Rings, Modules, Fields

7 Dimension and Integral extensions

Whenever \(R\subset S\) are rings, we say \(S\) is an

extension of \(R\) (sometimes written \(S/R\)). We would like to understand how the two rings of an extension relate to each other.

Definition 7.1. The height of a prime ideal \(\pp \) is the length of the maximal chain of ideals strictly contained in \(\pp \). The dimension of a ring is the supremum of the heights of all the primes.

Note that the height of a prime is the dimension of the localization at the prime.

Definition 7.2. Let \(S\) be an extension of \(R\). \(s \in S\) is called integral over \(R\) if it is the root of a monic polynomial with coefficients in \(R\). If \(R\) is a field, we can also call \(s\) algebraic. We say \(S\) is integral or algebraic over \(R\) if every element of it is.

Lemma 7.3. TFAE:
- 1. \(s\) is integral over \(R\).
- 2. \(s\) is contained in some ring \(T \supset R\) finitely generated as an \(R\)-module.

Proof. For \(1 \implies 2\) note that \(R[s]\) is generated as an \(R\) module by finitely many powers of \(s\). For \(2 \implies 1\) multiplication by \(s\) is an endomorphism, by the determinant trick, multiplication by some polynomial of \(s\) is identically \(0\), but then \(s\) is a root of that polynomial. □

Corollary 7.4. The elements of \(S\) integral over \(R\) form a ring, and if \(R\) is a field, they form a field. Finally, integrality is transitive.

Proof. For the first statement, note that \(R[s,t]\) is finitely generated as an \(R\)-module if \(s,t\) are integral, so that \(s+t,st\) are integral. If \(R\) is a field and \(s\neq 0\) satisfies the polynomial \(a_nx^n+\dots +a_1x + a_0\), \(s^{-1}\) satisfies \(a_0x^n+\dots + a_{n-1}x+a_n = 0\). Transitivity follows from the fact that if \(R\subset S \subset T\), and \(t_i\) are generators of \(T\) as an \(S\)-module, and \(s_i\) are generators of \(S\) as an \(R\)-module, then \(t_is_j\) are generators of \(T\) as an \(R\)-module. □

Thus we can make sense of the integral closure of \(R\) in \(S\) (denoted \(I_S(R)\)), namely the subring of integral elements. We say that \(R\) is integrally closed in \(S\) if it is its integral closure. If \(S\) is unspecified, and \(R\) a domain, we assume that \(S = \Frac (R)\).

There is an easy way to tell if an element is integral over an integrally closed domain.

Definition 7.5. The minimal polynomial of an element \(s\) algebraic over a field \(F\) is the unique monic generator of the ideal of \(F[x]\) for which \(s\) is a root.

Lemma 7.6. \(s\) is integral over a domain \(R\) iff \(s\) is algebraic over \(\Frac (R)\) and its minimal polynomial has coefficients in \(I(R)\).

Proof. If \(s\) is integral, it is algebraic. Now consider the minimal polynomial \(g\) of \(s\) over \(R\). If it is degree \(n\), then \(\Frac (R)[s]\) is an \(n\) dimensional \(\Frac [R]\) vector space, with basis \(1,s,\dots ,s^{n-1}\). Multiplication by \(s\) is an endomorphism whose matrix in this basis consists of elements integral. Then the characteristic polynomial of this is the minimal polynomial, and has integral coefficients. The converse is obvious. □

Lemma 7.7. UFDs are integrally closed.

Proof. Suppose \(x^n+\dots +a_0 \in R[x]\) has a root in \(\Frac (R)\), where \(R\) is a UFD. WLOG assume \(a_0 \neq 0\), and suppose the root is \(\frac {a}{b}\) with \(\gcd (a,b) = 1\). Then \(-a_0 = (\frac {a}{b})^n+\dots +a_1\frac a b\) has a \(b\) in the denominator, so \(b = 1\). □

Integral extensions behave nicely with respect to prime ideals.

Definition 7.8. If \(R\subset S\) is an extension, \(\pp \subset R\), \(\qq \subset S\) prime ideals, we say \(\qq \) lies above \(\pp \) if \(\qq \cap R = \pp \).

We can also write \(S_\pp \) to denote \((R-\pp )^{-1}S\).

Lemma 7.9. If \(S/R\) is an integral extension of a local ring and \(\aa \) is a proper ideal of \(R\), \(\aa S\) is a proper ideal of \(S\).

Proof. If \(1 \in \aa S\), write \(1 = \sum _1^n a_is_i\), \(a_i \in \aa , s_i \in S\). By Nakayama’s Lemma, it follows that \(R[s_i]=0\), a contradiction. □

Proposition 7.10 (Lying over). If \(R \subset S\) is an integral extension, then every prime \(\pp \) of \(R\) is of the form \(\qq \cap R\) for some prime \(\qq \subset Q\).

Proof. Note that \(S_{\pp }\) is an integral extension of \(R_{\pp }\), so that by Lemma 7.9 \(\pp S_{\pp }\) is a proper ideal of \(S_{\pp }\). Thus it is contained in a maximal ideal \(\qq S_\pp \), and \(\qq \cap R\) must be \(\pp \). □

Proposition 7.11 (Going up). If \(R \subset S\) is an integral extension, \(\pp _1\subsetneq \pp _2\subset R\) primes, and \(\qq _1 \subset R\) lies above \(S\), then there is a \(\qq _1 \subsetneq \qq _2\) such that \(\qq _2\) lies above \(\pp _2\).

Proof. Apply the previous proposition to \(\qq _2\), \(R/{\pp _1}\), and \(S/{\qq _1}\). □

Proposition 7.12 (Incomparability). If \(S/R\) is an integral extension, and \(\qq _1\subsetneq \qq _2\) are two primes in \(S\), then \(\qq _1 \cap R \subsetneq \qq _2 \cap R\).

Proof. Consider the extension \(S/\qq _1\) over \(R/\qq _1\cap R\). \(\qq _2/\qq _1\) is a nonzero ideal, so if \(0 \neq a \in \qq _2/\qq _1\) is a nonzero element that is the root of monic polynomial \(g\), we can assume \(g\) has nonzero constant term \(c\). Then \(0 \neq -c = (g-c)(a) \in \qq _2\cap R/\qq _1\cap R\), so \(\qq _2 \cap R \subsetneq \qq _1 \cap R\). □

Corollary 7.13. Integral extensions preserve dimension.

Proof. This follows from lying over, going up, and incomparability. □

The next nice property of integral extensions only holds for integrally closed rings.

Lemma 7.14. A localization \(M^{-1}R\) of an integrally closed domain \(R\) is an integrally closed domain.

Proof. WLOG, \(M^{-1}R\) is nontrivial, i.e. \(0 \notin M\). Then \(\Frac (R) = \Frac (M^{-1}R)\). Now suppose \(x \in \Frac (M^{-1}R)\) is integral over □

Proposition 7.15 (Going down). Let \(S/R\) be an integral extension, \(S\) a domain, and \(R\) integrally closed. Then if \(\pp _1 \subsetneq \pp _2\) are primes and \(\qq _2\) lies above \(\pp _2\), then there is a \(\qq _1 \subsetneq \qq _2\) lying above \(\pp _1\).

Proof.

□