), and thus \(s^n=-c'_{n-1}s^{n-1}-\cdots-c'_0\in\mathfrak{p}_1 B\subseteq\mathfrak{q}_2\), contradicting the fact that \(\mathfrak{q}_2\) is prime. Therefore \(a\in\mathfrak{p}_1\).

---

References

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

---