Commutative Algebra 40

More on Integrality

Lemma 1.

Let A\subseteq B be an integral extension. If \mathfrak b \subseteq B is an ideal and \mathfrak a = \mathfrak b \cap A, the resulting injection A/\mathfrak a \hookrightarrow B/\mathfrak b is an integral extension.


Any element of B/\mathfrak b can be written as x + \mathfrak b, x\in B. Then x satisfies a monic polynomial relation:

x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 = 0, \quad a_0, \ldots, a_{n-1} \in A.

Taking the relation modulo \mathfrak b gives a monic polynomial relation for x + \mathfrak b with coefficients in A/\mathfrak a. ♦

Lemma 2.

Let A\subseteq B be an integral extension and suppose C is the integral closure of A in B. If S \subseteq A is a multiplicative subset, then S^{-1}C is the integral closure of S^{-1}A in S^{-1}B.


Suppose \frac x s \in S^{-1}C where x\in C and s\in S. Since x is integral over A we have x^n + a_{n-1} x^{n-1} + \ldots + a_0 = 0 for some a_i \in A. Then in S^{-1}C we have the equality

(\frac x s)^n + \frac{a_{n-1}}s (\frac x s)^{n-1} + \ldots + \frac{a_0}{s^n} = 0, \quad \frac {a_i}{s^{n-i}} \in S^{-1}A

so \frac x s is integral over S^{-1}A.

Conversely, suppose \frac x s\in S^{-1}B is integral over S^{-1}A where x\in B and s\in S. Then \frac x 1\in S^{-1}B is also integral over S^{-1}A so

(\frac x 1)^n + \frac {a_{n-1}}{s_{n-1}} (\frac x 1)^{n-1} + \ldots + \frac{a_1}{s_1} (\frac x 1) + \frac{a_0}{s_0} = 0 for some \frac{a_i}{s_i} \in S^{-1}A.

Multiplying by s_0 \ldots s_{n-1}, there exist t\in S and a_0', \ldots, a_{n-1}' \in A such that

t(x^n + a_{n-1}' x^{n-1} + \ldots + a_1' x + a_0') = 0.

Multiplying by t^{n-1}, we see that tx \in B is integral over A so tx\in C and \frac x s \in S^{-1}C. ♦

Corollary 1.

  • If A\subseteq B is an integral extension, so is S^{-1}A \subseteq S^{-1}B.
  • If A is a normal domain, so is S^{-1}A.


For the second statement note that \mathrm{Frac} A = \mathrm{Frac} S^{-1}A. ♦

Exercise A

Prove that an integral domain A is normal if and only if A_{\mathfrak m} is normal for each maximal ideal \mathfrak m\subset A. Thus normality is a local property.


Spectra of Integral Extensions

The inclusion map A\hookrightarrow B induces \mathrm{Spec} B \to \mathrm{Spec} A. It turns out geometrically, such a map is like a finite-to-one map.

For example, let A = \mathbb C[X] and B = \mathbb C[X, Y]/(Y^2 - X^3 + X). The inclusion A\subset B corresponds to projection of the curve Y^2 = X^3 - X onto the X-axis. The map is generically two-to-one, except at the points X = -1, 0, +1 on the Y-axis. This corresponds to the following morphism W\to V.


[ Image edited from GeoGebra plot. ]

We start with the following.

Lemma 3.

If A\subseteq B is an integral extension of domains, then A is a field if and only if B is a field.


(⇒) Let A be a field and b\in B - \{0\}. We can find a_0, \ldots, a_{n-1} \in A such that b^n + a_{n-1} b^{n-1} + \ldots + a_0 = 0. If we assume n is minimal, then a_0 \ne 0 since B is a domain which gives

b(b^{n-1} + a_{n-1} b^{n-2} + \ldots + a_1)a_0^{-1} + 1 = 0

so b is a unit in B.

(⇐) Let B be a field and a\in A - \{0\}. Then b = a^{-1} exists in B. This is integral over A so we have b^n + a_{n-1} b^{n-1} + \ldots + a_0 = 0 for some a_i \in A. Multiplying throughout by a^n gives us

1 + a_{n-1}a + \ldots + a_1 a^{n-1} + a_0 a^n = 0 \implies a(\overbrace{a_{n-1} + \ldots + a_1 a^{n-2} + a_0 a^{n-1}}^{\in A}) = -1

so a is a unit in A. ♦

Exercise B

Find a counter-example when B is not an integral domain.

Corollary 2.

Let A\subseteq B be an integral extension of rings, \mathfrak q \in \mathrm{Spec} B and \mathfrak p = \mathfrak q \cap A. Then \mathfrak p is maximal if and only if \mathfrak q is maximal.


We get an inclusion of domains A/\mathfrak p \hookrightarrow B/\mathfrak q. By lemma 3, A/\mathfrak p is a field if and only if B/\mathfrak q is a field. ♦


Let f:A \to B be a homomorphism of any rings, which inducees

f^* : \mathrm{Spec} B \to \mathrm{Spec} A, \quad \mathfrak q \mapsto \mathfrak p = f^{-1}(\mathfrak q).

We say that \mathfrak q pulls back to \mathfrak p.



Proposition 1.

If A\subseteq B is integral, then \mathrm{Spec}B \to \mathrm{Spec} A is surjective.


Let \mathfrak p \subset A be prime and S = A-\mathfrak p; we get the following diagram


where the rows are injective. Since S^{-1}B is non-trivial it has a maximal ideal \mathfrak n, which pulls back to a prime ideal \mathfrak q\subset B. By corollary 2, \mathfrak n pulls back to a maximal ideal of A_{\mathfrak p} which must be \mathfrak p A_{\mathfrak p}; this pulls back to \mathfrak p \subset A. Hence \mathfrak q\subset B pulls back to \mathfrak p. ♦

Proposition 2.

If A\subseteq B is integral, then \mathrm{Spec B}\to \mathrm{Spec} A is a closed map, i.e. it takes closed subsets to closed subsets.


Let V(\mathfrak b) \subseteq \mathrm{Spec} B be a closed subset, for an ideal \mathfrak b\subseteq B. Let \mathfrak a = \mathfrak b \cap A so we get an injection A/\mathfrak a \hookrightarrow B/\mathfrak b. By proposition 1, we get a surjective map

V(\mathfrak b) \cong \mathrm{Spec} (B/\mathfrak b) \longrightarrow \mathrm{Spec} (A/\mathfrak a) \cong V(\mathfrak a)

so V(\mathfrak b) maps surjectively onto the closed subset V(\mathfrak a)\subseteq \mathrm{Spec} A. ♦


It follows that if f:A\to B is a ring homomorphism such that f(A)\subseteq B is integral, then f^* : \mathrm{Spec} B \to \mathrm{Spec} A is a closed map, since f(A) \cong A/\mathrm{ker} f.


Going Up Theorem

Proposition 3 (Going Up).

Let A\subset B be an integral extension. Suppose \mathfrak p_0 \subsetneq \mathfrak p_1 are prime ideals of A and \mathfrak q_0 is a prime ideal of B which pulls back to \mathfrak p_0.


Then there exists a prime ideal \mathfrak q_1 of B containing \mathfrak q_0 which pulls back to \mathfrak p_1.

Note that we must have \mathfrak q_0 \subsetneq \mathfrak q_1.


Since \mathfrak q_0 \cap A = \mathfrak p_0 we have an integral extension of domains A/\mathfrak p_0 \hookrightarrow B/\mathfrak q_0. By surjectivity of \mathrm{Spec} B/\mathfrak q_0 \to \mathrm{Spec} A/\mathfrak p_0, there is a prime ideal \mathfrak q_1 /\mathfrak q_0 of B/\mathfrak q_0 which pulls back to \mathfrak p_1/\mathfrak p_0, where \mathfrak q_1 is a prime ideal of B containing \mathfrak q_0. Then \mathfrak q_1 \in \mathrm{Spec} B pulls back to \mathfrak p_1 \in \mathrm{Spec} A. ♦

Corollary 3.

Suppose we have a chain of prime ideals \mathfrak p_0 \subsetneq \ldots \subsetneq \mathfrak p_n of A, and a prime ideal \mathfrak q_0 of B which pulls back to \mathfrak p_0.


Then there is a chain of prime ideals \mathfrak q_0 \subsetneq \mathfrak q_1 \subsetneq \ldots \subsetneq \mathfrak q_n of B such that each \mathfrak q_i lies over \mathfrak p_i.


Repeatedly apply proposition 3 to extend the bottom chain. ♦

The final piece of the puzzle is the following.

Proposition 4.

If \mathfrak q_0 \subsetneq \mathfrak q_1 are prime ideals of B and \mathfrak p_i = \mathfrak q_i \cap A for i=0,1, then \mathfrak p_0 \ne \mathfrak p_1.


Suppose \mathfrak p = \mathfrak p_0 = \mathfrak p_1; let S = A-\mathfrak p. Again we have


Since \mathfrak q_0 \cap S = \mathfrak q_1 \cap S = \emptyset, \mathfrak q_0, \mathfrak q_1 give us prime ideals \mathfrak q_0 (S^{-1}B) \subsetneq \mathfrak q_1(S^{-1}B) of S^{-1}B. They both pull back to \mathfrak p\in \mathrm{Spec} A and hence to \mathfrak p A_\mathfrak p \in \mathrm{Spec} A_{\mathfrak p}, the unique maximal ideal. By corollary 2, this means \mathfrak q_0 (S^{-1}B) and \mathfrak q_1(S^{-1}B) are both maximal, hence equal. So \mathfrak q_0 = \mathfrak q_1. ♦

As a result we have:

Main Theorem.

Let A\subseteq B be an integral extension of rings. Then \dim A = \dim B.


By Going Up and surjectivity of \mathrm{Spec}B \to \mathrm{Spec}A (proposition 1), any prime chain in A lifts to a prime chain in B. Conversely, any prime chain in B maps to a prime chain in A of the same length by proposition 4. ♦


Since \mathbb C[X] \subset \mathbb C[X, Y]/(Y^2 - X^3 + X) is a finite extension we have

\dim \mathbb C[X, Y]/(Y^2 - X^3 + X) = \dim \mathbb C[X] = 1,

since \mathbb C[X] is a PID.

Exercise C

1. Take A\subset B where A = \mathbb Z[X] and B = \mathbb Z[2i][X, Y]/(Y^2 - X^3 - 1). Lift the following prime chains of A to prime chains of B

0\subset (5) \subset (5, X-2), \quad 0 \subset (X-2) \subset (5, X-2).

2. Prove that if A\subseteq B is a finite extension with induced \phi : \mathrm{Spec} B \to \mathrm{Spec } A, for any \mathfrak p \in \mathrm{Spec } A, \phi^{-1}(\mathfrak p) is a finite subset of Spec B. We say that \phi has finite fibres.


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4 Responses to Commutative Algebra 40

  1. Vanya says:

    In the last example at the end “\mathbb C[X] \subset \mathbb C[X]/(Y^2 - X^3 + X) is a finite extension”. shouldn’t it be \mathbb C[X] \subset \mathbb C[X,Y]...?

  2. Vanya says:

    The last statement “We say that f^* has finite fibres” should be ” …\phi … $

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